Trig without Tears
or, How to Remember Trigonometric Identities
Copyright © 1997–2019 Stan Brown, BrownMath.com
or, How to Remember Trigonometric Identities
Copyright © 1997–2019 Stan Brown, BrownMath.com
Summary: Faced with the large number of trigonometric identities, students tend to try to memorize them all. That way lies disaster. When you memorize a formula by rote, you have no way to know whether you’re remembering it correctly. I believe it is much more effective (and, in the long run, much easier) to understand thoroughly how the trig functions work, memorize half a dozen formulas, and work out the rest as needed. That’s what these pages show you how to do.
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Trigonometry is fascinating! It started as the measurement (Greek metron) of triangles (Greek trigonon), but now it has been formalized under the influence of algebra and analytic geometry and we talk of trigonometric functions. not just sides and angles of triangles.
Trig is almost the ideal math subject. Big and complex enough to have all sorts of interesting odd corners, it is still small and regular enough to be taught thoroughly in a semester. (You can easily master the essential points in a week or so.) It has lots of obvious practical uses, some of which are actually taught in the usual trig course. And trig extends plenty of tentacles into other fields like complex numbers, logarithms, and calculus.
If you’d like to learn some of the history of trigonometry and peer into its dark corners, I recommend Trigonometric Delights by Eli Maor (Princeton University Press, 1998).
The computations in trigonometry used to be a big obstacle. But now that we have calculators, that’s no longer an issue.
Would you believe that when I studied trig, back when dinosaurs ruled the earth (actually, in the 1960s), to solve any problem we had to look up function values in long tables in the back of the book, and then multiply or divide those fiveplace decimals by hand? The “better” books even included tables of logs of the trig functions, so that we could save work by adding and subtracting fiveplace decimals instead of multiplying and dividing them. My College Outline Series trig book covered all of plane and spherical trigonometry in 188 pages—but then needed an additional 138 pages for the necessary tables!
Though calculators have freed us from tedious computation, there’s still one big stumbling block in the way many trig courses are taught: all those identities. They’re just too much to memorize. (Many students despair of understanding what’s going on, so they just try to memorize everything and hope for the best at exam time.) Is it tan² A + sec² A = 1 or tan² A = sec² A + 1? (Actually, it’s neither—see equation 39!)
Fortunately, you don’t need to memorize them. This paper shows you the few that you do need to memorize, and how you can produce the others as needed. I’ll present some ideas of my own, and a wonderful insight by W.W. Sawyer.
I wrote Trig without Tears to show that you need to memorize very little. Instead, you learn how all the pieces of trigonometry hang together, and you get used to combining identities in different ways so that you can derive most results on the fly in just a couple of steps.
You might like to read some ideas of mine on the pros and cons of memorizing.
To help you find things, I’ll number the most important equations and other facts. (Don’t worry about the gaps in the numbering. I’ve left those to make it easier to add information to these pages.)
A very few of those, which you need to memorize, will be marked “memorize”. Please don’t memorize the others. The whole point of Trig without Tears is to teach you how to derive them as needed without memorizing them. If you can’t think how to derive one, the boxes should make it easy to find it. But then please work through the explanation. I truly believe that if you once thoroughly understand how all these identities hang together, you’ll never have to memorize them again. (It’s worked for me since I first studied trig in 1965.)
Trig is really a mix of practical and theoretical. On one hand, you find distances or angles in triangles; that’s pretty concrete. Computers do it now, but when artillery was guided by humans they used trig to figure how to aim the cannon. Land surveying still needs trig, and so do plenty of other fields. If you take a highschool physics course, you’ll use plenty of trig in it. Parts 2 through 4 in this book have the practical stuff.
But as you go further in math, though there are usually some realworld problems there somewhere, you spend a greater and greater proportion of your time manipulating symbols. That makes sense: tougher problems take more thought and more calculation to solve.
Traditionally, trig courses are a mix of the practical stuff with the stuff that has no direct practical application but that you’ll need to do work in calculus or physics or other technical fields later. Nearly all the trig identities fall into that category. Part 5 and afterward are the groundwork for those future courses.
I don’t say you should stop at the end of Part 4, even if you’re never planning to take another course. “Is it useful to me?” isn’t always the right question to ask. (If people decided things that way, videogame manufacturers would go out of business tomorrow.) Better questions are, “Is this interesting? Would I like to pursue this just for the fun of stretching my mind?”
Yes, even “Is it beautiful?” Symmetry is a lot of how we judge beauty—if you doubt that, picture someone goodlooking, and then mentally change one side of their face. There are some beautiful symmetries in the first half of the book, but the second half has a lot more, plus answers to questions bright students have always asked, like What’s the logarithm of a negative number?
By the way, I love explaining things but sometimes I go on a bit too long. So I’ve put some interesting but nonessential notes at the end of most chapters and inserted hyperlinks to them at appropriate points. If you follow them (and I hope you will!), use your browser’s “back” command to return to the main text.
Much as it pains me to say so, if you’re pressed for time you can still get all the essential points by ignoring those side notes. In token of this, they’re labeled BTW. But you’ll miss some of the fun.
Trig without Tears deals exclusively with plane trigonometry, which is what’s taught today in nearly every first course. Spherical trigonometry is not part of this book.
I’m also restricting myself to real arguments to the functions and real values of the functions. I have to draw the line somewhere! (I do use realvalued functions with the polar form of complex numbers in the Notes.) For complex trigonometric functions, see chapter 14 of Eli Maor’s Trigonometric delights (Princeton University Press, 1998).
In Trig without Tears we’ll work with identities and solve triangles. A separate (and much shorter) page of mine explains how to solve trigonometric equations.
Let’s not take anything for granted. You probably remember some basic facts from earlier school, but—
In an hour, the minute hand of the clock travels all the way around the clock, through four right angles, so a full circle is 360°.
The famous Pythagorean Theorem or Theorem of Pythagoras, a² + b² = c², says that if you add the squares of the two legs you get the square of the hypotenuse.
If a triangle contains an obtuse angle (greater than 90° but less than 180°), it is called, naturally enough, an obtuse triangle. In a right triangle or obtuse triangle, the other two triangles must be acute; otherwise the three angles would total more than 180°.
Fractions other than ½ are written using the slash, such as a/b for a over b.
In talking about the domains and ranges of functions, it is handy to use interval notation. Thus instead of saying that x is between 0 and π, we can use the open interval (0, π) if the endpoints are not included, or the closed interval [0, π] if the endpoints are included.
You can also have a halfopen interval. For instance, the interval [0, 2π) is all numbers ≥ 0 and < 2π. You could also say it’s the interval from 0 (inclusive) to 2π (exclusive).
Starting with Part 5, I’ll be referring to the four quadrants of a circle, or of the xy plane. The division for a circle would be the two lines from 12 to 6 and from 3 to 9; on the plane, the division is the x and y axes.
The quadrants are usually given Roman numerals, starting at the upper right and going counterclockwise. So Quadrant I or Q I is the top right quarter, Q II is the top left, Q III is the bottom left, and Q IV is the bottom right.
These pages will show examples with both radians and degrees. The same theorems apply to either way of measuring an angle, and you need to practice with both.
A lot of students seem to find radians terrifying. But measuring angles in degrees and radians is no worse than measuring temperature in °F and °C. In fact, angle measure is easier because 0°F and 0°C are not the same temperature, but 0° and 0 radians are the same angle.
Just remember that a complete circle is 2π radians. Then, think of the twelve hours numbered around the circumference of a clock face. When the hour hand goes all the way around, it travels through 2π radians or 360°. Six hours is half of that, 2π/2 = π or 180°, one hour is π/6 or 30°, two hours is 2π/6 = π/3 or 60°, and so on. (Thanks are due to Jeffrey T. Birt for this suggestion.)
One technical note: Angles don’t actually have units — they’re dimensionless. If you say “π radians”, you could just as well say “π” and leave off the “radians”.
If you’re measuring in degrees, then you do need to use the degree mark. 180° = π, or π radians. In other words, that degree mark (°) just means “×π/180”. It’s the same sort of animal as the percent sign (%), which really just means “/100”.
So you can convert between degrees and radians exactly the same way you convert between inches and feet, or between centimeters and meters. (If conversions in general are a problem for you, you might like to consult my page on that topic.)
But even though you can convert between degrees and radians, it’s probably better to learn to think in both. Here’s an analogy.
When you learn a foreign language L, you go through a stage where you mentally translate what someone says in L into your own language, formulate your answer in your own language, mentally translate it into L, and then speak. Eventually you get past that stage, and you carry on a conversation in the other language without translating. Not only is it more fun, it’s a heck of a lot faster and easier.
You want to train yourself to work with radians for the same reason: it’s more efficient, and saving work is always good. Practice visualizing an angle of π/6 or 3π/4 or 5π/3 directly, without translating to degrees. You’ll be surprised how quickly it will become second nature!
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
Where possible, give an exact answer rather than a decimal approximation.
Dad sighed. “Kip, do you think that table was brought down from on high by an archangel?”
Robert A. Heinlein, in Have Space Suit—Will Travel (1958)
It’s not just that there are so many trig identities; they seem so arbitrary. Of course they’re not really arbitrary, since all can be proved; but when you try to memorize all of them they seem like a jumble of symbols where the right ones aren’t more obviously right than the wrong ones. For example, is it sec² A = 1 + tan² A or tan² A = 1 + sec² A ? I doubt you know off hand which is right; I certainly don’t remember. Who can remember a dozen or more like that, and remember all of them accurately?
Too many teachers expect (or allow) students to memorize the trig identities and parrot them on demand, much like a series of Bible verses. In other words, even if they’re originally taught as a series of connected propositions, they’re remembered and used as a set of unrelated facts. And that, I think, is the problem. The trig identities were not brought down by an archangel; they were developed by mathematicians, and it’s well within your grasp to redevelop them when you need to. With effort, we can remember a few key facts about anything. But it’s much easier if we can fit them into a context, so that the identities work together as a whole.
Why bother? Well, of course it will make your life easier in trig class. But you’ll also need the trig identities in later math classes, especially calculus, and in physics and engineering classes. In all of those, you’ll find the going much easier if you’re thoroughly grounded in trigonometry as a unified field of knowledge instead of a collection of unrelated facts.
This is why it’s easier to remember almost any song than an equivalent length of prose: the song gives you additional cues in the rhythm, common patterns of emphasis, and usually rhymes at the ends of lines. With prose you have only the general thought to hold it together, so that you must memorize it essentially as a series of words. With the song there are internal structures that help you, even if you’re not aware of them.
If you’re memorizing Lincoln’s Gettysburg Address, you might have trouble remembering whether he said “recall” or “remember” at a certain point; in a song, there’s no possible doubt which of those words is right because the wrong one won’t fit in the rhythm.
I’m not against all memorization. Some things have to be memorized because they’re a matter of definition. Others you may choose to memorize because you use them very often, you’re confident you can memorize them correctly, and the derivation takes more time than you’re comfortable with. Still others you may not set out to memorize, but after using them many times you find you’ve memorized them without trying to—much like a telephone number that you dial often.
I’m not against all memorization; I’m against needless memorization used as a substitute for thought. If you decide in particular cases that memory works well for you, I won’t argue. But I do hope you see the need to be able to rederive things on the spot, in case your memory fails. Have you ever dialed a friend’s telephone number and found you couldn’t quite remember whether it was 6821 or 8621? If you can’t remember a phone number, you have to look it up in the book. My goal is to free you from having to look up trig identities in the book.
Thanks to David Dixon for an illuminating exchange of notes on this topic. He made me realize that I was sounding more antimemory than I meant to, and in consequence I’ve added this note. But he may not necessarily agree with what I say here.
I wrote these pages to show you how to make the trig identities “fit” as a coherent whole, so that you’ll have no more doubt about them than you do about the words of a song you know well. The difference is that you won’t need to do it from memory. And you’ll gain the sense of power that comes from mastering your subject instead of groping tentatively and hoping to strike the right answer by good luck.
Summary: Every one of the six trig functions is just one side of a right triangle divided by another side. Or if you draw the triangle in a unit circle, every function is the length of one line segment. The easy way to remember all six definitions: memorize the definitions of sine and cosine and then remember the other four as combinations of sine and cosine, not as independent functions.
A picture is worth a thousand words (which is why it takes a thousand times as long to download). The trig functions are nothing more than lengths of various sides of a right triangle in various ratios. Since there are three sides, there are 3 × 2 = 6 different ways to make a ratio (fraction) of sides. That’s why there are six trig functions, no more and no less.
Of those six functions, three—sine, cosine, and tangent—get the lion’s share of the work. (The others are studied because they can be used to make some expressions simpler.) We’ll begin with sine and cosine, because they are really basic and the others depend on them.
Here is one of the conventional ways of showing a right triangle. A key point is that the lowercase letters a, b, c are the sides opposite to the angles marked with the corresponding capital letters A, B, C. Most books use this convention: lowercase letter for side opposite uppercase angle.
The two fundamental definitions are marked in the diagram. You must commit them to memory. In fact, they should become second nature to you, so that you recognize them no matter how the triangle is turned around. Always, always, the sine of an angle is equal to the opposite side divided by the hypotenuse (opp/hyp in the diagram). The cosine is equal to the adjacent side divided by the hypotenuse (adj/hyp).
(1) Memorize:
sine = (opposite side) / hypotenuse
cosine = (adjacent side) / hypotenuse
What is the sine of B in the diagram? Remember opp/hyp: the opposite side is b and the hypotenuse is c, so sin B = b/c. What about the cosine of B? Remember adj/hyp: the adjacent side is a, so cos B = a/c.
Do you notice that the sine of one angle is the cosine of the other? Since A + B + C = 180° for any triangle, and C is 90° in this triangle, A + B must equal 90°. Therefore A = 90° − B, and B = 90° − A. When two angles add to 90°, each angle is the complement of the other, and the sine of each angle is the cosine of the other. These are the cofunction identities:
(2) sin A = cos(90° − A) or cos(π/2 − A)
cos A = sin(90° − A) or sin(π/2 − A)
The definitions of sine and cosine can be rearranged a little bit to let you write down the lengths of the sides in terms of the hypotenuse and the angles. For example, when you know that b/c = cos A, you can multiply through by c and get b = c × cos A. Can you write another expression for length b, one that uses a sine instead of a cosine? Remember that opposite over hypotenuse equals the sine, so b/c = sin B. Multiply through by c and you have b = c × sin B.
Can you see how to write down two expressions for the length of side a? Please work from the definitions and verify that a = c × sin A = c × cos B.
Example: Given, a right triangle with angle A =52° and hypotenuse c = 150 m. What is the length of side b? Hint: draw a picture, and label A, c, and b.
Solution: Pictures are always good. You don’t have to obsess over getting the picture exactly right, but at least make it close. That will help you see when your answer is impossible, so you know you’ve made a mistake. In my little sketch, I set out to make angle A a bit more than 45°, but to my eyes it looks like a bit less. That’s okay.
You may notice that I marked side a, even though we don’t need it for the problem. I did that so I didn’t have to think about which side was b. Always remember the rule that the side with a given letter is opposite the angle with that letter. (And, conventionally, we always put C at the right angle, so that makes c the hypotenuse.)
Once you have the picture, solving the problem is pretty straightforward. You want something involving A, its adjacent side, and the hypotenuse; that has to be the cosine.
cos A = b/c
b = c × cos A = 150 × cos 52° = about 92.35 m.
Example: A guy wire is anchored in the ground and attached to the top of a 45foot flagpole. If it meets the ground at an angle of 63°, how long is the guy wire?
Solution: Presumably the flagpole is vertical, so this is a right triangle, with A = 63°, a = 45 ft, and hypotenuse c unknown. Which function involves the opposite side and the hypotenuse? It must be the sine. You know that
sin A = a/c
Therefore,
c = a/sin A = 45/sin 63° = about 50.5 ft.
You may be wondering how to find sides or angles of triangles when there is no right angle. We’ll get to that, under the topic of Solving Triangles.
One important special case comes up frequently. Suppose the hypotenuse c = 1; then we call the triangle a unit right triangle. You can see from the paragraphs just above that if c = 1 then a = sin A and b = cos A. In other words, in a unit right triangle the opposite side will equal the sine and the adjacent side will equal the cosine of the angle.
The triangle is often drawn in a unit circle, a circle of radius 1, as shown at right. The angle A is at the center of the circle, and the adjacent side lies along the x axis. If you do this, the hypotenuse is the radius, which is 1. The (x, y) coordinates of the outer end of the hypotenuse are the x and y legs of the triangle: (x, y) = (cos A, sin A). The unit circle is your friend: it can help you visualize lots of trig identities.
The other four functions have no real independent life of their own; they’re just combinations of the first two. You could do all of trigonometry without ever knowing more than sines and cosines. But knowing something about the other four, especially the tangent, can often save you some steps in a calculation—and your teacher will expect you to know about them for exams.
I find it easiest to memorize (sorry!) the definition of the tangent in terms of the sine and cosine:
(3) Memorize:
tan A = (sin A) / (cos A)
You’ll use the tangent (tan) function a lot more than the last three functions. (I’ll get to them in a minute.)
There’s an alternative way to remember the meaning of the tangent. Remember from the diagram that sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse. Plug those into equation 3, the definition of the tan function, and you have tan A = (opposite/hypotenuse) / (adjacent/hypotenuse) or
(4) tangent = (opposite side) / (adjacent side)
Notice this is not marked “memorize”: you don’t have to memorize it because it flows directly from the definition equation 3, and in fact the two statements are equivalent. I’ve chosen to present them in this order to minimize the jumble of opp, adj, and hyp among sin, cos, and tan. However, if you prefer you can memorize equation 4 and then derive the equivalent identity equation 3 whenever you need it.
Example: A guy wire is anchored in the ground and attached to the top of a 45foot flagpole. How far is the anchor from the base of the flagpole, if the wire meets the ground at an angle of 63°?
Solution: This is a variation on the previous example. This time, you want to know the side adjacent to angle A, not the hypotenuse. As before, assume the flagpole is vertical, so this is a right triangle, with A = 63°, a = 45 ft, and adjacent side b unknown. Which function involves the adjacent side and the opposite side? It’s the tangent. You know that
tan A = a/b
Therefore,
b = a/tan A = 45/tan 63° = about 22.9 ft.
Now, I said you could do all of trig with just sines and cosines. How would that work for this problem? Well, sine and cosine both need the hypotenuse, so you’d have
sin A = a/c ⇒ c = a/sin A and
cos A = b/c ⇒ c = b/cos A. Therefore,
b/cos A = a/sin A
b = a × cos A/sin A = 45 × cos 63°/sin 63° = about 22.9 ft.
You got to the same place in the end, but the journey was longer. So, although it’s not strictly necessary, the tangent can make your work easier.
The other three trig functions—cotangent, secant, and cosecant—are defined in terms of the first three. They’re less often used, but they do simplify some problems in calculus. In practical problems, not involving calculus, you’ll pretty much never need them.
(5) Memorize:
cot A = 1 / (tan A)
sec A = 1 / (cos A)
csc A = 1 / (sin A)
Guess what! That’s the last trig identity you have to memorize.
(You’ll probably find that you end up memorizing certain other identities without even intending to, just because you use them frequently. But equation 5 makes the last ones that you’ll have to sit down and make a point of memorizing just on their own.)
Unfortunately, the definitions in equation 5 aren’t the easiest thing in the world to remember. Does the secant equal 1 over the sine or 1 over the cosine? Here are two helpful hints: Each of those definitions has a cofunction on one and only one side of the equation, so you won’t be tempted to think that sec A equals 1/sin A. And secant and cosecant go together just like sine and cosine, so you won’t be tempted to think that cot A equals 1/sin A.
For an alternative approach to remembering the above identities, you might like:
You can immediately notice an important relation between tangent and cotangent. Each is the cofunction of the other, just like sine and cosine:
(6) tan A = cot(90° − A) or cot(π/2 − A)
cot A = tan(90° − A) or tan(π/2 − A)
If you want to prove this, it’s easy from the definitions and equation 2:
cot A = 1 / tan A
Apply the definition of tan:
cot A = 1 / (sin A / cos A)
Simplify the fraction:
cot A = cos A / sin A
Apply equation 2:
cot A = sin(90° − A) / cos(90° − A)
Finally, recognize that this fraction fits the definition of the tan function, equation 3:
cot A = tan(90° − A)
Tangent and cotangent are cofunctions just like sine and cosine. By doing the same sort of substitution, you can show that secant and cosecant are also cofunctions:
(7) sec A = csc(90° − A) or csc(π/2 − A)
csc A = sec(90° − A) or sec(π/2 − A)
You saw earlier how the sine and cosine of an angle are the sides of a triangle in a unit circle. It turns out that all six functions can be shown geometrically in this way.
unit circle (radius = AB = 1)
sin θ = BC;
cos θ = AC;
tan θ = ED
csc θ = AG;
sec θ = AE;
cot θ = FG
Graphic courtesy of TheMathPage
In the illustration at right, triangle ABC has angle θ at the center of a unit circle (AB = radius = 1). You already know that BC = sin θ and AC = cos θ.
What about tan θ? Well, since DE is tangent to the unit circle, you might guess that its length is tan θ, and you’d be right. Triangles ABC and AED are similar, and therefore
ED / AD = BC / AC
ED / 1 = sin θ / cos θ
ED = tan θ
More information comes from the same pair of similar triangles:
AE / AB = AD / AC
AE / 1 = 1 / cos θ
AE = sec θ
The lengths that represent cot θ and csc θ will come from the other triangle, GAF. That triangle is also similar to triangle AED. (Why? FG is perpendicular to FA, and FA is perpendicular to AD; therefore FG and AD are parallel. In beginning geometry you learned that when parallel lines are cut by a third line, the corresponding angles—marked θ in the diagram—are equal. Thus FG is a tangent to the unit circle, and therefore angles G and θ are equal. )
Using similar triangles GAF and AED,
FG / FA = AD / ED
FG / 1 = 1 / tan θ
FG = cot θ
That makes sense: FG is tangent to the unit circle, and is the tangent of the complement of angle θ, namely angle GAF. Therefore, FG is the cotangent of the original angle θ (or angle GAD).
Finally, using the same pair of similar triangles again, you can also say that
AG / FA = AE / ED
AG / 1 = sec θ / tan θ
AG = ( 1 / cos θ ) / ( sin θ / cos θ )
AG = 1 / sin θ
AG = csc θ
This one diagram beautifully depicts the geometrical meaning of all six trig functions when the angle θ is drawn at the center of a unit circle:
sin θ = BC; cos θ = AC; tan θ = ED
csc θ = AG; sec θ = AE; cot θ = FG
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
From the picture, it’s obvious why the name “tangent” makes sense: the tangent of an angle is the length of a segment tangent to the unit circle. But what about the sine function? How did it get its name?
Please look at the picture again, and notice that sin θ = BC is half a chord of the circle. The Hindu mathematician Aryabhata the elder (about 475–550) used the word “jya” or “jiva” for this halfchord. In Arabic translation the word was unchanged, but in the Arabic system of writing “jiva” was written the same way as the Arabic word “jaib”, meaning bosom, fold, or bay. The Latin word for bosom, bay, or curve is “sinus”, or “sine” in English, and beginning with Gherardo of Cremona (about 1114–1187) that became the standard term.
Edmund Gunter (1581–1626) seems to have been the first to publish the abbreviations sin and tan for sin and tangent.
My source for this history is Eli Maor’s Trigonometric Delights (1998, Princeton University Press), pages 35–36. I urge you to consult the book for a fuller account.
Trig without Tears Part 3:
Copyright © 1997–2019 Stan Brown, BrownMath.com
Summary: You need to know the function values of certain special angles, namely 30° (π/6), 45° (π/4), and 60° (π/3). You also need to be able to go backward and know what angle has a sine of ½ or a tangent of −√3. While it’s easy to work them out as you go (using easy right triangles), you really need to memorize them because you’ll use them so often that deriving them or looking them up every time would really slow you down.
Look at this 454590° triangle, which means sides a and b are equal. By the Pythagorean Theorem,
a² + b² = c²
But a = b and c = 1; therefore
2a² = 1
a² = 1/2
a = 1/√2 = (√2)/2
Since a = sin 45°,
sin 45° = (√2)/2
Also, b = cos 45° and b = a; therefore
cos 45° = (√2)/2
Use the definition of tan A, equation 3 or equation 4:
tan 45° = a/b = 1
(14) sin 45° = cos 45° = (√2)/2
tan 45° = 1
Now look at this diagram. I’ve drawn two 306090° triangles back to back, so that the two 30° angles are next to each other. Since 2×30° = 60°, the big triangle is a 606060° equilateral triangle. Each of the small triangles has hypotenuse 1, so the length 2b is also 1, which means that
b = ½
But b also equals cos 60°, and therefore
cos 60° = ½
You can find a, which is sin 60°, by using the Pythagorean Theorem:
(½)² + a² = c² = 1
1/4 + a² = 1
a² = 3/4 ⇒ a = (√3)/2
Since a = sin 60°, sin 60° = (√3)/2.
Knowing the sine and cosine of 60°, you can easily use the cofunction identities (equation 2) to get the cosine and sine of 30°:
cos 30° = sin(90°−30°) = sin 60° = (√3)/2
sin 30° = cos(90°−30°) = cos 60° = 1/2
As before, use the definition of the tangent to find the tangents of 30° and 60° from the sines and cosines:
tan 30° = sin 30° / cos 30°
tan 30° = (1/2) / ((√3)/2)
tan 30° = 1 / √3 = (√3)/3
and
tan 60° = sin 60° / cos 60°
tan 60° = ((√3)/2) / (1/2)
tan 60° = √3
The values of the trig functions of 30° and 60° can be summarized like this:
(15) sin 30° = ½, sin 60° = (√3)/2
cos 30° = (√3)/2, cos 60° = ½
tan 30° = (√3)/3, tan 60° = √3
Incidentally, the sines and cosines of 0, 30°, 45°, 60° and 90° display a pleasing pattern:
(16) For angle A = 0, 30° 45° 60° 90° — or 0, π/6, π/4, π/3, π/2 —
sin A = √0/2, √1/2, √2/2, √3/2, √4/2
cos A = √4/2, √3/2, √2/2, √1/2, √0/2
tan A = 0, √3/3, 1, √3, undefined
It’s not surprising that the cosine pattern is a mirror image of the sine pattern, since sin(90°−A) = cos A.
If A + B = 90°, then angles A and B are complements of each other. For example, the complement of a 40° angle is 90° − 40° = 50°.
If A + B = 180°, then angles A and B are supplements of each other. For example, the supplement of a 40° angle is 180° − 40° = 140°.
You already know about trig functions of complementary angles: sin(90° − A) = cos A, tan(90° − A) = cot A, sec(90° − A) = csc A, and vice versa. Your mnemonic is the “co” in cofunction and complementary angle.
But what about supplementary angles? Is there any relation between sin 40° and sin 140°? As a matter of fact, there is: the sine of an angle equals the sine of its supplement. And cosines? the cosine of an angle equals minus the cosine of its supplement. You can look ahead, if you want, to see why that’s true, or for now you can just take it as a given, while we work some practical problems in the next part of this book.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Special advice: Don’t be afraid to draw a picture of a 454590° or 306090° triangle if you need to, especially while you’re first getting used to the functions of the special angles.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
(b) Draw a triangle with all sides equal to 1. If all sides are equal, all angles must be equal. Knowing that they add to 180°, fill in the value of each angle. Now drop a perpendicular from the top of the triangle to the middle of the opposite side. You now have two 306090° triangles with hypotenuse of 1. Fill in the exact lengths of the short sides of those two triangles.
For the rest of these problems, refer to these sketches if you need to. Give exact answers, not decimal approximations.
Prove: tan(180° − A) = −tan A.
For all of this, you need only two tools, the Law of Sines and the Law of Cosines. The Law of Sines relates any two sides and the angles opposite them, and the Law of Cosines relates all three sides and one angle.
Let’s look at a specific example to start with. Suppose you have a triangle where one side has a length of 180, an adjacent angle is 42°, and the opposite angle is 31°. You’re asked to find the other angle and the other two sides.
It’s always a good idea to draw a rough sketch, like this one. Not only does it help you organize your solution process better, but it can help you check your work. For instance, since the 31° angle is the smallest, you know that the opposite side must also be the shortest. If you were to come up with an answer of, say, 110 for one of the other sides, you’d know at once that you had made a mistake somewhere because 110 is < 180 and the other two sides must both be > 180.
How would you go about solving this problem? It’s not immediately obvious, I agree. But maybe we can get some help from some useful general techniques in problem solving:
We’ve already got the diagram, but let’s see if those other techniques will be helpful. (By the way, they’re not original with me, but are from a terrific book on problemsolving techniques that I think you should know about.)
“Can you use what you already know to solve a piece of this problem?” For example, if this were a right triangle you’d know right away how to write down the lengths of sides in terms of sines or cosines.
But it’s not a right triangle, alas. Is there any way to turn it into a right triangle? Not exactly, but if you construct a line at right angles to one side and passing through the opposite vertex, you’ll have two right triangles. Maybe solving those right triangles will show how to solve the original triangle.
This diagram shows the same triangle after I drew that perpendicular. I’ve also used another principle (“Can you solve a more general problem?”) and replaced the specific numbers with the usual letters for sides and angles. Dropping perpendicular CD in the diagram divides the big triangle (which you don’t know how to solve) into two right triangles ACD and BCD, with a common side CD. And you can solve those right triangles.
We’re going to use this simple diagram to develop two important tools for solving triangles: the Law of Sines and the Law of Cosines. Just drawing this one perpendicular line will show you how to solve not just the triangle we started with, but any triangle. (Some trig courses teach other laws like the Law of Tangents and the Law of Segments. I’m ignoring them because you can solve triangles just fine without them.)
The Law of Sines is simple and beautiful and easy to derive. It’s useful when you know two angles and any side of a triangle, or two angles and the area, or (sometimes) two sides and one angle.
Let’s start by writing down things we know that relate the sides and angles of the two right triangles in the diagram above. You remember how to write down the lengths of the legs of a right triangle? The leg is always equal to the hypotenuse times either the cosine of the adjacent angle or the sine of the opposite angle. (If that looks like just empty words to you, or even if you’re not 100% confident about it, please go back and review that section until you feel confident.)
In the diagram, look at triangle ADC at the left: the right angle is at D and the hypotenuse is b. We don’t know how much of original angle C is in this triangle, so we can’t use C to find the lengths of any sides. What can we write down using angle A? By using its cosine and sine we can write the lengths of both legs of the triangle:
AD = b cos A and CD = b sin A
By the same reasoning, in the other triangle you have
DB = a cos B and CD = a sin B
This is striking: you see two different expressions for the length CD. But things that are equal to the same thing are equal to each other. That means that
b sin A = a sin B
Divide through by sin A and you have the solution for the general case:
b = a sin B / sin A
How does that apply to the triangle we started with? Well, plug in the values and you get the length of the side next to the 31° angle (or opposite the 42° angle):
b = 180 × sin 42° / sin 31° ≈ 234
What about the third angle, C, and the third side, c? Well, when you have two angles of a triangle you can find the third one easily:
A + B + C = 180°
C = 180° − A − B
In this case, C = 180° − 31° − 42° = 107°.
For the third side, there are a couple of ways to go. You wrote expressions above for AD and DB, and you know that c = AD+DB, so you could compute c = b cos A + a cos B.
But that’s two multiplies and an add, a bit more complicated than the one multiply and one divide to find side b. I’m lazy, and I like to reduce the amount of tapping I do on my calculator. Is there an easier way, even if just slightly easier? Yes, there is. Go back a step, to
a sin B = b sin A
Divide left and right by (sin A)(sin B) to get
a/sin A = b/sin B
But there’s nothing special about the two angles A and B. You could just as well have dropped a perpendicular from A to BC or from B to AC. Shown at right is the result of dropping a perpendicular from B to line CD.
Because C > 90°, this perpendicular happens to be outside the triangle and the two right triangles ABD and CBD overlap. But this won’t affect the algebra. By the way, the angle in triangle CBD is not C but 180° − C, the supplement of C. Angle C belongs to the original triangle ABC.
You can write the length of the common side BD as
BD = c sin A (in triangle ABD)
and
BD = a sin(180° − C) (in triangle CBD)
But sin(180° − C) = sin C, so you have
BD = a sin C (in triangle CBD)
Set the two computed lengths of BD equal to each other, and divide by (sin A)(sin C):
a sin C = c sin A
a/sin A = c/sin C
But we already figured out earlier that
a/sin A = b/sin B
Combining these two equations you have the Law of Sines:
(28) Law of Sines—First Form:
a/sin A = b/sin B = c/sin C
This is very simple and beautiful: for any triangle, if you divide any side by the sine of the opposite angle, you’ll get the same result. This law is valid for any triangle.
You can derive the Law of Sines at need, so I don’t specifically recommend memorizing it. But it’s so simple and beautiful that it’s pretty hard not to memorize if you use it at all. It’s also pretty hard to remember it wrong: there are no alternating plus and minus signs or combinations of different functions.
Coming back to our original triangle, we can compute the length of the third side:
a/sin A = c/sin C
a (sin C)/(sin A) = c
c = 180 × (sin 107°)/(sin 31°) ≈ 334
The Law of Sines is sometimes given upside down:
(29) Law of Sines—Second Form:
(sin A)/a = (sin B)/b = (sin C)/c
Of course that’s the same law, just as 2/3 = 6/9 and 3/2 = 9/6 are the same statement. Work with it either way, and you’ll come up with the same answers.
In most cases where you use the Law of Sines, you get a unique solution. But sometimes you get two solutions (or none) in the sidesideangle case, where you know two sides and an angle that’s not between them. Please see the Special Note below, after the table.
The Law of Sines is fine when you can relate sides and angles. But suppose you know three sides of the triangle—for instance a = 180, b = 238, c = 340—and you have to find the three angles. The Law of Sines is no good for that, because it relates two sides and their opposite angles. If you don’t know any angles, you have an equation with two unknowns and you can’t solve it.
But a triangle can be solved when you know all three sides; you just need a different tool. And knowing me, you can be sure I’m going to help you develop one! It’s called the Law of Cosines.
Let’s look back at that generic triangle with a perpendicular dropped from vertex C. You may remember that when we first looked at this picture, we pulled out information using both the sine and the cosine of the two angles. We used the sine information to develop the Law of Sines, but we never went anywhere with the cosine information, which was
AD = b cos A and DB or BD = a cos B
Let’s see where that can lead us. You remember that the way we came up with the Law of Sines was to write two equations that featured the length of the construction line CD, and then combine the equations to eliminate CD. Can we do anything like that here?
Well, we know the other two sides of those right triangles, so we can write an expression for the height CD using the Pythagorean theorem—actually, two expressions, one for each triangle.
a² = (CD)² + (BD)² ⇒ (CD)² = a² − (BD)²
b² = (CD)² + (AD)² ⇒ (CD)² = b² − (AD)²
and therefore
a² − (BD)² = b² − (AD)²
Substitute the known values BD = a cos B and AD = b cos A, and you have
a² − a² cos² B = b² − b² cos² A
Bzzt! No good! That uses two sides and two angles, but we need an equation in three sides and one angle, so that we can solve for that angle. Let’s back up a step, to a² − (BD)² = b² − (AD)², and see if we can go in a different direction.
Maybe the problem is in treating BD and AD as separate entities when actually they’re parts of the same line. Since BD + AD = c, we can write
BD = c − AD
BD = c − b cos A.
Notice that this brings in the third side, c, and angle B drops out. Substituting, we now have
a² − (BD)² = b² − (AD)²
a² − (c − b cos A)² = b² − (b cos A)²
This looks worse than the other one, but actually it’s better because it’s what we’re looking for: an equation for the three sides and one angle. We can solve it with a little algebra:
a² − c² + 2bc cos A − b²cos² A = b² − b²cos² A
a² − c² + 2bc cos A = b²
2bc cos A = b² + c² − a²
cos A = (b² + c² − a²) / 2bc
We were a long time getting there, but finally we made it. Now we can plug in the lengths of the sides that I mentioned in the first paragraph, and come up with a value for cos A, which in turn will tell us angle A:
cos A = (238² + 340² − 180²) / (2 × 238 × 340)
cos A ≈ 0.864088
A ≈ 30.2°
Do the same thing to find the second angle (or use the Law of Sines, since it’s less work), then subtract the two known angles from 180° to find the third angle.
You can find the Law of Cosines for the other angles by following the same process using the other two perpendiculars.
(30) Law of Cosines—First Form:
cos A = (b² + c² − a²) / 2bc
cos B = (a² + c² − b²) / 2ac
cos C = (a² + b² − c²) / 2ab
Just for fun, let’s find the other two angles of that triangle:
cos C = (a² + b² − c²) / 2ab
cos C = (180² + 238² − 340²) / (2 × 180 × 238)
cos C ≈ −0.309944
C ≈ 108.1°
Notice that the Law of Cosines automatically handles acute and obtuse angles. Remember from the diagram in Functions of Any Angle that cos A is negative when A is between 90° and 180°. Because the cosine has unique values all the way from 0° to 180°, you never have to worry about multiple solutions of a triangle when you use the Law of Cosines.
There’s another wellknown form of the Law of Cosines, which may be a bit easier to remember. Start with the above form, multiply through by 2ab, and isolate c on one side:
cos C = (a² + b² − c²) / 2ab
2ab cos C = a² + b² − c²
c² = a² + b² − 2ab cos C
You can play the same game to solve for the other two sides:
(31) Law of Cosines—Second Form:
a² = b² + c² − 2bc × cos A
b² = a² + c² − 2ac × cos B
c² = a² + b² − 2ab × cos C
Typically you’ll use the Law of Cosines in the first form for finding an angle and the second form for finding a side.
Probably you don’t want to try to remember that, but it’s not as hard as it looks. I think of it this way: the square of one side is the sum of the squares of the other two, like Pythagoras, but with a “correction factor” of 2 times those same sides times the cosine of the opposite angle.
With just the definitions of sine, cosine, and tangent, you can solve any right triangle. If you’ve got the Law of Sines and the Law of Cosines under your belt, you can solve any triangle that exists. (Some sets of givens lead to an impossible situation, like a “triangle” with sides 349.)
Really, it’s pretty straightforward. Whenever you have to solve a triangle, think about what you have and then think about which formula you can use to get what you need. (When you have two angles, you can always find the third by A + B + C = 180°.)
Many people find it easier to think about the known elements of a triangle as a “case”. For instance, if you know two angles and the side between them, that’s case ASA; if you know two angles and a side that’s not between them, that’s case AAS, and so on.
I’m not presenting the following table for you to memorize. Instead, what I hope to do is show you that between the Law of Sines and the Law of Cosines you can solve any triangle, and that you simply pick which law to use based on which one has just one unknown and otherwise uses information you already have.
Most cases can be solved with the Law of Sines. But if you have three sides (SSS), or two sides and the angle between them (SAS), you must begin with the Law of Cosines.
If you know this...  You can solve the triangle this way...  

three angles, AAA  There’s not enough information. Without at least one side you have the shape of the triangle, but no way to scale it correctly. For example, the same angles could give you a triangle with sides 71213, 356065, or any other multiple.  
two angles and a side, AAS or ASA  Find the third angle by subtracting from 180°. Then use the Law of Sines (28)★ twice to find the second and third sides.  
two sides and ...  the included angle, SAS  Use the Law of Cosines (31)★ to find the third side. Then use either the Law of Sines (29)★ or the Law of Cosines (30)★ to find the second angle. 
a nonincluded angle, SSA  Use the Law of Sines (29)★ to get the
second angle, and the Law of Sines (28)★
to get the third side.
But...


three sides, SSS  Find one angle with the Law of Cosines (30). Use that angle and its opposite side in the Law of Sines (29) to find the second angle, then subtract to find the third angle.  
two angles and the area  See Given: Area and Two Angles, below.
Find the third angle. Next, find a side using a = √2 × area × sin A/(sin B sin C) Then, proceed as in the ASA case, above. 

two sides and the area  See Given: Area and Two Sides, below.
Find the included angle with sin A = 2 × area/(b c) Then, proceed as in the SAS case, above. 

★ If a 90° angle is given, the Law of Sines and the Law of Cosines are overkill. Just apply the definitions of the sine and cosine (equation 1) and the tangent (equation 4) to find the other sides and angles. 
For most sets of facts, either there’s a unique solution or they’re obviously absurd. (If you don’t see why a “triangle” with sides 5060200 is absurd, try to sketch it.) But the SSA case can be tricky.
Suppose you know acute angle B and sides a and b. Given those facts, there are two different ways you could draw the triangle, as shown in the picture. How can this be? Well, you use the Law of Sines to find the sines of angles A and C. Let’s say you find sin C = 0.5. That means C could be either 30° or the supplement, 150°. Remember that the sine of any angle and the sine of its supplement are the same.
This is the infamous ambiguous case. You can see the problem from the picture: the known opposite side b can take either of two positions that satisfy the given the lengths of a and b. Those two positions give rise to two different values for angle A, two different values for angle C, and two different values for side c. Think about it for a while, and you’ll see that this ambiguity can arise only when the known angle is acute, and the adjacent side is longer than the opposite side, and the opposite side is greater than the height.
Here’s a complete rundown of all the possibilities with the SSA case:
Possibilities within the SSA Case  

known angle < 90°  known angle ≥ 90°  
adjacent side < opposite side  one solution  one solution 
adjacent side = opposite side  one solution  no solution (Angles that are opposite equal sides must be equal, but you can’t have two angles both ≥ 90° in a triangle.) 
adjacent side > opposite side  Compute the triangle height h (adjacent side times sine of known
angle).

no solution (The conditions violate the theorem that the longest side is always opposite the largest angle.) 
For heaven’s sake, don’t try to memorize that table! Instead, always draw a picture. If you can draw two pictures that both fit all the available facts, you have two legitimate solutions. If only one picture fits all the facts, it will show you which angle (if any) is > 90°. And if you can’t make any picture that fits the facts, the triangle has no solution.
If you do have two solutions, what do you do? If you have no other information to go on, of course you report both solutions. But check the situation carefully. Maybe you’re told explicitly which is the largest angle, or it’s implied by other facts you know. In that case your solution is constrained, and you reject the solution that doesn’t meet the constraints.
Example: Suppose you are asked to solve a triangle with B = 36.9°, a = 75.3, and b = 51.3. How do you proceed?
Solution: Start with a sketch, like the one shown at right. This helps you assign the numbers to the right elements of the triangle.
This is the sidesideangle case: you know two sides a and b, and a nonincluded angle B. The adjacent side to angle B, a = 75.3, is larger than the opposite side, a = 51.3, so you have to compute the height, h = 75.3 sin 36.9° ≈ 45.2. The opposite side, a = 51.3, is larger than this, so there are two solutions.
Use the Law of Sines, equation 29, to get the second angle:
(sin A)/a = (sin B)/b
sin A = (a/b) sin B
sin A = (75.3 / 51.3) sin 36.9° ≈ 0.8813
A = 61.8° or 180° − 61.8° = 118.2°
If A = 61.8° ...  If A = 118.2° ... 

Angle C = 180 − A − B C = 180 − 61.8 − 36.9 = 81.3° Use the Law of Sines, equation 28, for the third side: c/(sin C) = b/(sin B) c = b sin C / sin B c = 51.3 sin 81.3° / sin 36.9° ≈ 84.5 All six elements of the triangle, in order, are A=61.8°, c=84.5, B=36.9°, a=75.3, C=81.3°, b=51.3. 
Angle C = 180 − A − B C = 180 − 118.2 − 36.9 = 24.9°
Use the Law of Sines, equation 28, for the third side: c / sin C = b / sin B c = b sin C / sin B c = 51.3 sin 24.9° / sin 36.9° ≈ 36.0 All six elements of the triangle, in order, are A=118.2°, c=36.0, B=36.9°, a=75.3, C=24.9°, b=51.3. 
In April 2016, Caroline McKnoe asked me how to solve a triangle if you have two angles and the area. I hadn’t run across that one before, but it’s doable with the standard ploy of dropping a perpendicular.
Recall that the area of a triangle is base × height/2. Here the base is c and the height (CD) is b sin A. (CD also equals a sin B, but for this solution it doesn’t matter which expression you use.) That gives you
area = (c b sin A)/2
We know angle A—even if A isn’t one of the two givens we can easily find it by subtracting the other two from 180°—but there are two unknown sides in that equation. How can we eliminate one of them? We need some second equation that involves b and c but no other unknowns. The answer is in the Law of Sines:
b/sin B = c/sin C ⇒ b = c sin B/sin C
Substitute that in the equation for area:
area = (c b sin A)/2
area = (c² sin B sin A)/(2 sin C)
Solve for side c:
c² = 2 area sin C/(sin A sin B)
(32) c = √2 × area × sin C/(sin A sin B)
Finally, use the Law of Sines to find sides a and b.
After solving a triangle given the area and two angles, it’s natural to wonder if you can do it given the area and two sides. The answer is yes, and it’s even a bit easier than the case where you know the area and two angles.
In the previous section, we found a formula for area in terms of two sides and the included angle:
area = (c b sin A)/2
We couldn’t use that directly when we knew two angles and the area, but if we know two sides and the area then this formula is exactly what we want. Just solve for sin A:
(33) sin A = 2 × area/(b c)
Next, use the Law of Cosines to find side a. Finally, use the Law of Sines or Law of Cosines to find a second angle, and subtract those angles from 180° to find the third angle.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
How long must the monorail and the foot bridge be?
Bonus question: If the river has the same width all along the stretch from A to B, how wide is it?
You’re now driving along a straight side road. At the end of 9.8 miles on the side road, you turn 135° to the right, on a third road. (If you’re visualizing this from above, the 135° change of direction corresponds to an angle of 180° − 135° = 45° in the triangle.)
Assuming that road continues in the same direction, how far must you drive to reach your starting point?
I have to recommend a terrific little book, How To Solve It by G. Polya. Most teachers aren’t very good at teaching you how to solve problems and do proofs. They show you how they do them, and expect you to pick up their techniques by a sort of osmosis. But most of them aren’t very good at explaining the thought process that goes into doing a geometrical proof, or solving a dreaded “story problem”.
Polya’s book does a great job of teaching you how to solve problems. He shows you the kinds of questions you should ask yourself when you see a problem. In other words, he teaches you how to get yourself over the hum, past the floundering that most people do when they see an unfamiliar problem. And he does it with lots of examples, so that you can develop confidence in your techniques and compare your methods with his. The techniques I’ve mentioned above are just three out of the many in his book.
There’s even a handy checklist of questions you can ask yourself whenever you’re stuck on a problem.
How To Solve It was first published in 1945, and it’s periodically in and out of print. If you can’t get it from your bookstore, go to the library and borrow a copy. You won’t be sorry.
Summary: Up till now we’ve been working with the trig functions in terms of the sides of triangles. But now, we’ll extend the definitions for any angle—actually, for any number. That’s a little bit more abstract, but it makes the trig functions useful for all sorts of problems that don’t involve triangles at all.
The trig functions are sometimes called circular functions because they’re intimately associated with circles. You’ve already seen that, with all six functions in a complicated diagram, but let’s reduce it to the essentials.
Take a right triangle, and place one of the two acute angles at the center of a circle, with the adjacent leg along the x axis. The hypotenuse then runs from the center of the circle to a point on the circumference, so the hypotenuse is a radius of the circle. Or, you could say that it runs from the origin (0, 0) to the point (x, y).
You already know the lengths of the two legs, in terms of the hypotenuse r and A:
y = r sin A
x = r cos A
This comes right from the original definitions of sine and cosine of an angle:
sin A = opposite/hypotenuse = y/r
cos A = adjacent/hypotenuse = x/r
But here’s the thing. If you take away the triangle, nothing else really needs to change. Think about an angle as a rotation around the center of a circle, and you can extend the trig functions to work for any angle.
The angle is always measured as a rotation from the positive x axis. Positive angles are counterclockwise rotations, and negative angles are clockwise rotations. This puts the acute angles, 0° to 90°, in Quadrant I, and the obtuse angles, 90° to 180°, in Quadrant II.
In the diagram, the general angle A is drawn in standard position, meaning that the vertex of the angle is at the origin, and one side lies along the positive x axis. I could have drawn any angle, but I just happened to draw one in Q II.
You’re probably used to thinking of angles in terms of degrees, but really they make more sense in radian measure. Why is that? You probably remember that the circumference of a circle is 2πr, where r is the radius. The arc length for any angle A is s = Ar, or the angle is A = s/r. If you sweep out a full rotation, all the way around the circle, you sweep out an arc of the whole circumference, 2πr, and the angle must be 2πr/r, or just 2π.
Fractions of a full rotation are fractions of 2π. For example, the 12 o’clock position is a quarter of the way around the circle from the positive x axis, so the angle is 2π/4, which is π/2. The 6’oclock position is 3/4 of the way around, going counterclockwise, so the angle is (3/4) × 2π = 3π/2. But it’s also a quarter of the way around, going clockwise, which is the negative direction, so it’s also (−1/4) × 2π = −π/2. (We’ll get into that more in Periodic Functions, below.)
Half a rotation is 2π/2 = π, and of course it’s also 360°/2 = 180°. Since π = 180°, π/180° = 1 and 180°/π = 1. This is why, if you need to convert between degrees and radians, you multiply by π/180° or 180°/π—you’re multiplying by 1, so you’re changing the form of the number but not its value. Incidentally, this technique of multiplying by a carefully chosen form of the number 1 lets you convert almost any units, like meters per second into miles per hour, for instance.
Not only do you not really need a triangle, you don’t even really need an angle. Just think of a point moving along the circle, in either direction. The “angle” is just the distance the point has traveled, divided by the radius: A = s/r. Counterclockwise motion is positive; clockwise is negative.
But it can get even simpler. Set the radius to 1, and you have a unit circle. The moving point starts at (1, 0) and travels a distance A. The point has coordinates (x, y), which depend only on the distance (counterclockwise or clockwise) from the point (1, 0).
The circle’s radius is 1—not 1 meter or 1 foot or 1 mile, just plain old 1. Likewise, A is just a pure number. So really you can say that the arguments of the trig functions are just pure numbers. And A can be any number, positive or negative, so the trig functions can take any real number as their arguments.
(21) On the unit circle:
sin A = y
cos A = x
This doesn’t change anything you’ve already learned. It just extends the sine and cosine—instead of functions of an angle 0° to 180°, they are functions of any real number.
The other function definitions don’t change at all. From equation 3, you still have
tan A = sin A / cos A
which means that
tan A = y/x
and the other three functions are still defined as reciprocals (equation 5).
Once again, there’s nothing new here: we’ve just extended the original definitions to a larger domain. sin A and cos A still have a range of [−1, +1], just as they always did.
It turns out that all kinds of physical processes vary in terms of sines and cosines as functions of time: length of the day over the course of a year; vibrations of a spring, or of atoms, or of electrons in atoms; voltage and current in an AC circuit; pressure of sound waves, electric and magnetic field strength as a beam of light propagates. Nearly every periodic process can be described in terms of sines and cosines.
Even static processes show sines and cosines: when forces are operating at an angle, for instance. There’s also Snell’s Law, which governs how light rays bend when passing from water to air, or between any two media: n_{1} sin θ_{1} = n_{2} sin θ_{2}, where the n’s are characteristics of each material called an index of refraction, and the angles are measured from a perpendicular to the interface.
We won’t get too far into all that in these pages, but this is why you need a solid grounding in trig to study physics, engineering, and other fields.
In the olden days, before calculators, every trig book had a table of function values for acute angles, and if you needed the function value for any other angle you would have to work it out using a reference angle. Now, of course, we use an app or a pocket calculator to get the function values, but the concept of a reference angle is still useful in simplifying expressions.
The reference angle is the acute angle between the x axis and the terminal side of the original angle.
Take a look at angle A, which is in Quadrant II. The terminal side crosses the unit circle at (x, y), where x = cos A (which is negative) and y = sin A (which is positive).
A vertical line from that point to the x axis has length y, and it strikes the x axis −x units to the left of the origin, at coordinates (x, 0). Why do I say −x units left of the origin? Because x itself is negative, but all distances are positive, so the positive distance must be minus the negative x coordinate.
The reference angle is 180° − A, or π − A. Why? Because the two angles together equal 180° (π).
The diagram shows that y is not only sin A but also sin(180° − A). Therefore sin(180° − A) or sin(π − A) = sin A.
What about x? Well, cos(180° − A) = −x, using the right triangle in the diagram. But cos A = x (which is negative). Therefore cos(180° − A) or cos(π − A) = −cos A.
There’s nothing special about angle A; sin(π − A) = sin A and cos(180° − A) = −cos A for all angles A or all numbers A. We can get even more general: the six function values for any angle equal the function values for its reference angle, give or take a minus sign.
What’s this “give or take” business? That’s what the next section is about.
Remember the extended definitions from equation 21:
sin A = y, cos A = x
Therefore the signs of sine and cosine are the same as the signs of y and x. But you know which quadrants have positive or negative y and x, so you know which angles (or numbers) have positive or negative sines and cosines. And since the other functions are defined in terms of the sine and cosine, you also know where they are positive or negative.
Spend a few minutes thinking about it, and draw some sketches. For instance, is cos 300° positive or negative? Answer: 300° is in Q IV, which is in the righthand half of the circle. Therefore x is positive, and the cosine must be positive as well. The reference angle is 60° (draw it!), so cos 300° equals cos 60° and not −cos 60°.
You can check your thinking against the chart that follows. Don’t memorize the chart! Its purpose is to show you how to reason out the signs of the function values whenever you need them, not to make you waste storage space in your brain.
Signs of Function Values  

Q I 0 to 90° 0 to π/2 
Q II 90 to 180° π/2 to π 
Q III 180 to 270° π to 3π/2 
Q IV 270 to 360° 3π/2 to 2π 

x  +  −  −  + 
y  +  +  −  − 
sin A (= y)
csc A (= 1/y) 
+  +  −  − 
cos A (= x)
sec A (= 1/x) 
+  −  −  + 
tan A (= y/x)
cot A (= x/y) 
+  −  +  − 
Though I told you not to memorize the chart, I have to share a cute mnemonic I ran across: All Students Take Calculus.
What about other angles? Well, 420° = 360° + 60°, and therefore 420° ends in the same position in the circle as 60°—it’s just going once around the circle and then an additional 60°. So 420° is in Q I, just like 60°.
You can analyze negative angles the same way. Take −45°. That occupies the same place on the circle as 360° − 45°, which is +315°. −45° is in Q IV.
As you’ve seen, for any function you get the numeric value by considering the reference angle and the positive or negative sign by looking where the angle is.
Example: What’s cos 240°?
Solution: Draw the angle and see that the reference angle is 60°; remember that the reference angle always goes to the x axis, even if the y axis is closer. cos 60° = ½, and therefore cos 240° will be ½, give or take a minus sign. The angle is in Q III, where x is negative, and therefore cos 240° is negative. cos 240° = −½.
Example: What’s tan(−225°)?
Solution: Draw the angle and find the reference angle of 45°. tan 45° = 1. But −225° is in Q II, where x is negative and y is positive; therefore y/x is negative. tan(−225°) = −1.
The techniques we worked out above can be generalized into a set of identities. For instance, if two angles are supplements then you can write one as A and the other as 180° − A or π − A. You know that one will be in Q I and the other in Q II, and you also know that one will be the reference angle of the other. Therefore you know at once that the sines of the two angles will be equal, and the cosines of the two will be numerically equal but have opposite signs.
This diagram may help:
Here you see a unit circle (r = 1) with four identical triangles. Their angles A are at the origin, arranged so that they’re mirror images of each other, and their hypotenuses form radii of the unit circle. Look at the triangle in Quadrant I. Since its hypotenuse is 1, its other two sides are cos A and sin A.
The other three triangles are the same size as the first, so their sides must be the same length as the sides of the first triangle. But you can also look at the other three radii as belonging to angles 180° − A in Quadrant II, 180° + A in Quadrant III, and −A or 360° − A in Quadrant IV. The thin arcs near the center of the circle trace the rotations. All of them have a reference angle equal to A. From the symmetry, you can immediately see things like sin(180° + A) = −sin A and cos(−A) = cos A.
The relations are summarized below. Don’t memorize them! Just draw a diagram whenever you need them—it’s easiest if you use a hypotenuse of 1. Soon you’ll find that you can quickly visualize the triangles in your mind and you won’t even need to draw a diagram. The identities for tangent are easy to derive: just divide sine by cosine as usual.
sin(180° − A) = sin A
sin(π − A) = sin A 
cos(180° − A) = −cos A
cos(π − A) = −cos A 
tan(180° − A) = −tan A
tan(π − A) = −tan A 
(22) 
sin(180° + A) = −sin A
sin(π + A) = −sin A 
cos(180° + A) = −cos A
cos(π + A) = −cos A 
tan(180° + A) = tan A
tan(π + A) = tan A 

sin(−A) = −sin A  cos(−A) = cos A  tan(−A) = −tan A 
The formulas for the other functions aren’t needed very often, but when you do need them they drop right out of the definitions in equation 3 and equation 5, plus the formulas just above this paragraph. Two examples:
cot(180° + A) = 1/tan(180° + A) = 1/tan A ⇒ cot(180° + A) = cot A
csc(−A) = 1/sin(−A) = 1/(−sin A) = −1/sin A ⇒ csc(−A) = −csc A
One final comment: Drawing pictures is helpful to avoid memorizing things, but you might not consider it a rigorous proof of equation 22. Later, in Sum and Difference Formulas, you’ll see how to prove these identities with algebra, independent of any picture.
Once you think of the trig functions as based on the (x, y) coordinates of a point on a circle, you can see that adding 2π (360°) to an angle or subtracting 2π (360°) is just moving once around the circle. But if you move all the way around a circle, in either direction, you end up where you started. So, for example, sin 495° = sin(360° + 135°). (If you want to, you can go further by noticing that 495° or 135° is in Quadrant II, and the reference angle is 45°, so sin 495° = sin 45°.)
But if you can go around the circle once, you can go around the circle any number of times. So adding or subtracting any multiple of 2π (360°) doesn’t change the coordinates (x, y), and therefore doesn’t change the value of the sine or cosine.
For this reason we say that sine and cosine are periodic functions with a period of 360° or 2π. Their values repeat over and over again. Of course secant and cosecant, being reciprocals of cosine and sine, must have the same period.
Example: Express cos(−85π/12) using the smallest possible positive angle.
Solution: −85π/12 = −7π − π/12. Because multiples of 2π don’t change anything, you want to break out an even multiple of 2π, so subtract π and add 12π/12 to get
−85π/12 = −8π + 11π/12 ⇒cos(−85π/12) = cos(11π/12)
(Yes, you could write it as −6π − 13π/12, but I think that makes it harder to visualize. I like to keep any minus sign to the multipleof2π part, which gets thrown away.)
Nearly there now! But instructions were to use the smallest positive angle, and you can do better than 11π/12. The reference angle is π/12—sketch the angles if you need to! 11π/12 is in Quadrant II, where cosines are negative, so cos 11π/12 = −cos π/12, and therefore cos(−85π/12) = −cos π/12.
What about tangent and cotangent? They are periodic too, but their period is 180° or π: they repeat twice as fast as the others. You can see this from equation 22: tan(180° + A) = tan A says that the function values repeat every 180° or π radians.
Summarizing all of this:
(23) For any integer k:
sin(A+2πk) = sin A
cos(A+2πk) = cos A
tan(A+πk) = tan A
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
You may want to skip this section, especially the first time you read the chapter. I find the equations of periodic motion interesting, but I have to admit that if you need them you’ll get them in your other classes. In that case, if you find something hard to follow, you may want to come back here for another approach.
Almost every repetitive process is governed by sines and cosines. You’ll learn about this in physics or engineering classes, or you can look at one of the many Web sites out there. I like Waves in Tom Henderson’s Physics Classroom site.
Sine and cosine have periods of 2π, but obviously the great majority of physical processes have different periods.
How can we modify a sine or cosine function to have any desired period? Well, suppose you want a process that repeats every second. That needs to go faster than sin t, which only repeats every 2π seconds (about 6.3 s). To speed things up, multiply t by some factor greater than 1. In this case, use f(t) = sin(2πt) for a period of 1 second.
Take a look at the plot at right. The blue curve is sin t, and the red curve is sin(2πt). The dots are spaced at 1 second horizontally, with a 2πsecond “yardstick” added. You can see that sin t repeats its cycle once in 2π seconds, where sin(2πt) repeats once per second. (You can click the image to enlarge it.)
That was just one specific example, but here’s the general rule: the period of a sine function is 2π over the coefficient of the variable. So the period of sin(2πt) is 2π divided by the coefficient, which is also 2π: 2π/(2π) = 1, and the period is 1 second. What would a function with a period of 5 seconds look like? Well, 2π/5 = 0.4π, so the coefficient needs to be 0.4π: sin(0.4πt). At right I’ve graphed sin t in blue again, and sin(0.4πt) in purple. You see that sin(0.4πt) varies a bit more rapidly than sin t, with a period of 5 seconds (5 tick marks) versus about 6.3 seconds.
Period answers the question, how long does it take one complete waveform to pass a given point? The flip side of that, frequency, answers the question, how many waveforms pass a given point in one unit of time? If the period is 1/4 second, then 4 waveforms pass each second, and the frequency is 4 Hz (Hertz), or in older US lingo 4 cycles per second. If the period is 5 seconds, then 1/5 of a waveform passes in 1 second, and the frequency is 1/5 = 0.2 Hz.
Frequency is 1/period. Therefore, in sin at, where the period is 2π/a, the frequency is a/2π.
The sine function varies between −1 and +1, but of course most physical processes vary between other numbers. We use the word amplitude for half the difference between the low point of a wave and its high point (trough and crest). The amplitude of a sine function is the coefficient of the function (not the variable). Suppose you have a waveform measuring 4″ (10 cm) from trough to crest. The amplitude is therefore 2″ (5 cm), and the function is 2 sin(something) in inches, or 5 sin(something) in centimeters. (For this example I’m ignoring the period of the wave.)
At the right I’ve graphed sin t in blue again, and 3 sin t in green. You can see that the two waveforms have the same period of 2π, but the green one varies three times as much in value. The amplitude is a measure of the energy in a wave; specifically, the energy in a wave is proportional to the square of its amplitude, and the proportion depends on the material that’s carrying the wave, as well as other factors.
Finally, a waveform can have phase, meaning that it’s shifted left or right from the normal position, or forward or backward in time. Take a look at the plot to the right. The blue curve is our old friend sin t, and the black curve is sin(t + π/3). The two are out of phase by π/3 or 1/6 of a cycle. We can also say that the phase of the black curve is −π/3, because the black curve lags behind the blue curve by π/3 units.
The second view is the same curves, just displaced vertically so that you can see each curve more easily. Notice the crests of the two curves, their troughs, the ascending crossings of the x axis, and the descending crossings. In each case, the black curve is π/3 behind the blue curve.
You might have noticed that I gradually stopped talking about sine and cosine curves, and focused only on the sine curve. Why is that? Phase is the key.
Look at the plot. The blue is sin t, displaced downward two units. The brown is sin(t + π/2), and the yellow is none other than our neglected friend cos t (displaced upward two units). The cosine curve is just the sine curve, with a phase shift. Pretty cool, huh?
Is this just an accident? No, and it’s easy to prove. From equation 22,
sin(π − A) = sin A
Putting π/2 − t for A,
sin(π − (π/2−t)) = sin(π/2 − t)
sin(π − π/2 + t) = sin(π/2 − t)
sin(t + π/2) = sin(π/2 − t)
But from equation 2,
cos t = sin(π/2 − t)
And therefore,
cos t = sin(t + π/2)
(24) If f(t) =
a + b sin(ct + d), then:
frequency = c/2π,
period = 2π/c,
phase = d,
amplitude = b,
vertical shift = a.
Example: My bicycle tire has a diameter of 622 mm (about 24.5 in), and my normal cruising speed is around 10 mph (16.21 km/h) Suppose there’s a spot on the outer rim of the tire. Find the height of that spot above the ground, as a function of time, assuming that the spot is touching the ground at time 0.
Solution: I find it easiest to build up these functions piece by piece.
The diameter of the wheel is 622/1000 = 0.622 m. The radius is half that, 0.311 m. The circumference of the wheel is 0.622π m ≈ 1.954 m (76.9 in).
The bicycle is moving at 16.21 km/h (10 mph) = 16,210 m/h = 16,210/3,600 m/s ≈ 4.503 m/s.
The wheel revolves 4.503/1.954 ≈ 2.304 times per second, so its frequency is 2.304 Hz. (The period is 1/2.304 = 0.434 s, by the way.) Frequency is the coefficient of t over 2π, so the coefficient must be 2π times the frequency. 2π × 2.304 ≈14.48.
The amplitude is the radius of the wheel, 0.311 m (12.24 in). Our tentative function is
h(t) = 0.311 sin 14.48t in meters, 12.24 sin 14.48t in inches
That doesn’t take account of phase or vertical shift, but it’s a start.
Vertical shift is easy. The center of the wheel isn’t on the ground; it’s 0.311 m (12.24 in) above the ground, so now our function is
h(t) = 0.311 + 0.311 sin 14.48t in meters, 12.24 + 12.24 sin 14.48t in inches
The last thing to consider is phase. sin 0 = 0, so with a phase of 0 the spot would have to start at 0.311 m (12.24 in) above the ground—the same height as the center of the wheel, and toward the rear of the bicycle. But we’re told that the spot starts at ground level: h(0) = 0, not 0.311 m. This means that the spot is lagging by π/2, a quarter of a revolution. So our function becomes
h(t) = 0.311 + 0.311 sin(14.48t − π/2) ⇒
h(t) = 0.311 + 0.311 sin(14.48t − 1.571) in meters
h(t) = 12.24 + 12.24 sin(14.48t − 1.571) in inches
Let’s check that against equation 24. Frequency is 14.48/2π = 2.304 Hz, check. Phase is −1.571, which is a lag of π/2, a quarter cycle, check. Amplitude is 0.311 m (radius of the wheel), check. And vertical shift is also the radius of the wheel, 0.331 m, check.
Summary: This chapter begins exploring trigonometric identities. Three of them involve only squares of functions. These are called Pythagorean identities because they’re just the good old Theorem of Pythagoras in new clothes. Learn the really basic one, namely sin² A + cos² A = 1, and the others are easy to derive from it in a single step.
Students seem to get bogged down in the huge number of trigonometric identities. As I said earlier, I think the problem is that students are expected to memorize all of them. But really you don’t have to, because they’re all just forms of a very few basic identities. The next couple of chapters will explore that idea.
For example, let’s start with the really basic identity:
(38) sin² A + cos² A = 1
That one’s easy to remember: it involves only the basic sine and cosine, and you can’t get the order wrong unless you try.
But you don’t have to remember even that one, since it’s really just another form of the Pythagorean Theorem. (You do remember that, I hope?) Just think about a right triangle with a hypotenuse of one unit, as shown at right.
First convince yourself that the figure is right, that the lengths of the two legs are sin A and cos A. (Check back in the section on lengths of sides, if you need to.) Now write down the Pythagorean Theorem for this triangle. Voilà! You’ve got equation 38.
What’s nice is that you can get the other “squared” or Pythagorean identities from this one, and you don’t have to memorize any of them. Just start with equation 38 and divide through by either sin² A or cos² A.
For example, what about the riddle we started with, the relation between tan² A and sec² A? It’s easy to answer by a quick derivation—easier than memorizing, in my opinion.
If you want an identity involving tan² A, remember equation 3: tan A is defined to be sin A/cos A. Therefore, to create an identity involving tan² A you need sin² A/cos² A. So take equation 38 and divide through by cos² A:
sin² A + cos² A = 1
sin² A/cos² A + cos² A/cos² A = 1/cos² A
(sin A/cos A)² + 1 = (1/cos A)²
which leads immediately to the final form:
(39) tan² A + 1 = sec² A
You should be able to work out the third identity (involving cot² A and csc² A) easily enough. You can either start with equation 39 above and use the cofunction rules (equation 6 and equation 7), or start with equation 38 and divide by something appropriate. Either way, check to make sure that you get
(40) cot² A + 1 = csc² A
It may be easier for you to visualize these two identities geometrically. Start with the sin A, cos A, 1 right triangle above. Divide all three sides by cos A and you get the first triangle below; divide by sin A instead and you get the second one. You can then just read off the Pythagorean identities.
From the first triangle, tan² A + 1 = sec² A; from the second triangle, cot² A + 1 = csc² A.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
Trig without Tears Part 7:
Copyright © 1997–2019 Stan Brown, BrownMath.com
Summary: Continuing with trig identities, this page looks at the sum and difference formulas, namely sin(A ± B), cos(A ± B), and tan(A ± B). Remember one, and all the rest flow from it. There’s also a beautiful way to get them from Euler’s formula.
Formulas for cos(A + B), sin(A − B), and so on are important but hard to remember. Yes, you can derive them by strictly trigonometric means. But such proofs are lengthy, too hard to reproduce when you’re in the middle of an exam or of some long calculation.
This brings us to W.W. Sawyer’s marvelous idea, as expressed in chapter 15 of Mathematician’s Delight (1943; reprinted 1991 by Penguin Books). He shows how you can derive the sum and difference formulas by ordinary algebra and one simple formula.
The ordinary algebra is simply the rules for combining powers:
(46) x^{a} x^{b} = x^{a+b}
(x^{a})^{b} = x^{a b}
(If you’re a bit rusty on the laws of exponents, you may want to review them.)
You may already know the “simple formula” that I mentioned above. It’s
(47) Memorize:
cos x + i sin x = e^{i x}
The formula is not Sawyer’s, by the way; it’s commonly called Euler’s formula. I don’t even know whether the idea of using Euler’s formula to get the sine and cosine of sum and difference is original with Sawyer. But I’m going to give him credit, since his explanation is simple and clear and I’ve never seen it explained in this way anywhere else.
You’ll sometimes see cos x + i sin x abbreviated as cis x for brevity.
I’ve marked Euler’s formula “memorize”. Although it’s not hard to derive (and Sawyer does it in a few steps by means of power series), you have to start somewhere. And that formula has so many other applications that it’s well worth committing to memory. For instance, you can use it to get the roots of a complex number and the logarithm of a negative number.
Okay, back to Sawyer’s idea. What happens if you substitute x = A + B in equation 47 above? You get
cos(A + B) + i sin(A + B) = e^{iA + iB}
Hmmm, this looks interesting. It involves exactly what we’re looking for, cos(A + B) and sin(A + B). Can you simplify the righthand side? Yes, use equation 46 and then equation 47 to rewrite it:
cos(A + B) + i sin(A + B) = e^{iA + iB}
cos(A + B) + i sin(A + B) = e^{iA} e^{iB}
cos(A + B) + i sin(A + B) = (cos A + i sin A) (cos B + i sin B)
Multiply out the righthand side, and group real and imaginary terms separately:
cos(A + B) + i sin(A + B) = cos A cos B + i sin A cos B + i cos A sin B + i² sin A sin B
cos(A + B) + i sin(A + B) = cos A cos B + i sin A cos B + i cos A sin B − sin A sin B
cos(A + B) + i sin(A + B) = (cos A cos B − sin A sin B) + i (sin A cos B + cos A sin B)
Now here’s the sneaky part. If I tell you a+bi = 7−9i and ask you to solve for a and b, you know immediately that a = 7 and b = −9, right? More formally, if two complex numbers are equal, their real parts must be equal and their imaginary parts must be equal. So the above equation in sines and cosines is actually two equations, one for the real part and one for the imaginary part. (I’m showing the imaginary part first in the box below, to put sine before cosine.)
(48) sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
In just a few short steps, the formulas for cos(A + B) and sin(A + B) flow right from equation 47, Euler’s equation for e^{i x}. No more need to memorize which one has the minus sign and how all the sines and cosines fit on the righthand side: all you have to do is a couple of substitutions and a multiply.
Example: What’s the exact value of cos 75° or cos(5π/12)?
Solution: 75° = 45°+30° (5π/12 = π/4+π/6). Using equation 48,
cos 75° = cos(45°+30°)
cos 75° = cos 45° cos 30° − sin 45° sin 30°
cos 75° = (√2/2)×(√3/2) − (√2/2)×(1/2)
cos 75° = √6/4 − √2/4
What about the formulas for the differences of angles? You can write them down at once from equation 48 by substituting −B for B and using equation 22. Or, if you prefer, you can get them by substituting x = A−B in equation 47 above. Either way, you get
(49) cos(A − B) = cos A cos B + sin A sin B
sin(A − B) = sin A cos B − cos A sin B
I personally find the algebraic reasoning given above very easy to follow, though you do have to remember Euler’s formula.
If you prefer geometric derivations of sin(A ± B) and cos(A ± B), you’ll find a beautiful set by Len and Deborah Smiley. (Phil Kenny drew my attention to this page’s original version and to the link at the University of Alaska.) Eric’s Treasure Trove of Mathematics has smaller versions of the pictures.
The fallback position is the standard proof: draw a diagram and use the distance formula or Pythagorean Theorem to prove the formula for cos(A − B).
Sometimes—though not very often—you have to deal with the tangent of the sum or difference of two angles. I have only a vague idea of the formula, but it’s easy enough to work out “on the fly”:
tan(A + B) = sin(A + B) / cos(A + B)
tan(A + B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)
What a mess! There’s no way to factor that and remove common terms—or is there? Suppose you start with a vague idea that you’d like to know tan(A+B) in terms of tan A and tan B rather than all those sines and cosines. The numerator and denominator contain sines and cosines, so if you divide by cosines you’d expect to end up with sines or perhaps sines over cosines. And sine/cosine is tangent, so this seems like a promising line of attack. Since you’ve got cosines of angles A and B to contend with, try dividing the numerator and denominator of the fraction by cos A cos B:
tan(A + B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)
tan(A + B) = [sin A/cos A + sin B/cos B] / [1 − (sin A/cos A)(sin B/cos B)]
Success! Simplify it using the definition of tan x, and you have
(50) tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
Now if you replace B with −B, you have the formula for tan(A − B). (Take a minute to review why tan(−x) = −tan x.)
(51) tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Example: What’s the exact value of tan 15° or tan(π/12)?
Solution: 15° = 60°−45° (π/12 = π/3 − π/4). Therefore
tan(π/12) = tan(π/3 − π/4)
tan(π/12) = [tan(π/3) − tan(π/4)] / [1 + tan(π/3) tan(π/4)]
tan(π/12) = (√3− 1) / (1 + √3×1)
tan(π/12) = (√3− 1) / (√3 + 1)
If you like, you can rationalize the denominator:
tan(π/12) = (√3− 1)² / (√3 + 1)(√3 − 1)
tan(π/12) = (3 − 2√3 + 1) / (3 − 1)
tan(π/12) = (4 − 2√3) / 2
tan(π/12) = 2 − √3
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
(a) cos(−A) = cos A (I’ve done the first step for you.)
cos(−A) = cos(0 − A)
(b) tan(π + A) = tan A
(c) sin(π − A) = sin A
Sometimes you need to simplify an expression like cos 3x cos 5x. Of course it’s not equal to cos(15x²), but can it be simplified at all? The answer is yes, and in fact you need this technique for calculus work. There are four formulas that can be used to break up a product of sines or cosines.
These producttosum formulas come from equation 48 and equation 49 for sine and cosine of A ± B. First let’s develop one of these formulas, and then we’ll look at an application before developing the others.
Take the two formulas for cos(A ± B) and add them:
cos(A − B) = cos A cos B + sin A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) + cos(A + B) = 2 cos A cos B
½ [cos(A − B) + cos(A + B)] = cos A cos B
Example: Suppose you need to graph the function
f(x) = cos 2x cos 3x,
or perhaps you need to find its integral. Both of these are rather hard to do with the function as it stands. But you can use the producttosum formula, with A = 2x and B = 3x, to rewrite the function as a sum:
f(x) = cos 2x cos 3x
f(x) = ½ [cos(2x − 3x) + cos(2x + 3x)]
f(x) = ½ [cos(−x) + cos 5x]
You know that cos(−x) = cos x, and therefore
f(x) = ½ [cos x + cos 5x]
f(x) = ½ cos x + ½ cos 5x
This is quite easy to integrate. And while it’s not exactly trivial to graph, it’s much easier than the original, because cos x and cos 5x are easy to graph.
The other three producttosum formulas come from the other three ways to add or subtract the formulas in equation 48 and equation 49. If you subtract the two cosine formulas instead of adding:
cos(A − B) = cos A cos B + sin A sin B
−cos(A + B) = −cos A cos B + sin A sin B
you get
cos(A − B) − cos(A + B) = 2 sin A sin B
½ [cos(A−B) − cos(A + B)] = sin A sin B
To get the other two productto sum formulas, add the two sine formulas from equation 48 and equation 49, or subtract them. Here are all four formulas together:
(52) cos A cos B = ½ cos(A − B) + ½ cos(A + B)
sin A sin B = ½ cos(A − B) − ½ cos(A + B)
sin A cos B = ½ sin(A + B) + ½ sin(A − B)
cos A sin B = ½ sin(A + B) − ½ sin(A − B)
The fourth one of those formulas really isn’t needed, because you can always evaluate cos p sin q as sin q cos p. But it’s traditional to present all four formulas.
There are also formulas that combine a sum or difference into a product. Heon Joon Choi, a physics student from Cornell, has kindly told me of an application: “superposing two waves and trying to figure out the nodes is much easier if they are multiplied, rather than added.” This makes sense: solving most equations is easier once you’ve factored them. The sumtoproduct formulas are also used to prove the Law of Tangents, though that itself is no longer used in solving triangles.
Here’s how to get the sumtoproduct formulas. First make these definitions:
A = ½(u + v), and B = ½(u − v)
Then you can see that
A + B = u, and A − B = v
Now make those substitutions in all four formulas of equation 52, and after simplifying you will have the sumtoproduct formulas:
(53) cos u + cos v = 2 cos(½(u + v)) cos(½(u − v))
cos u − cos v = −2 sin(½(u + v)) sin(½(u − v))
sin u + sin v = 2 sin(½(u + v)) cos(½(u − v))
sin u − sin v = 2 sin(½(u − v)) cos(½(u + v))
Euler’s formula (equation 47) is easily proved by means of power series. Start with the formulas
You can click the image to see a larger version. Or if you can’t see the image at all, here are the formulas in linear text:
e^{x} = SUM [ x^{k} / k! ] = 1 + x + x^{2}/2! + x^{3}/3! + ⋯
cos x = SUM [ (−1)^{k} x^{2k} / (2k)! ] = 1 − x^{2}/2! + x^{4}/4! − x^{6}/6! + ⋯
sin x = SUM [ (−1)^{k} x^{2k+1} / (2k+1)! ] = x − x^{3}/3! + x^{5}/5! − x^{7}/7! + ⋯
(These are how the function values are actually calculated, by the way. All three series converge quickly, meaning that you get quite an accurate result from computing just the first few terms. If you want to know the value of e^{2}, you just substitute 2 for x in the formula and compute until the additional terms fall within your desired accuracy.)
Now we have to find the value of e^{ix}, where i = √−1. Use the first formula to find e^{ix}, by substituting ix for x:
e^{ix} = SUM [ (ix)^{k} / k! ]
e^{ix} = 1 + (ix) + (ix)^{2}/2! + (ix)^{3}/3! + (ix)^{4}/4! + (ix)^{5}/5! + (ix)^{6}/6! + (ix)^{7}/7! + ...
Simplify the powers of i, using i² = −1:
e^{ix} = 1 + ix − x^{2}/2! − ix^{3}/3! + x^{4}/4! + ix^{5}/5! − x^{6}/6! − ix^{7}/7! + ...
Finally, group the real and imaginary terms separately:
e^{ix} = [1 − x^{2}/2! + x^{4}/4! − x^{6}/6! + ...] + i[x − x^{3}/3! + x^{5}/5! − x^{7}/7! + ...]
Those should look familiar. If you refer back to the power series at the start of this section, you’ll see that the first group of terms is just cos x and the second group is just sin x. So you have
e^{ix} = cos x + i sin x
which is Euler’s formula, as advertised!
You may wonder where the series for cos x, sin x, and e^{x} come from. The answer is that they are the Taylor series expansions of the functions. (You’ll probably study Taylor series in second or thirdsemester calculus.)
Trig without Tears Part 8:
Copyright © 1997–2019 Stan Brown, BrownMath.com
Summary: Very often you can simplify your work by expanding something like sin(2A) or cos(½A) into functions of plain A. Sometimes it works the other way and a complicated expression becomes simpler if you see it as a function of half an angle or twice an angle. The formulas seem intimidating, but they’re really just variations on equation 48 and equation 50.
With equation 48, you can find sin(A + B). What happens if you set B = A?
sin(A + A) = sin A cos A + cos A sin A
But A + A is just 2A, and the two terms on the righthand side are equal. Therefore:
sin 2A = 2 sin A cos A
The cosine formula is just as easy:
cos(A + A) = cos A cos A − sin A sin A
cos 2A = cos² A − sin² A
Though this is valid, it’s not completely satisfying. It would be nice to have a formula for cos 2A in terms of just a sine or just a cosine. Fortunately, we can use sin² x + cos² x = 1 to eliminate either the sine or the cosine from that formula:
cos 2A = cos² A − sin² A = cos² A − (1 − cos² A) = 2 cos² A − 1
cos 2A = cos² A − sin² A = (1 − sin² A) − sin² A = 1 − 2 sin² A
On different occasions you’ll have occasion to use all three forms of the formula for cos 2A. Don’t worry too much about where the minus signs and 1s go; just remember that you can always transform any of them into the others by using good old sin² x + cos² x = 1.
(59) sin 2A = 2 sin A cos A
cos 2A = cos² A − sin² A = 2 cos² A − 1 = 1 − 2 sin² A
There’s a very cool second proof of these formulas, using Sawyer’s marvelous idea. Also, there’s an easy way to find functions of higher multiples: 3A, 4A, and so on.
To get the formula for tan 2A, you can either start with equation 50 and put B = A to get tan(A + A), or use equation 59 for sin 2A / cos 2A and divide top and bottom by cos² A. Either way, you get
(60) tan 2A = 2 tan A / (1 − tan² A)
What about the formulas for sine, cosine, and tangent of half an angle? Since A = (2A)/2, you might expect the doubleangle formulas equation 59 and equation 60 to be some use. And indeed they are, though you have to pick carefully.
For instance, sin 2A isn’t much help. Put A = B/2 and you have
sin B = 2 sin(B/2) cos(B/2)
That’s true enough, but there’s no easy way to solve for sin(B/2) or cos(B/2).
There’s much more help in equation 59 for cos 2A. Put 2A = B or A = B/2 and you get
cos B = cos² (B/2) − sin² (B/2) = 2 cos² (B/2) − 1 = 1 − 2 sin² (B/2)
Use just the first and last parts of that:
cos B = 1 − 2 sin² (B/2)
Rearrange a bit:
sin² (B/2) = (1 − cos B) / 2
and take the square root
sin(B/2) = ± √(1 − cos B)/2
You need the plus or minus sign because sin(B/2) may be positive or negative, depending on B. For any given B (or B/2) there will be only one correct sign, which you already know from the diagram that we explored back in Functions of Any Angle.
Example: If B = 280°, then B/2 = 140°, and you know that sin 140° is positive because the angle is in Quadrant II (above the axis).
To find cos(B/2), start with a different piece of the cos 2A formula from equation 59:
cos 2A = 2 cos² A − 1
As before, put = B/2:
cos B = 2 cos² (B/2) − 1
Rearrange and solve for cos(B/2):
cos² (B/2) = (1 + cos B)/2
cos(B/2) = ± √(1 + cos B)/2
You have to pick the correct sign for cos(B/2) depending on the value of B/2, just as you did with sin(B/2). But of course the sign of the sine is not always the sign of the cosine.
(61) sin(B/2) = ± √(1 − cos B)/2
cos(B/2) = ± √(1 + cos B)/2
Example: Find sin 75°, which is sin 5π/12.
Solution: 75° is half of 150°, and you know the functions of 150° exactly because they are the same as the functions of 30°, give or take a minus sign.
sin 75° = sin(150°/2) = ±√(1 − cos 150°)/2
Here, cos 150° is negative because 150° is to the left of the origin, in Quadrant II, and 180° − 150° = 30°, so
cos 150° = −cos 30° = −(√3)/2
sin 75° must be positive, because 75° is in Quadrant I. Therefore,
sin 75° = √(1 − (−√3/2))/2
sin 75° = √(2 + √3)/4
sin 75° = √2 + √3 / 2
The expression √2 + √3 is called a nested radical. Some of these can be decomposed to a simpler √a + √b form, but some cannot. Desnesting Radicals (or Unnesting Radicals) explains how you can tell whether a particular one can be unnested, and gives an easy method to unnest it. This one can be unnested, leading to
sin 75° = (√6 + √2)/4
You can find tan(B/2) in the usual way, dividing sine by cosine from equation 61:
tan(B/2) = sin(B/2)/cos(B/2) = ± √(1 − cos B)/(1 + cos B)
In the sine and cosine formulas we couldn’t avoid the square roots, but in this tangent formula we can. Multiply top and bottom by √1+cos B:
tan(B/2) = √(1 − cos B) / (1 + cos B)
tan(B/2) = √(1 − cos B) (1 + cos B) / (1 + cos B)²
tan(B/2) = √1−cos² B / (1+cos B)
Then use equation 38, your old friend: sin² x + cos² x = 1;
tan(B/2) = √sin² B / (1+cos B) = sin B / (1+cos B)
If you multiply top and bottom by √1−cos B instead of √1+cos B, you get another form of the halfangle tangent formula:
tan(B/2) = sin(B/2)/cos(B/2) = ± √(1−cos B) / (1+cos B)
tan(B/2) = √(1−cos B)² / ((1+cos B)(1−cos B))
tan(B/2) = (1−cos B) / √1−cos² B
tan(B/2) = (1−cos B) / √sin² B = (1−cos B) / sin B
The halfangle tangent formulas can be summarized like this:
(62) tan(B/2) = (1 − cos B) / sin B = sin B / (1 + cos B)
You may wonder what happened to the plus or minus sign in tan(B/2). Luckily for us, it drops out. Since cos B is always between −1 and +1, (1 − cos B) and (1 + cos B) are never negative for any B. And the sine of any angle always has the same sign as the tangent of the corresponding halfangle.
Don’t take my word for that last statement, please. There are only four possibilities, and they’re easy enough to work out in a table. (Review interval notation if you need to.)
B/2  Q I, (0°;90°)  Q II, (90°;180°)  Q III, (180°;270°)  Q IV, (270°;360°) 

tan(B/2)  +  −  +  − 
B  (0°;180°), Q I or II  (180°;360°), Q III or IV  (360°;540°), Q I or II  (540°;720°), Q III or IV 
sin B  +  −  +  − 
Of course, you can ignore the whole matter of the sign of the sine and just assign the proper sign when you do the computation.
Another question you may have about equation 62: what happens if cos B = −1, so that (1 + cos B) = 0? Don’t we have division by zero then? Well, take a little closer look at those circumstances. The angles B for which cos B = −1 are ±180°, ±540°, and so on. But in this case the half angles B/2 are ±90°, ±270°, and so on: angles for which the tangent is not defined anyway. So the problem of division by zero never arises.
And in the other formula, sin B = 0 is not a problem. Excluding the cases where cos B = −1, this corresponds to B = 0°, ±360°, ±720°, etc. But the half angles B/2 are 0°, ±180°, ±360°, and so on. For all of them, tan(B/2) = 0, as you can verify from the second half of equation 62.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
(This is actually done, in a later section, by using a different method.)
I can’t resist pointing out another cool thing about Sawyer’s marvelous idea: you can also use it to prove the doubleangle formulas directly. From Euler’s formula for e^{ix} you can immediately obtain the formulas for cos 2A and sin 2A without going through the formulas for sums of angles. Here’s how.
Remember the laws of exponents: x^{ab} = (x^{a})^{b}. One important special case is that x^{2b} = (x^{b})². Use that with Euler’s formula:
cos 2A + i sin 2A = e^{i(2A)}
cos 2A + i sin 2A = (e^{iA})²
cos 2A + i sin 2A = (cos A + i sin A)²
cos 2A + i sin 2A = cos² A + 2i sin A cos A + i²sin² A
cos 2A + i sin 2A = cos² A + 2i sin A cos A − sin² A
cos 2A + i sin 2A = (cos² A − sin² A) + i (2 sin A cos A)
Since the real parts on left and right must be equal, you have the formula for cos 2A. Since the imaginary parts must be equal, you have the formula for sin 2A. That’s all there is to it.
The above technique is even more powerful for deriving formulas for functions of 3A, 4A, or any multiple of angle A. To derive the formulas for nA, expand the nth power of (cos A + i sin A), then collect real and imaginary terms.
This was inspired by a May 2009 reading of Paul Nahin’s An Imaginary Tale: The Story of √−1 (Princeton, 1998).
Just to show the method, I’ll derive the functions of 3A and 4A. You can try the bruteforce approach of cos(A + 2A), sin(2A + 2A), and so forth, and see how much effort my method saves.
De Moivre’s theorem tells us that
cos 3A + i sin 3A = (cos A + i sin A)³
Expand the righthand side via the binomial theorem, remembering that i² = −1 and i³ = −i:
cos 3A + i sin 3A = cos³ A + 3 i cos² A sin A − 3 cos A sin² A − i sin³ A
Collect real and imaginary terms:
cos 3A + i sin 3A = (cos³ A − 3 cos A sin² A) + i (3 cos² A sin A − sin³ A)
Set the real part on the left equal to the real part on the right to find the formula for cos 3A:
cos 3A = cos³ A − 3 cos A sin² A
It would be nice to have a formula that involves only cos A, not sin A. To get that, factor and then use sin² A = 1 − cos² A:
cos 3A = cos A (cos² A − 3 sin² A)
cos 3A = cos A (cos² A − 3(1 − cos² A))
cos 3A = cos A (cos² A − 3 + 3 cos² A)
cos 3A = cos A (4 cos² A − 3)
Going back to the formula with collected terms, set the imaginary part on the left equal to the real part on the right to find the formula for sin 3A:
sin 3A = 3 cos² A sin A − sin³ A
sin 3A = sin A (3 cos² A − sin² A)
sin 3A = sin A (3(1 − sin² A) − sin² A)
sin 3A = sin A (3 − 4 sin² A)
The tangent formula is easy to get: just divide. But it turns out to be easier if you don’t divide the final forms, but rather the “raw” collected terms from above:
sin 3A = 3 cos² A sin A − sin³ A
cos 3A = cos³ A − 3 cos A sin² A
tan 3A = sin 3A / cos 3A
tan 3A = (3 cos² A sin A − sin³ A) / (cos³ A − 3 cos A sin² A)
Factor top and bottom, then divide both by cos² A:
tan 3A = (sin A (3 cos² A − sin² A)) / (cos A (cos² A − 3 sin² A))
tan 3A = (sin A (3 − tan² A)) / (cos A (1 − 3 tan² A))
tan 3A = (sin A / cos A) (3 − tan² A) / (1 − 3 tan² A)
tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)
(63) sin 3A = sin A (3 − 4 sin² A)
cos 3A = cos A (4 cos² A − 3)
tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)
It’s no more work to find the functions of 4A. Since the technique is similar, I’ll just run through the steps without commentary. First, De Moivre’s theorem:
cos 4A + i sin 4A = (cos A + i sin A)^{4}
cos 4A + i sin 4A = cos^{4} A + 4 i cos³ A sin A − 6 cos² A sin² A − 4 i cos A sin³ A + sin^{4} A
cos 4A + i sin 4A = (cos^{4} A − 6 cos² A sin² A + sin^{4} A) + i (4 cos³ A sin A − 4 cos A sin³ A)
Set the real parts equal to get cos 4A:
cos 4A = cos^{4} A − 6 cos² A sin² A + sin^{4} A
cos 4A = cos^{4} A − 6 cos² A sin² A + (sin² A)²
cos 4A = cos^{4} A − 6 cos² A (1 − cos² A) + (1 − cos² A)²
cos 4A = cos^{4} A − 6 cos² A + 6 cos^{4} A + 1 − 2 cos² A + cos^{4} A
cos 4A = 8 cos^{4} A − 8 cos² A + 1
Set the imaginary parts equal to get sin 4A:
sin 4A = 4 cos³ A sin A − 4 cos A sin³ A
sin 4A = 4 sin A cos A (cos² A − sin² A)
sin 4A = 4 sin A cos A (1 − sin² A − sin² A)
sin 4A = 4 sin A cos A (1 − 2 sin² A)
Unfortunately, the first power of cos A can’t be changed to something involving only sin A. Or rather, it can, but at a cost. cos A would be replaced with ±√1 − sin² A, and we’d have to figure out the proper sign every time.
Now the tangent formula:
tan 4A = sin 4A / cos 4A
tan 4A = (4 cos³ A sin A − 4 cos A sin³ A) / (cos^{4} A − 6 cos² A sin² A + sin^{4} A)
Divide top and bottom by cos^{4} A:
tan 4A = (4 tan A − 4 tan³ A) / (1 − 6 tan² A + tan^{4} A)
There are several possibilities for simplifying this formula, but no one that’s clearly superior to the others, so I’ll just factor tan A from the top of the fraction:
tan 4A = 4 tan A (1 − tan² A) / (1 − 6 tan² A + tan^{4} A)
(64) sin 4A = 4 sin A cos A (1 − 2 sin² A)
cos 4A = 8 cos^{4} A − 8 cos² A + 1
tan 4A = 4 tan A (1 − tan² A) / (1 − 6 tan² A + tan^{4} A)
Summary: The inverse trig functions (also called arcfunctions) are similar to any other inverse functions: they go from the function value back to the angle (or number). Their ranges are restricted, by definition, because an inverse function must not give multiple answers. Once you understand the inverse functions, you can simplify composite functions like sin(Arctan x).
An inverse is the math equivalent of an undo. For example, if you have an angle A = 40°, you can find sin A ≈ 0.64. But you have to go the other direction when you’re solving a triangle. For instance, you might get sin B = 0.82 and have to find the angle B. You’re not asking what’s the sine of some angle, but rather, “What angle has a sine equal to 0.82?”
A shorter way to ask that question is, “what’s arcsin(0.82)?” This uses the name “arcsin” for the inverse of the sine function — not going from angle to its sine, but from the sine to the angle.
Is arcsin a function? Well, look at the graph of y = 0.82 against y = sin x. Where they cross is arcsin(0.82), and obviously there are many possible answers. So although sin is a function, arcsin is not. Your calculator will give you an answer of around 55°, but that’s just one out of infinitely many. You know from equation 22 that sin(180° − x) = sin x, and since 180° − 55° = 125°, sin 125° = sin 55. But all the trig functions are periodic, repeating every 360° forever, so infinitely many angles (or numbers) have a sine of 0.82:
arcsin(0.82) ≈ (55 + 360k)° and (125 + 360k)°, where k is any integer.
arcsin(0.82) ≈ (0.96 + 2πk) and (2.18 + 2πk), where k is any integer.
So arcsin(0.82) is all the numbers (or angles) whose sine is 0.82. But if we want an inverse sine function, we have to have a single answer. That single answer is called the principal value, and is written Arcsin(0.82):
Arcsin(0.82) ≈ 55°, or 55π/180 = 0.96 in radian measure.
Arcsin(x), with the capital letter, is the principal value of arcsin(x). Lowercase arcsin(x) is all possible numbers or angles whose sine is x. The same convention applies to the other five functions. (See Notation, below, for other ways of writing the inverse relation and the inverse function.)
Now all we need is a rule for picking the principal values of all the inverse trig functions. We want a continuous interval, with no gaps, and we want that interval to include the range from 0° to 90° (0 to π/2). It turns out that the six functions can’t all have the same range.
For Arcsin, the only possibility that meets those requirements is that Arcsin x must return a number in the interval [−π/2, +π/2], which is the same as [−90°, +90°].
The inverse tangent, Arctan, is almost the same. But since there’s no value for tan(±90°) or tan(±π/2), the range of Arctan is the open interval (−π/2, +π/2) or (−90°, +90°).
What about Arccos? The cosine is positive in both Quadrant I and Quadrant IV, so the arccosine of a negative number must fall in Quadrant II or Quadrant III. Thomas (Calculus and Analytic Geometry, 4th edition) resolves this in a neat way. Remember from equation 2 that
cos A = sin(π/2 − A)
It makes a nice symmetry to write
Arccos x = π/2 − Arcsin x
And that is how Thomas defines the inverse cosine function. Since the range of Arcsin is the closed interval [−π/2, +π/2], the range of Arccos is π/2 minus that, [0, π] or [0°, 180°].
Once the range for Arctan is defined, there’s really only one sensible way to define Arccot:
cot x = tan(π/2 − x) ⇒ Arccot x = π/2 − Arctan x
which gives the single open interval (0, π) or (0°, 180°) as the range.
Thomas defines the Arcsec and Arccsc functions using the reciprocal relationships from equation 5:
sec x = 1/(cos x) ⇒ Arcsec x = Arccos(1/x)
csc x = 1/(sin x) ⇒ Arccsc x = Arcsin(1/x)
This means that Arcsec and Arccsc have the same ranges as Arccos and Arcsin, respectively.
Here are the domains (inputs) and ranges (outputs) of all six inverse trig functions:
function  derived from  domain  range 

Arcsin  inverse of sine function  [−1, +1]  Q IV, I: [−π/2, +π/2] 
Arccos  Arccos x = π/2 − Arcsin x  [−1, +1]  Q I, II: [0, π] 
Arctan  inverse of tangent function  all reals  Q IV, I: (−π/2, +π/2) 
Arccot  Arccot x = π/2 − Arctan x  all reals  Q I, II: (0, π) 
Arcsec  Arcsec x = Arccos(1/x)  (−∞, −1] and [1, ∞)  Q I, II: [0, π] 
Arccsc  Arccsc x = Arcsin(1/x)  (−∞, −1] and [1, ∞)  Q IV, I: [−π/2, +π/2] 
Remember that the inverse relations arcsin etc. are manyvalued, not limited to the above ranges of the functions. If you see the capital A in the function name, you know you’re talking about the function; otherwise you have to depend on context.
In this book, the manyvalued inverse relation is arcsin with lowercase a, and the inverse function is Arcsin with capital A. That’s a common choice in the U.S., but it’s not the only choice.
Because there are so many conventions, authors generally explain which notation they’re using, so watch for that.
Advice to
the reader:
The methods in this section aren’t
really very useful in trigonometry itself, but are used in integral calculus
and some physics or engineering courses.
You may wish to skip them, especially on a first reading. On the other
hand, they are pretty cool.
Sometimes you have to evaluate expressions like
cos(Arctan x)
That looks scary, but actually it’s a piece of cake. You can simplify any trig function of any inverse trig function in two easy steps, using this method:
Think of the inner arcfunction as an angle. Draw a right triangle and label that angle and the two relevant sides.
Use the Pythagorean Theorem to find the third side of the triangle, then write down the value of the outer function according to its definition.
It may be helpful to read the expression out in words: “the cosine of Arctan x.” Doesn’t help much? Well, remember what Arctan x is. It’s the (principal) angle whose tangent is x. So what you have to find reads as “the cosine of the angle whose tangent is x.” And that suggests your plan of attack: first identify that angle, then find its cosine.
Let’s give a name to that “angle whose”. Call it A:
A = Arctan x
from which you know that
tan A = x
Now all you have to do is find cos A, and that’s easy if you draw a little picture.
Start by drawing a right triangle, and mark one acute angle as A.
Using the definition of A, write down the lengths of two sides of the triangle. Since tan A = x, and the definition of tangent is opposite side over adjacent side, the simplest choice is to label the opposite side x and the adjacent side 1. Then, by definition, tan A = x/1 = x, which we needed, because A = Arctan x.
The next step is to find the third side. Here you know the two legs, so you use the theorem of Pythagoras to find the hypotenuse, √1+x². (For some problems, you’ll know one leg and the hypotenuse, and you’ll use the theorem to find the other leg.)
Once you have all three sides’ lengths, you can write down the value of any function of A. In this case you need cos A, which is adjacent side over hypotenuse:
cos A = 1/√(1+x²)
But cos A = cos(Arctan x). Therefore
cos(Arctan x) = 1/√(1+x²)
and there’s your answer.
Read this as “the cosine of the angle A whose sine is x”. Draw your triangle, and label angle A. (Please take a minute and make the drawing.) You know from equation 1 that
sin A = x = opposite/hypotenuse
and therefore you label the opposite side x and the hypotenuse 1.
Next, solve for the third side, which is √(1−x²), and write that down. Now you need cos A, which is the adjacent side over the hypotenuse, which is √(1−x²)/1. Answer:
cos(Arcsin x) = √(1−x²)
There you go: quick and painless.
This looks similar to Example 1, but as you’ll see there’s an additional wrinkle. (Thanks to Brian Scott, who raised the issue, albeit inadvertently, in his article “Re: Expression” on 10 Dec 2000 in alt.algebra.help.)
Begin in the regular way by drawing your triangle. Since A = Arctan(1/x), or tan A = 1/x, you make 1 the length of the opposite side and x the length of the adjacent side. The hypotenuse is then √(1+x²).
Now you can write down cos A, which is adjacent over hypotenuse:
cos A = x/√(1+x²)
cos(Arctan 1/x) = x/√(1+x²)
But this example has a problem that does not occur in the earlier examples.
Suppose x is negative, say −√3. Then Arctan(−1/√3) = −π/6, and cos(−π/6) = +(√3)/2. But the answer above, x/√(1+x²), yields −√3/√(1+3) = −(√3)/2, which has the wrong sign.
What went wrong? The trouble is that Arctan yields values in (−π/2, +π/2), which is Quadrants IV and I. But the cosine is always positive on that interval. Therefore cos(Arctan x) always yields a positive result. Remember also from equation 22 that cos(−A) = cos A. To ensure this, use the absolute value sign, and the true final answer is
cos(Arctan(1/x)) =  x /√(1+x²)
Why doesn’t every example have this problem? The earlier examples involved only the square of a variable, which is naturally nonnegative. Only here, where we have an odd power, does it matter. Yes, that applies to the first power, even though the exponent ^{1} isn’t written.
Summary: When your answer contains an odd power (1, 3, 5, etc.) of a variable, you must add a third step to the process: carefully examine the signs and adjust your answer so that it has the correct sign for both positive and negative values of the variable.
Advice to
the reader:
The methods in this section are
for the really hardcore trig fan. They aren’t
really very useful in trigonometry itself, but are used in integral calculus
and some physics or engineering courses.
You may wish to skip them, especially on a first reading.
After the previous section, you may be wondering about the insideout versions, taking the arcfunction of a function. Some of these expressions can be solved algebraically, on a restricted domain at least, but some cannot. (I am grateful to David Cantrell for help with analysis of these problems in general and Example 6 in particular.)
We can say at once that there will be no pure algebraic equivalent to an arcfunction of a trig function. This means there will be no nice neat procedure as there was for functions of arcfunctions
Why? The six trig functions are all periodic, and therefore any function of any of them must also be periodic. But no algebraic functions are periodic, except trivial ones like f(x) = 2, and therefore no function of a trig function can be represented by purely algebraic operations. As we will see, some can be represented if we add nonalgebraic functions like mod and floor.
This is the angle whose cosine is sin u. To come up with a simpler form, set x equal to the desired expression, and solve the equation by taking cosine of both sides:
x = Arccos(sin u)
cos x = sin u
This could be solved if we could somehow transform it to sin(something) = sin u or cos x = cos(something else). In fact, we can use equation 2 to do that. It tells us that
sin u = cos(π/2 − u)
and combining that with the above we have
cos x = cos(π/2 − u)
Now if x is in Quadrant I, which is the interval [0, π/2], then u will be in Quadrant I also, and we can write
x = π/2−u
and therefore
Arccos(sin u) = π/2−u for u in Quadrant I
But this solution does not work for all quadrants. For instance, try a number from Quadrant II:
Arccos(sin(5π/6)) = Arccos(½) = π/6
but
π/2 − 5π/6 = −π/3
Obviously π/2−u isn’t a general solution for Arccos(sin u). Try graphing Arccos(sin x) and π/2−x and you’ll see the problem: one is a sawtooth and the other is a straight line.
Sparing you the gory details, π/2−u is right only in Quadrants IV and I. We have to “decorate” it rather a lot to make it match Arccos(sin u) in the other quadrants, and also to account for the repetition of values every 2π. The first modification is not too hard: On the interval [−π/2, +3π/2], the absolutevalue expression  π/2−u  matches the sawtooth graph of Arccos(sin u).
The repetition every 2π is harder to reflect, but this manages it:
Arccos(sin u) =  π/2 − u + 2π*floor[(u+π/2)/2π] 
where “floor” means the greatest integer less than or equal to. Messy, eh? (Note also that “floor” is not an algebraic function.)
It could be made a bit shorter with mod (which is also not algebraic):
Arccos(sin u) =  π − mod(u+π/2, 2π) 
where mod(a, b) is the nonnegative remainder when ais divided by b.
This one, the angle whose secant is cos u, has a very odd solution. Try the solution method from Example 4 and you get
x = Arcsec(cos u)
sec x = cos u
But sec x = 1/cos x, and therefore
1/(cos x) = cos u
Now think about that equation. The cosine’s values are all between −1 and +1. So the only way one cosine can be the reciprocal of another is if they’re both equal to 1 or both equal to −1; no other solutions exist.
First case: If cos u = 1, then u is an even multiple of π, or in other words a multiple of 2π. But Arcsec 1 = 0, and therefore
Arcsec(cos u) = 0 when u = 2kπ
Second case: If cos u = −1, then u is an odd multiple of π. But Arcsec(−1) = π, and therefore
Arcsec(cos u) = π when u = (2k+1)π
If u is not a multiple of π, cos u will be less than 1 and greater than −1. The Arcsec function is not defined for such values, and therefore
Arcsec(cos u) does not exist when u is not a multiple of π
The graph of Arcsec(cos u) is rather curious: single points at the ends of an infinite sawtooth: ..., (−3π, π), (−2π, 0), (−π, π), (0, 0), (π, π), (2π, 0), (3π, π), ...
Proceeding in the regular way, we have
x = Arctan(sin u)
tan x = sin u
The most likely approach is the one from Example 4: try to transform the above into tan(x) = tan(something) or sin(something else) = sin u.
If there is any trig identity or combination that can be used to do that, it is unknown to me. I suspect strongly that Arctan(sin u) can’t be converted to an algebraic expression, even with the use of mod or floor, but I can’t prove it.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
Trig without Tears Part 10:
Copyright © 1997–2019 Stan Brown, BrownMath.com
Summary: Trig can answer some of the questions your algebra teacher wouldn’t answer, like what’s √i? and what’s the log of a negative number?
As you may remember, complex numbers like 3+4i and 2−7i are often plotted on a complex plane. This makes it easier to visualize adding and subtracting. The illustration at right shows that
(3+4i) + (2−7i) = 5−3i
It turns out that for multiplying, dividing, and finding powers and roots, complex numbers are easier to work with in polar form. This means that instead of thinking about the real and imaginary components (the “−3” and “5” in −3+5i), you think of the length and direction of the line.
The length is easy, just the good old Pythagorean theorem in fact:
r = √(3² + 5²) = √34 ≈ 5.83
(The length r is called the absolute value or modulus of the complex number.)
But what about the direction? You can see from the picture that θ is about 120° or about 2 radians (measured counterclockwise from the positive real axis, which points east), but what is it exactly? Well, you know that tan θ = −5/3. If you take Arctan(−5/3) on your calculator, you get −1.03, and adding π to get into the proper quadrant gives θ = 2.11 radians. (In degrees, Arctan(−5/3) = −59.04, and adding 180 gives θ = 120.96°.) You can write it this way:
−3+5i ≈ 5.83 e^{2.11i} or 5.83[cos 2.11 + i sin 2.11] or 5.83 cis 2.11 or 5.83∠120.96°
There’s a nice trick that finds the angle in the correct quadrant automatically:
θ = 2 Arctan(y/(x + r))
This builds in the adjustment to the correct quadrant, so you never have to worry about whether to add or subtract 180° or π. If you test 2 Arctan(5/(−3 + √34)) on your calculator, you get 2.11 radians or 120.96° as before.
This formula is particularly handy when you’re writing a program or spreadsheet formula to find θ. It flows directly from the formulas for tangent of half an angle, where the functions are expressed in terms of x, y, and r. A July 2003 article by Rob Johnson, archived here, inspired me to add this section.
One of the applications of Euler’s formula (equation 47) is finding any power or root of any complex number. (Sawyer doesn’t do this, or at least not in the same book.) It’s not hard to develop the formula, using just Euler’s formula and the laws of exponents.
Start by writing down Euler’s formula, multiplied left and right by a scale factor r:
r (cos x + i sin x) = r e^{ix}
Next, raise both sides to the nth power:
[r (cos x + i sin x)]^{n} = [r e^{ix}]^{n}
Let’s leave the lefthand side alone for a while, and simplify the righthand side. First, by the laws of exponents
[r e^{ix}]^{n} = r^{n} e^{inx}
Apply Euler’s formula to the righthand side and you have
[r e^{ix}]^{n} = r^{n} (cos nx + i sin nx)
Connect that righthand side to the original lefthand side and you have DeMoivre’s theorem:
(82) [r(cos x + i sin x)]^{n} = r^{n} (cos nx + i sin nx)
This tells you that if you put a number into polar form, you can find any power or root of it. (Remember that the nth root of a number is the same as the 1/n power.)
For instance, you know the square roots of −1 are i and −i, but what’s the square root of i? You can use DeMoivre’s theorem to find it.
First, put i into e^{ix} form (polar form), using the above technique. i = 0+1i. cos x = 0 and sin x = 1 when x = 90° or π/2. Therefore
i = cos(π/2) + i sin(π/2)
√i = i^{½} = [cos(π/2) + i sin(π/2)]^{½}
And by equation 82
√i = cos(½×π/2) + i sin(½×π/2)
√i = cos(π/4) + i sin(π/4)
√i = 1/√2 + i×1/√2
√i = (1+i)/√2
The other square root is minus that, as usual.
You can find any root of any complex number in a similar way, but usually with one preliminary step.
For instance, suppose you want the cube roots of 3+4i. The first step is to put that number into polar form. The absolute value is √(3²+4²) = 5, and you use the Arctan technique given above to find the angle θ = 2 Arctan(4/(3+5)) = 2 Arctan(½) ≈ 53.13° or 0.9273 radians. The polar form is
3+4i ≈ 5(cos 53.13° + i sin 53.13°)
To take a cube root of that, use equation 82 with n = 1/3:
cube root of (3+4i) = 5^{1/3} [cos(53.13°/3) + i sin(53.13°/3)]
The cube root of 5 is about 1.71. Dividing the angle θ by 3, 53.13°/3 ≈ 17.71°; the cosine and sine of that are about 0.95 and 0.30. Therefore
cube root of (3+4i) ≈ 1.71 × (cos 17.71° + i sin 17.71°)
cube root of (3+4i) ≈ 1.63+0.52i
You may have noticed that I talked about “the cube roots [plural] of 3+4i” and what we found was “a cube root”. Even with the square root of i, I waved my hand and said that the “other” square root was minus the first one, “as usual”.
You already know that every positive real has two square roots. In fact, every complex number has n nth roots.
How can you find them? It depends on the periodic nature of the cosine and sine functions. To start, look back at Euler’s formula,
e^{ix} = cos x + i sin x
What happens if you add 2π or 360° to x? You have
e^{i(x+2π)} = cos(x + 2π) + i sin(x + 2π)
But taking sine or cosine of 2π plus an angle is exactly the same as taking sine or cosine of the original angle. So the righthand side is equal to cos x + i sin x, which is equal to e^{ix}. Therefore
e^{i(x+2π)} = e^{ix}
In fact, you can keep adding 2π or 360° to x as long as you like, and never change the value of the result. Symbolically,
e^{i(x+2πk)} = e^{ix} for all integer k
When you take an nth root, you simply use that identity.
Getting back to the cube roots of 3+4i, recall that that number is the same as 5e^{iθ}, where θ is about 53.13° or 0.9273 radians. The three cube roots of e^{iθ} are
e^{i(θ+2πk)/3} for k = 0,1,2
Expanding that, you have
e^{iθ/3}, e^{i(θ+2π)/3}, and e^{i(θ+4π)/3}
or
e^{iθ/3}, e^{i(θ/3+2π/3)}, and e^{i(θ/3+4π/3)}
Compute those as cos x + i sin x in the usual way, and then multiply by the (principal) cube root of 5. I get these three roots:
cube roots of 3+4i ≈ 1.63+0.52i, −1.26+1.15i, −0.36−1.67i
More generally, we can extend DeMoivre’s theorem (equation 82) to show the n nth roots of any complex number:
(83) [r (cos x + i sin x)]^{(1/n)} = r^{(1/n)} [cos(x/n+2πk/n) + i sin(x/n+2πk/n)] for k = 0,1,2,...,n−1
Another application flows from a famous special case of Euler’s formula. Substitute x = π or 180° in equation 47. Since sin 180° = 0, the imaginary term drops out. And cos 180° = −1, so you have the famous formula
(84) −1 = e^{iπ}
It’s interesting to take the natural log of both sides:
ln(−1) = ln e^{iπ}
ln(−1) = iπ
It’s easy enough to find the logarithm of any other negative number. Since
ln ab = ln a + ln b
then for all real a you have
ln(−a) = ln(a × −1) = ln a + ln(−1) = ln a + iπ
(85) ln(−a) = ln a + iπ
I don’t honestly know whether all of this has any practical application. But if you’ve ever wondered about the logarithm of a negative number, now you know.
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
Solutions:
(a) (π/6) × (180°/π) = 30°
(b) 2π × (180°/π) = 360°
(c) 1 × (180°/π) = (180/π)° ≈ 57.3°
Answer: (0°, 90°) is 0 to 90 degrees excluding 0° and 90°; [0°, 90°] is 0 to 90 degrees including 0° and 90°. Acute angles are between 0° and 90° exclusive, so the answer is (a) (0°, 90°).
Solution: The inside angles of a triangle must always add to 180°. 80° + 40° = 120°, so to make the full 180° the third angle must be 60°.
Solution: Cue the Pythagorean Theorem!
c² = a² + b²
c² = 5² + 12²
c² = 25 + 144 = 169
c = √169 = 13
Where possible, give an exact answer rather than a decimal approximation.
Solutions:
(a) 60° + (π/180°) = π/3.
(b) 126° × (π/180°) ≈ 2.20
(c) 45° × (π/180°) = π/4
Notice that you don’t have to say “radians” when giving an angle in radian measure, though it wouldn’t be wrong. In this book, angles in degrees have the degree mark (°), so I’ll only say “radians” when it’s necessary to avoid confusion.
Answer: It was the Scarecrow, in the movie The Wizard of Oz (1939). And no, it sounds mathy but it’s bosh. It can’t possibly be true for any triangle, isosceles or not. (Can you see why?)
Answers: Quadrant I: 12 and 3; Quadrant II: 9 and 12; Quadrant III: 6 and 9; Quadrant IV: 3 and 6.
Solution: First off, you need the length of the horizontal side. You remember the theorem of Pythagoras: 1² + b² = 2², from which you get b = √3. After that, it’s just a matter of remembering the definitions. If you need a refresher, you’ll find sine and cosine at equation 1, tangent at equation 4, and the others at equation 5.
sin 30° = 1/2
cos 30° = (√3)/2
tan 30° = 1/√3 or (√3)/3
cot 30° = 1/(1/√3) = √3
sec 30° = 1/(√3/2) = 2/√3 or (2 √3)/3
csc 30° = 1/(1/2) = 2
sin 60° = (√3)/2
cos 60° = 1/2
tan 60° = √3/1 = √3
Since 60° = 90° − 30°, notice that sin 60° = cos 30°, cos 60° = sin 30°, and tan 60° = cot 30°.
Solution:
sin A = 3/5 or 0.6
sin B = 4/5 or 0.8
tan A = 3/4 or 0.75
tan B = 4/3 ≈ 1.33
Incidentally, A ≈ 36.87°, and B ≈ 53.13°.
Solution: You have the base (5), so you just need the height. But sin A = h/3, so h = 3 × sin A. The area therefore is (5 × 3 × sin A)/2 ≈ 5.99999.
(b) Draw a triangle with all sides equal to 1. If all sides are equal, all angles must be equal. Knowing that they add to 180°, fill in the value of each angle. Now drop a perpendicular from the top of the triangle to the middle of the opposite side. You now have two 306090° triangles with hypotenuse of 1. Fill in the exact lengths of the short sides of those two triangles.
For the rest of these problems, refer to these sketches if you need to. Give exact answers, not decimal approximations.
Solution: Check your sketches against the ones in the chapter.
Answers: tan 45° = 1; cos 45° = 1/√2 or √2/2; sin 90° = 1; cos 30° = √3/2; sin 30° = 1/2; cos 90° = 0.
Answers: sin(π/4) = √2/2 or 1/√2; cos(π/6) = √3/2; tan(π/3) = √3
Answers:
sin A = 0 ⇒ A = 0° or 0 [radians]
cos B = √3/2 ⇒ B = 30° or π/6
sin C = 1/2 ⇒ C = 30° or π/6
sin D = 1 ⇒ D = 90° or π/2
tan E = 1 ⇒ E = 45° or π/4
cos F = 1/2 ⇒ F = 60° or π/3
tan G = 0 ⇒ G = 0° or 0 [radians]
tan H = √3 ⇒ H = 60° or π/3
cos I = 1 ⇒ I = 0° or 0 [radians]
cos J = 0 ⇒ J = 90° or π/2
Solutions: sec 60° = 1/(cos 60°) = 1/(1/2) ⇒ sec 60° = 2
cot 30° = 1/(tan 30°) = 1/(√3/3) = 3/√3 = √3 ⇒ cot 30° = √3
You could also do the second one using equation 6:
cot 30° = tan(90°−30°) = tan 60° = √3
Solution:
sin 120° = sin(180° − 120°) = √3/2.
cos 120° = −cos(180° − 120°) = −1/2.
What about the tangent? You don’t have a supplement rule for it, but you have another way to find the answer:
tan 120° = sin 120° / cos 120° = (√3/2) / (−1/2) = −√3
Prove: tan(180° − A) = −tan A.
Proof:
tan(180° − A) = sin(180° − A) / cos(180° − A)
tan(180° − A) = sin A / (−cos A)
tan(180° − A) = −sin A / cos A
tan(180° − A) = −tan A QED
Solution: tan 150° = −tan(180° − 150°) = −tan 30° = −√3/3
Always start with a sketch. From the sketch, you can see right away that this is the SAS case, or sideangleside. To get the third side, you need the Law of Cosines, equation 31.
c² = a² + b² − 2ab × cos C
c² = 37² + 88² − 2 × 37 × 88 × cos 90°
Notice that when the included angle C is 90°, cos C = 0 and you’ve just got the Pythagorean Theorem.
c² = 37² + 88²
c = √37² + 88²
c ≈ 95.5
To find the two angles, you could use the Law of Sines, but why not take advantage of the right angle and use the tangent?
tan A = 37/88 ⇒ A ≈ 22.8°
Of course tan B = 88/37, but you also know that
A + B = 90° ⇒ B = 90° − A ⇒ B ≈ 67.2°
How long must the monorail and the foot bridge be?
Bonus question: If the river has the same width all along the stretch from A to B, how wide is it?
In the sketch, b is the monorail, a is the foot bridge, and w is the width of the river. Since you know two angles and the included side, this is the ASA case. Although you don’t really care about angle C, you have to find it so that you can use the Law of Sines to get sides a and b:
C = 180° − 67° − 38° = 75°
Obviously, when I eyeballed the angles A and B I didn’t estimate them very well! But that’s okay—the sketch is close enough to be useful.
How long is the foot bridge? From the Law of Sines,
a/sin A = c/sin C ⇒ a = c sin A/sin C
a = 1800 × sin 67°/sin 75°
Foot bridge: a ≈ 1715 m
How long is the monorail? From the Law of Sines,
b/sin B = c/sin C ⇒ b = c sin B/sin C
b = 1800 × sin 38°/sin 75°
Monorail: b ≈ 1147 m
How wide is the river?
w = b sin A
w ≈ 1147 sin 67°
River width: w ≈ 1056 m
This is the SSA (or ASS) case; refer to the table of possibilities for SSA. This problem belongs in row 1, column 2: the opposite side is longer than the adjacent side, and the known angle is > 90°. So there is one and only one solution.
You can use the Law of Sines to find angle A:
(sin A)/a = (sin B)/b
sin A = (a/b) sin B
sin A = (16/25) sin 117° ≈ 0.57024
A ≈ 34.8°
Find the third angle by subtracting:
C = 180° − A − B
C ≈ 180° − 34.8° − 117°
C ≈ 28.2°
Finally, use the Law of Sines to find side c:
c/sin C = b/sin B
c = b sin C/sin B
c ≈ 25 sin 28.2°/sin 117°
c ≈ 13.3 cm
This is the SSS case. First, use the Law of Cosines to find angle A:
cos A = (b² + c² − a²) / 2bc
cos A = (9² + 12² − 6²)/(2 × 9 × 12) = 0.875
A ≈ 29.0°
Next, use the Law of Sines to find angle B:
(sin B)/b = (sin A)/a ⇒ sin B = (b/a) sin A
sin B ≈ (9/6) sin 29.0° ≈ 0.72618
B ≈ 46.6°
Finally, subtract the two angles from 180° to find the third angle:
C = 180° − A − B
C ≈ 180° − 29.0° − 46.6°
C ≈ 104.5°
That’s not a typo, by the way. A and B both happened to round up, but I used the unrounded values to find C. You should never use rounded numbers in further calculations.
The third angle is 180° − 30° − 40°, so C = 110°. Make your sketch using those three angles. (I did this one without measuring the angles, so it’s not perfect. But sketches don’t need to be perfect, just reasonably close.)
Find side c by using equation 32:
c = √2 area × sin C/(sin A sin B)
c = √2 × 25 × sin 110°/(sin 30° sin 40°)
c ≈ 12.1 ft
Then the Law of Sines gets you the other two sides:
a = c sin A/sin C
a ≈ 12.1 sin 30°/sin 110°
a ≈ 6.4 ft
And
b = c sin B/sin C
b ≈ 12.1 sin 40°/sin 110°
b ≈ 8.3 ft
You’re now driving along a straight side road. At the end of 9.8 miles on the side road, you turn 135° to the right, on a third road. (If you’re visualizing this from above, the 135° change of direction corresponds to an angle of 180° − 135° = 45° in the triangle.)
Assuming that road continues in the same direction, how far must you drive to reach your starting point?
This is a difficult sketch to draw, because you don’t know the angle of the first turn. But the description gives you two sides and a nonincluded angle; this is the problematic SSA case. You don’t know exactly where side c will meet side a. To be more precise, you don’t even know if they will meet.
Is it possible for them to meet? Referring back to the table of possibilities within SSA, we see that we’re in the third row, first column: the adjacent side (9.8) is longer than the opposite side (6.0), and the known angle (45°) is < 90°. Compute
h = b sin A
h = 9.8 sin 45° ≈ 6.9 miles
What’s the significance of this? The shortest distance from point C to side c is the segment that meets side c in a right angle. In other words, to get from point C to side c, the shortest possible distance is 6.9 miles. But side a is only 6.0 miles long, so it can never meet side c.
This problem has no solution.
You know two sides and the angle between them. You can use the Law of Cosines to get the third side:
a² = b² + c² − 2bc × cos A
a² = 60² + 40² − 2 × 60 × 40 × cos 22°
c²≈ √749.52 ⇒ c ≈ 27.4 m
Next, for angle B you can use the Law of Sines:
(sin B)/b = (sin A)/a ⇒ sin B = (b/a) sin A
sin B ≈ (60/27.4) sin 22° ≈ 0.82099
Your calculator gives about 55.2° as the angle whose sine is 0.82099, but that looks wrong from the sketch. Clearly B needs to be an obtuse angle, so you remember that sin(180° − x) = sin x, and you subtract 180° − 55.2° to get
B ≈ 124.8°
See how important a sketch is? Of course your sketch probably isn’t perfectly accurate, so you treat it as a device to point out that something might be wrong, but then you look for a way to confirm it. In this case, you have two ways to confirm it:
cos B = (a² + c² − b²)/(2ac)
cos B ≈ (27.4² + 40² − 60²)/(2 × 27.4 × 40)
cos B ≈ −0.57095 ⇒ B ≈ 124.8°
Now turn to angle C:
(sin C)/c = (sin A)/a ⇒ sin C = (c/a) sin A
sin C ≈ (40/27.4) sin 22° ≈ 0.54732
C ≈ 33.2°
Solution: Your task always is to cast out multiples of ±360° or ±2π, so that you’re left with a positive angle between 0° and 360° (0 and 2π).
−868° = −1080° + 212°. And 212° is between 180° and 270°, so −868° occurs in Q III.
42 radians is about 13.37π, or 12π + 1.37π. 1.37π is obviously between π and 3π/2 (1.5π), so 42 (radians) is in Q III.
In Q III, x and y are both negative. Therefore In Q III, sine and cosine are negative and tangent is positive.
Solution: (a) 700° = 360° + 340°, so sin 700° = sin 340°. 340° is in Q IV, where y is negative; therefore the sine is negative. The reference angle is 20°, so sin 700° = −sin 20°.
(b) 780° = 720° + 60°, so tan 780° = tan 60°.
(c) −390.5 ≈ −124.3π, or −126π + 1.7π; therefore cos(−390.5) ≈ cos 1.7π. The angle 1.7π radians is between 1.5π and 2π, so it’s in Q IV, where x is positive and therefore the cosine is positive. The reference angle is about 2π − 1.7π ≈ 0.3π or 0.94, so cos(−390.5) ≈ cos 0.94.
Solution: (a) 720° is a multiple of 360°, so cos(720° − A) = cos(−A). But cos(−A) = cos A, and therefore cos(720° − A) = cos A.
(b) sin(43π + A) = sin(42π + π + A) = sin(π + A). sin(π + A) = −sin A, so sin(43π + A) = −sin A.
Solution:
sin² A + cos² A = 1
cos² A = 1 − sin² A = 1 − (3/4)² = 1 − 9/16
cos² A = 7/16
cos A = ±√7/16 ⇒ cos A = √7/4 or −√7/4
Did you remember the ± sign? Just as x² = 9 has two solutions, 3 and −3, so any equation in cos² A has two solutions.
This makes sense in terms of the functions. There are two angles in the interval [0, 2π) or [0°, 360°) where sin A = 3/4, one acute and one obtuse. The acute one has a positive cosine, and the obtuse one has a negative cosine.
Solution: Since this chapter is about the “squared” identities, you can be pretty sure that one exists that connects tan x and sec x. But suppose you met this problem in a different context?
Well, you know two identities involving the tangent function. The definition, tan B = (sin B)/cos B, doesn’t do you a lot of good because it’s got sine and cosine mixed together. You have a dim memory of a squared identity (if you’re like me, it’s dim), but you also know you can recreate it easily, from the one squared identity that you can’t possibly forget:
sin² B + cos² B = 1
You have sin² B, which will become tan² B if you divide both sides by cos² B.
(sin² B)/cos² B + (cos² B)/cos² B = 1/cos² B
tan² B + 1 = sec² B
Success! You have an identity that connects the tangent and secant functions. Now you can proceed to solve the problem.
(−2√2)² + 1 = sec² B
sec² B = (4 × 2) + 1 = 9
sec B = ±3
You need the ± sign because both (x)² and (−x)² are x². But does it make sense in terms of trig? Yes, because the tangent is negative in Q II and Q IV, while the secant—which has the same sign as the cosine, being 1 over the cosine—is positive in Q IV but negative in Q II.
Solution: Wait, what? You don’t have an identity connecting tangent and cosine. But you do have one connecting tangent and secant, and you know that the secant is 1 over the cosine, so you can do this one too.
Start with the squared identity from the previous problem:
tan² C + 1 = sec² C
(√15)² + 1 = 15 + 1 = 16 ⇒ sec² C = 16
sec C = ±4 ⇒ cos C = 1/4
Solution: Okay, in the previous problem you got from tangent to cosine. But you already know how to get from cosine to sine, so you just have one more link in the chain.
tan² D + 1 = sec² D
sec² D = (√15)² + 1 = 16
sec D = 4 ⇒ cos D = ±1/4
sin² D + cos² D = 1
sin² D = 1 − cos² D = 1 − (±1/4)² =1 − 1/16 =15/16
sin D = ±√15/4
Proof:
tan x = tan x
tan x cos x = tan x cos x
sin x = tan x / sec x
sin² x = tan² x / sec² x
But tan² x + 1 = sec² x, so substituting you have
sin² x = tan² x / (tan² x + 1) QED
Many other proofs are possible. For example, you could replace tan² x with sin² x / cos² x in the identity you were asked to prove, and then simplify the fraction. As long as every step in your proof is valid both backwards and forwards, your proof is fine.
If your proof starts with an obvious identity and then works forward to the identity you we asked to prove, as the above proof did, then it only matters that all the steps are valid in that forward direction.
Solution: Start with −15° = 30° − 45°
sin(−15°) = sin(30°−45°)
sin(−15°) = sin 30° cos 45° − cos 30° sin 45°
sin(−15°) = (1/2)(√2/2) − (√3/2)(√2/2) = √2/4 − √6/4
sin(−15°) = (√2 − √6)/4
Alternative solution: You could also do this using 45°−60°:
sin(−15°) = sin(45°−60°)
sin(−15°) = sin 45° cos 60° − cos 45° sin 60°
sin(−15°) = (√2/2)(1/2) − (√2/2)(√3/2) = √2/4 − √6/4 = (√2 − √6)/4
Solution: 105° = 60° + 45°
tan 105° = tan(60° + 45°)
tan 105° = (tan 60° + tan 45°)/(1 − tan 60° tan 45°)
tan 105° = (√3 + 1)/(1 − (√3)(1)) = (√3 + 1)/(1 − √3)
Multiply top and bottom by 1 + √3 to rationalize the denominator:
tan 105° = (√3 + 1)²/((1 − √3) (1 + √3))
tan 105° = (3 + 2√3 + 1)/(1 − 3) = (4 + 2√3)/(−2) = −2 − √3
Proof:
cos 2A = cos(A + A)
cos 2A = cos A cos A − sin A sin A
cos 2A = cos² A − sin² A
But sin² A = 1 − cos² A
cos 2A = cos² A − (1 − cos² A) = cos² A − 1 + cos² A
cos 2A = 2 cos² A − 1 QED
(a) cos(−A) = cos A (I’ve done the first step for you.)
cos(−A) = cos(0 − A)
cos(−A) = cos 0 cos A + sin 0 sin A
cos(−A) = 1 cos A + 0 sin A
cos(−A) = cos A QED
(b) tan(π + A) = tan A
tan(π + A) = (tan π + tan A) / (1 − tan π tan A)
tan(π + A) = (0 + tan A) / (1 − 0 tan A)
tan(π + A) = (tan A) / 1
tan(π + A) = tan A QED
(c) sin(π − A) = sin A
sin(π − A) = sin π cos A − cos π sin A
sin(π − A) = 0 cos A − (−1) sin A
sin(π − A) = sin A QED
Solution: : 90° is half of 180°. The sine and cosine are positive or zero at 90°, so the ± signs in the formulas can be treated as positive.
sin(180°/2) = √(1 − cos 180°)/2 = √(1 − (−1))/2 = √2/2
sin 90° = 1
cos(180°/2) = √(1 + cos 180°)/2 = √(1 + (−1))/2 = √0/2
cos 90° = 0
Why didn’t I ask you to do the same for tan 90°?
Solution: 120° is double 60°. sin 60° = (√3)/2, cos 60° = 1/2, and tan 60° = √3.
sin(2×60°) = 2 sin 60° cos 60° = 2 (1/2) (√3)/2
sin 120° = (√3)/2
cos(2×60°) = 2 cos² 60° − 1 = 2 × (1/2)² − 1 = (1/2) − 1
cos 120° = −1/2
tan(2×60°) = 2 tan 60° / (1 − tan² 60°) = 2 √3 / (1 − ((√3)²) = 2 √3 / (1 − 3) = 2 √3 / (−2)
tan 120° = −√3
Alternative solution: You could also divide:
tan 120° = sin 120° / cos 120° = ((√3)/2) / (−1/2) = −√3
(This is actually done, in a later section, by using a different method.)
Solution: First the sine:
sin(2A + A) = sin 2A cos A + cos 2A sin A
sin 3A = (2 sin A cos A) cos A + (1 − 2 sin² A) sin A
sin 3A = 2 sin A cos² A + sin A − 2 sin³ A
sin 3A = 2 sin A (1 − sin² A) + sin A − 2 sin³ A
sin 3A = 2 sin A − 2 sin³ A + sin A − 2 sin³ A
sin 3A = 3 sin A − 4 sin³ A, or (3 − 4 sin² A) sin A
Now the cosine:
cos(2A + A) = cos 2A cos A − sin 2A sin A
cos 3A = (2 cos² A − 1) cos A − (2 sin A cos A) sin A
cos 3A = 2 cos³ A − cos A − 2 sin² A cos A
cos 3A = 2 cos³ A − cos A − 2 (1 − cos² A) cos A
cos 3A = 2 cos³ A − cos A − 2 cos A + 2 cos³ A
cos 3A = 4 cos³ A − 3 cos A, or (4 cos² A − 3) cos A
Solution:
tan 3A = sin 3A / cos 3A
tan 3A = ((3 − 4 sin² A) sin A) / ((4 cos² A − 3) cos A)
tan 3A = (sin A / cos A) (3 − 4 sin² A) / (4 cos² A − 3)
tan 3A = tan A (3 − 4 sin² A) / (4 cos² A − 3)
Divide top and bottom by cos² A. That at least gets rid of sin² A and cos² A, even though it introduces secant functions. (You remember that 1/cos A = sec A, right?)
tan 3A = tan A (3 sec² A − 4 tan² A) / (4 − 3 sec² A)
This may not look better, but it is—before, you had to get rid of two unwanted functions; now you have only one unwanted function, even though it occurs twice. And it’s sec² A. Isn’t there some sort of Pythagorean identity involving sec² A? Yes, there is! sin² A + cos² A = 1 ⇒ tan² A +1 = sec² A.
tan 3A = tan A (3 (tan² A + 1) − 4 tan² A) / (4 − 3 (tan² A + 1))
tan 3A = tan A (3 tan² A + 3 − 4 tan² A) / (4 − 3 tan² A − 3)
tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)
Check:
tan(2A+A) = (tan 2A + tan A) / (1 − tan 2A tan A)
From equation 60, tan 2A = 2 tan A / (1 − tan² A).
tan 3A = (2 tan A/(1 − tan² A) + tan A) / (1 − (2 tan A/(1 − tan² A)) tan A)
Well, that’s a mess! Clean it up by multiplying top and bottom by (1 − tan² A).
tan 3A = (2 tan A + (1 − tan² A) tan A) / ((1 − tan² A) − 2 tan A tan A)
tan 3A = (3 tan A − tan³ A) / (1 − 3 tan² A)
Factor out tan A from the top of the fraction, and it’s the same as what you got with the first method:
tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)
Solution: π/8 (22½°) is half of π/4 (45°), so you want the halfangle formulas. And π/8 is in Quadrant I, so all function values will be positive.
sin π/8 = √(1 − cos π/4)/2
sin π/8 = √(1 − √2/2)/2 = √(2 − √2)/4 = (½)√2 − √2
This expression contains nested radicals. Though some nested radicals can be denested, following the technique in Desnesting Radicals (or Unnesting Radicals), this one unfortunately cannot.
cos π/8 = √(1 + cos π/4)/2
cos π/8 = √(1 + √2/2)/2 = √(2 + √2)/4 = (½)√2 + √2
Now turn to the tangent. You can use the halfangle formula, or simply divide sine by cosine.
tan π/8 = (sin π/8) / (cos π/8)
tan π/8 = √2 − √2 / √2 + √ 2
This one can be denested, if you multiply top and bottom by √2 − √2. You’d normally do that anyway, to rationalize the denominator, but it’s a nice bonus that that happens to clean up the numerator also.
tan π/8 = √(2 − √2) (2 − √2) / √(2 + √ 2) (2 − √2)
tan π/8 = (2 − √2) / √(4 − 2) = (2 − √2) / √2 = √2 − 1
Solution: Arctan x is an angle whose tangent is x, so the possible values of Arctan x are angles in Q IV and Q I whose tangents can be taken. But tan −π/2 and tan π/2 don’t exist, so they are not possible values of Arctan x.
Solution: If angle A is Arcsin x, then we can make x the opposite side and 1 the hypotenuse. By the theorem of Pythagoras, that makes the adjacent side √(1 − x²). sec A is therefore sec(Arcsin x) = 1/√(1 − x²).
There are no odd powers of x in the answer, so we don’t have to worry about the sign of x.
Let’s check that with x = −0.7: Arcsin −0.7 ≈ −44.43°, and sec −44.3° = 1/cos −44.3° ≈ 1.40. 1/√(1 − 0.7²) ≈ 1.40 also, so the formula is right, at least for this test case.
Solution: If angle A is Arccos 1/x, we can set the hypotenuse to x and the adjacent side to 1 so that cos A = 1/x as required. The third side must be √(x² − 1).
What’s sin A? It must be √(x² − 1) / x. But that expression has an odd power of x, so we need to check the signs. If x is negative, then Arccos x will be in Q II. The sine function has all positive values in Q II, so we should have a positive answer. But as written, the fraction is negative if x is negative, so it needs an absolutevalue sign, just as in Example 3. Our final answer is
sin(Arccos 1/x) = √(x² − 1) /  x 
Let’s check that with x = −1.3: Arccos(1/−1.3) ≈ 140.28°, in Q II as expected, and sin 140.28° ≈ 0.64, a positive number like the sines of all angles in Q II. √((−1.3)² − 1) /  −1.3  ≈ 0.64 also. So the formula is right, at least for this test case — even with negative x, putting the angle in Q II, the formula returns a positive number, as it needs to.
Solutions: (a) 62(cos(240°) + i sin(240°)) ≈ −31−53.69i
(b) is apparently in radian measure, since there’s no degree mark. 100(cos(1.17) + i sin(1.17)) ≈ 39.02+92.08i
Solutions: (a) r = √((−42)² + 17²) ≈ 45.31. θ = 2 Arctan(17/(−42+45.31)) ≈ 2.76; multiply by 180°/π to convert to 157.96°. Answers: 45.31 cis 157.96° or 45.31 cis 2.76. Of course you could use any of the other forms shown in the chapter.
(b) and (c) are “gimmes”, since you don’t have
to compute the radius or the angle.
(b) 100i = 100∠90° or
100e^{iπ/2}
(c) −14.7 = 14.7∠180° or
14.7 e^{iπ}
Solution: First, put −i into polar form, which is easy enough since −i is on the negative y axis: −i = 1 cis 3π/2. Then apply the formula, equation 83:
(1 cis 3π/2)^{1/3} = 1^{1/3} cis (π/2 + 2πk/3) for k = 0, 1, 2
The three angles are π/2, π/2 + 2π/3 = 7π/6, and π/2 + 4π/3 = 11π/6 = 3π/2.
(1 cis 3π/2)^{1/3} = 1 cis π/2, 1 cis 7π/6, 1 cis 11π/6.
You know exact sines and cosines for any multiple of π/6, so you don’t need your calculator to convert back to a+bi form:
Cube roots of −i: i, −(√3)/2−(1/2)i, (√3/2)−(1/2)i.
Incidentally, not only is i a cube root of −i, but −i is a cube root of i. I’ll let you find the other two cube roots of i yourself.
Solution: First, put that number into polar form: r = √(1.04²+0.10²) ≈ 1.044796631. Then, find θ = 2 Arctan(−0.10/(1.04+1.044796631)) ≈ −0.0958591471. (My calculator happens to be in radian mode, but it doesn’t matter because I’ll be converting back to rectangular format anyway.) So
(1.04−0.10i)≈ 1.04796631 cis −0.0958591471
And therefore
(1.04−0.10i)^{16}≈ 1.04796631^{16} cis (−16&215;0.0958591471)
≈ 2.016082111 cis −1.533746354
≈ 0.0746787−2.01469853i
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