Trig without Tears Part 9:
Trig without Tears Part 9:
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).
Sometimes you have a sine or cosine or tangent and need to find the associated angle. For instance, this happens whenever you solve a triangle. When you have a sine function value and find the corresponding angle, we say you are finding the arcsine or inverse sine of that value, and similarly for the other functions.
Different books use different notation: sin with a superscript −1, or arcsin. I prefer the “arc” forms because the superscript −1 looks too much like an exponent.
You may be less intimidated by the arcfunctions if you pronounce them as the angle whose. For instance, arccos 0.6 becomes “the angle whose cosine is 0.6”.
What is arcsin(0.5)? You probably recognize that 0.5 is ½, and it must be a sine of one of the special angles. In fact, equation 15 tells you that sin(30°) = ½. So you can say that arcsin(0.5) = 30° or π/6.
But wait, there’s more! You know from equation 22 that
sin(30°) = sin(150°) = sin(390°) = sin(−210°)
and so on, and therefore they all equal ½. In fact,
sin(30° + 360°k) = sin(150° + 360°k) = 0.5 for all integer k
So which of this infinite number of values is the arcsine?
To make an arcsine function, we have to restrict the range (the possible answers given by each function) so that each number has at most one arcsine. (Why do I say “at most one” and not “one and only one”? The sine of any angle is between −1 and +1 inclusive; therefore only those numbers have arcsines. There is o angle whose sine is 2, so 2 has no arcsines and arcsin 2 is a meaningless noise.)
The arcsine is defined so that its range is the interval [−π/2;+π/2], which is the same as [−90°;+90°]. Some books use a capital letter for the function Arcsin distinguish it from the multi-valued relation arcsin. So we could say
arcsin(0.5) = π/6+2kπ or 5π/6+2kπ for all integer k
Arcsin(0.5) = π/6
Why the particular range −π/2 to +π/2? To start with, it seems tidy that any arcfunction of a positive number should fall in Quadrant I, [0;+π/2]. So the only real question is arcfunctions of negative numbers. If we prefer the numerically smallest values for the arcsine function, then Arcsin(−0.5) = −30° = −π/6 fits that rule, and a negative number’s arcsine (and arctangent, too) will be in Quadrant IV, [−π/2;0].
What about the arccosine? 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, or [0;π].
You can remember the range of the arctangent function this way: the only two choices that make sense are 0 to π and −π/2 to +π/2. But tan(π/2) is undefined, so if you picked [0;π] for the range of Arctan it would have an ugly hole in it at π/2. Therefore the range is defined to be (−π/2;+π/2), an open interval but with no points missing inside.
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) as the range.
Thomas defines the arcsecant and arccosecant 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]||[−π/2; +π/2]|
|Arccos||Arccos x = π/2 − Arcsin x||[−1; +1]||[0; π]|
|Arctan||inverse of tangent function||all reals||(−π/2; +π/2)|
|Arccot||Arccot x = π/2 − Arctan x||all reals||(0; π)|
|Arcsec||Arcsec x = Arccos(1/x)||(−∞; −1], [1; ∞)||[0; π]|
|Arccsc||Arccsc x = Arcsin(1/x)||(−∞; −1], [1; ∞)||[−π/2; +π/2]|
Remember that the relations 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.
Advice to the reader: The formulas 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.
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 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.
(If the answer contains variables raised to odd powers, you may need to add some absolute value signs. See Example 3.)
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 is how we needed to set up angle A.
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, sqrt(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/sqrt(1+x²)
But cos A = cos(Arctan x). Therefore
cos(Arctan x) = 1/sqrt(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 sqrt(1−x²), and write that down. Now you need cos A, which is the adjacent side over the hypotenuse, which is sqrt(1−x²)/1. Answer:
cos(Arcsin x) = sqrt(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 sqrt(1+x²).
Now you can write down cos A, which is adjacent over hypotenuse:
cos A = x / sqrt(1+x²)
cos(Arctan 1/x) = x / sqrt(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 in the previous paragraph, x/sqrt(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 | / sqrt(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.
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 formulas 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.
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, 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 a is 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. 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.