Trig without Tears Part 5:
Functions of Any Angle
Copyright © 1997–2024 by Stan Brown, BrownMath.com
Trig without Tears Part 5:
Copyright © 1997–2024 by Stan Brown, BrownMath.com
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.
next: 6/Squared Identities
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