If you restrict to the interval , the function
*increases*:

This implies that the function is *one-to-one*, and hence it
has an inverse. The inverse is called the * inverse
sine* or * arcsine function*, and is denoted
or . Note that
in the second case *does not
mean* " "!

Note: "Arcsine" (and ) are older terms, and there is similar terminology for the other inverse trig functions (so "arctangent" and for the inverse tangent function, and so on). I'll use the inverse function terminology instead.

In word, is *the angle whose sine is
x*. Another way of saying this is:

The fact that and are inverse functions can be expressed by the following equations:

Since the restricted takes angles in the range and produces numbers in the range , takes numbers in the range and produces angles in the range .

* Example.* Compute and .

Sine and arcsine are inverses, so they undo one another --- but you have to be careful!

can't be , because always returns an angle in the range .

* Example.* Find .

First, let . This means that . Now , so I get the following picture:

I got the adjacent side using Pythagoras: .

Using the triangle, I have

You can find a derivative formula for using implicit differentiation. Let . This is equivalent to . Differentiate implicitly:

I'd like to express the result in terms of x. Here's the right triangle that says :

I found the other leg using Pythagoras. You can see that . Hence, . That is,

Every derivative formula gives rise to a corresponding antiderivative formula:

Before I do some calculus examples, I want to mention some of the
other inverse trig functions. I'll discuss the *
inverse cosine*, * inverse tangent*, and * inverse secant* functions.

(a) You get the inverse cosine by inverting , restricted to .

(b) You get the inverse tangent by inverting , restricted to .

(c) You get the inverse secant by inverting , restricted to together with .

As with and , the domains and ranges of these functions and their inverses are "swapped":

* Example.* Compute and .

You can derive the derivative formulas for the other inverse trig functions using implicit differentiation, just as I did for the inverse sine function.

For example, I'll derive the formula for .

The derivation starts out like the derivation for . Let , so . Differentiating implicitly, I get

There are two cases, depending on whether or .

Suppose . Then is in the interval , as illustrated in the first diagram above. You can see from the picture that

Therefore,

, so x is positive, and . Therefore,

Now suppose that . Then is in the interval , as illustrated in the second diagram above. Since x is negative, the hypotenuse must be , since it must be positive and since must equal x. In this case,

Therefore,

, so x is negative, and . Therefore,

This proves that in all cases.

* Example.* Compute:

(a) .

(b) .

(c) .

(a)

(b)

(c)

I don't need absolute values in the last example, because is always positive.

* Example.* Prove the identity

Hence,

A function with zero derivative is constant, so

But when ,

Therefore,

Here are the integration formulas for some of the inverse trig functions. I'm giving extended versions of the formulas --- with " " replacing the "1" that you'd get if you just reversed the derivative formulas --- in order to save you a little time in doing problems.

For instance, here's how to derive the extended integral formula from the formula using substitution:

* Example.* Compute and .

Using the formula with ,

Using the formula with ,

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

Copyright 2018 by Bruce Ikenaga