Evaluating the Composition of Inverse Trig Functions

In this discussion, we learn how to evaluate the composition of inverse trig functions.

For example, we have

$$y={\sin }^{-1}[\sin (\frac{4\pi }{3}\left)\right]$$

Note that the inner function is a trig function and the outer function is an inverse trig function.

Find the Value of the Inner Trig Function First

The first thing is to work from the inside out.

Inside the brackets, we have

$$\sin \left(\frac{4\pi }{3}\right)$$

Here is a visual example; otherwise, content continues below.

Observe the Unit Circle

On the unit circle, there are coordinates in the form (a,b).

At the angle $\frac{4\pi }{3}$, the terminal points are,

$$(a,b)=(-\frac{1}{2},-\frac{\sqrt{3}}{2})$$

Find the Value of the Trig Function

Remember, sine takes the “b” coordinate.

Thus,

$$\sin \left(\frac{4\pi }{3}\right)=-\frac{\sqrt{3}}{2}.$$

Replace the Inner Trig Function with the Value

Now, the original problem is

$$y={\sin }^{-1}[\sin (\frac{4\pi }{3}\left)\right].$$

Replace $\sin \left(\frac{4\pi }{3}\right)$

with $-\frac{\sqrt{3}}{2}$

to get $y={\sin }^{-1}[-\frac{\sqrt{3}}{2}].$

Observe the Domain of the Inverse Trig Function

Recall that the domain of ${\sin }^{-1}\left(x\right)$ is $[-1,1]$.

Since we have $x=-\frac{\sqrt{3}}{2}$, we know that our domain is satisfied.

Now take the sine of both sides to get
$\sin \left(y\right)=-\frac{\sqrt{3}}{2}.$

Observe the Unit Circle

Looking at the unit circle, we search for all the values of y such that

$\sin \left(y\right)=-\frac{\sqrt{3}}{2}.$

Thus, when

$y=\frac{5\pi }{3},-\frac{\pi }{3},\frac{4\pi }{3},\text{and}-\frac{2\pi }{3}$

we get:

$\sin \left(\frac{5\pi }{3}\right)=-\frac{\sqrt{3}}{2}$

$\sin (-\frac{\pi }{3})=-\frac{\sqrt{3}}{2}$

$\sin \left(\frac{4\pi }{3}\right)=-\frac{\sqrt{3}}{2}$

$\sin (-\frac{2\pi }{3})=-\frac{\sqrt{3}}{2}$

Observe the Range of the Inverse Trig Function

Now, we cannot choose any value for y for the final answer.

We must remember to stay within the range of sin^-1(x).

The range of sin^-1(x) is
[-pi/2,pi/2],
which is the right region of the unit circle.

Choose the Value of the Inverse Trig Function

Looking at the values mentioned above, we see that the only choice we have is
-pi/3,
because all the other values are outside of the range.

Thus,
sin(-pi/3) = -sqrt(3)/2,

which means that
-pi/3 = sin^-1(-sqrt(3)/2)

and we say that
y = -pi/3

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