Trig without Tears Part 9:
Copyright © 1997–2023 by Stan Brown, BrownMath.com
Trig without Tears Part 9:
Copyright © 1997–2023 by Stan Brown, BrownMath.com
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). Lower-case 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:
|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 many-valued, 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 many-valued inverse relation is arcsin with lower-case 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.
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
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, gives −√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.
the reader: The methods in this section are for the really hard-core 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 inside-out 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 non-algebraic 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
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
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 absolute-value 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.
next: 10/Complex Numbers
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