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# Composition of Trig Functions and Their Inverses

## Application of sine, cosine, tangent, or their inverses and then another function.

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Composition of Trig Functions and Their Inverses

You've considered trigonometric functions, and you've considered inverse functions, and now it's time consider how to compose trig functions and their inverses. If someone were to ask you to apply the inverse of a trig function to a different trig function, would you be able to do this? For example, can you find sin1(cos(3π2))\begin{align*}\sin^{-1} (\cos (\frac{3\pi}{2}))\end{align*}?

### Trigonometric Functions and Their Inverses

In other sections, you learned that for a function f(f1(x))=x\begin{align*}f(f^{-1}(x)) = x\end{align*} for all values of x\begin{align*}x\end{align*} for which f1(x)\begin{align*}f^{-1}(x)\end{align*} is defined. If this property is applied to the trigonometric functions, the following equations will be true whenever they are defined:

sin(sin1(x))=xcos(cos1(x))=xtan(tan1(x))=x\begin{align*}\sin(\sin^{-1}(x)) = x && \cos(\cos^{-1}(x)) = x && \tan(\tan^{-1}(x)) = x\end{align*}

As well, you learned that f1(f(x))=x\begin{align*}f^{-1}(f(x)) = x\end{align*} for all values of x\begin{align*}x\end{align*} for which f(x)\begin{align*}f(x)\end{align*} is defined. If this property is applied to the trigonometric functions, the following equations that deal with finding an inverse trig function of a trig function, will only be true for values of x\begin{align*}x\end{align*} within the restricted domains.

sin1(sin(x))=xcos1(cos(x))=xtan1(tan(x))=x\begin{align*}\sin^{-1}(\sin(x)) = x && \cos^{-1}(\cos(x)) = x && \tan^{-1}(\tan(x)) = x\end{align*}

These equations are better known as composite functions. However, it is not necessary to only have a function and its inverse acting on each other. In fact, it is possible to have composite function that are composed of one trigonometric function in conjunction with another different trigonometric function. The composite functions will become algebraic functions and will not display any trigonometry. Let’s investigate this phenomenon.

When solving these types of problems, start with the function that is composed inside of the other and work your way out. Use the following problems as a guideline.

1. Find \begin{align*}\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )\end{align*}.

We know that \begin{align*}\sin^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}\end{align*}, within the defined restricted domain. Then, we need to find \begin{align*}\sin \frac{\pi}{4}\end{align*}, which is \begin{align*}\frac{\sqrt{2}}{2}\end{align*}. So, the above properties allow for a short cut. \begin{align*}\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right ) = \frac{\sqrt{2}}{2}\end{align*}, think of it like the sine and sine inverse cancel each other out and all that is left is the \begin{align*}\frac{\sqrt{2}}{2}\end{align*}.

2. Without using technology, find the exact value of each of the following:

a. \begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right )\end{align*}

\begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right )\end{align*}: First find \begin{align*}\tan^{-1} \sqrt{3}\end{align*}, which is \begin{align*}\frac{\pi}{3}\end{align*}. Then find \begin{align*}\cos \frac{\pi}{3}\end{align*}. Your final answer is \begin{align*}\frac{1}{2}\end{align*}. Therefore, \begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right ) = \frac{1}{2}\end{align*}.

b. \begin{align*}\tan \left ( \sin^{-1}\left( -\frac{1}{2} \right ) \right )\end{align*}

\begin{align*}\tan \left ( \sin^{-1} \left ( -\frac{1}{2} \right ) \right ) = \tan \left ( -\frac{\pi}{6} \right ) = -\frac{\sqrt{3}}{3}\end{align*}

c. \begin{align*}\cos (\tan^{-1} (-1))\end{align*}

\begin{align*}\cos (\tan^{-1} (-1)) = \cos^{-1} \left ( -\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}\end{align*}.

d. \begin{align*}\sin \left ( \cos^{-1}\frac{\sqrt{2}}{2} \right )\end{align*}

\begin{align*}\sin \left ( \cos^{-1} \frac{\sqrt{2}}{2} \right ) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to solve \begin{align*}\sin^{-1} (\cos (\frac{3\pi}{2}))\end{align*}.

To solve this problem: \begin{align*}\sin^{-1} (\cos (\frac{3\pi}{2}))\end{align*}, you can work outward.

First find:

\begin{align*}\cos (\frac{3\pi}{2}) = 0\end{align*}

Then find:

\begin{align*}\sin^{-1} 0 = 0\end{align*}

or

\begin{align*}\sin^{-1} 0 = \pi\end{align*}

#### Example 2

Find the exact value of \begin{align*}\cos^{-1} \frac{\sqrt{3}}{2} \end{align*}, without a calculator, over its restricted domain.

\begin{align*}\frac{\pi}{6}\end{align*}

#### Example 3

Evaluate: \begin{align*}\sin \left ( \cos^{-1} \frac{5}{13} \right )\end{align*}

\begin{align*}\cos \theta & = \frac{5}{13}\\ \sin \left ( \cos^{-1} \left ( \frac{5}{13} \right ) \right ) & = \sin \theta\\ \sin \theta & = \frac{12}{13}\end{align*}

#### Example 4

Evaluate: \begin{align*}\tan \left ( \sin^{-1} \left( -\frac{6}{11} \right) \right )\end{align*}

\begin{align*}\tan \left ( \sin^{-1} \left ( -\frac{6}{11} \right ) \right ) \rightarrow \sin \theta = -\frac{6}{11}\end{align*}.

The third side is \begin{align*}b = \sqrt{121-36} = \sqrt{85}\end{align*}.

\begin{align*}\tan \theta = -\frac{6}{\sqrt{85}} = -\frac{6\sqrt{85}}{85}\end{align*}

### Review

Without using technology, find the exact value of each of the following.

1. \begin{align*}\sin \left ( \sin^{-1} \frac{1}{2} \right )\end{align*}
2. \begin{align*}\cos \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right )\end{align*}
3. \begin{align*}\tan \left ( \tan^{-1} \sqrt{3} \right )\end{align*}
4. \begin{align*}\cos \left ( \sin^{-1} \frac{1}{2} \right )\end{align*}
5. \begin{align*}\tan \left ( \cos^{-1} 1 \right )\end{align*}
6. \begin{align*}\sin \left ( \cos^{-1} \frac{\sqrt{2}}{2} \right )\end{align*}
7. \begin{align*}\sin^{-1} \left ( \sin \frac{\pi}{2} \right )\end{align*}
8. \begin{align*}\cos^{-1} \left ( \tan \frac{\pi}{4} \right )\end{align*}
9. \begin{align*}\tan^{-1} \left ( \sin \pi \right )\end{align*}
10. \begin{align*}\sin^{-1} \left ( \cos \frac{\pi}{3} \right )\end{align*}
11. \begin{align*}\cos^{-1} \left ( \sin -\frac{\pi}{4} \right )\end{align*}
12. \begin{align*}\tan \left ( \sin^{-1} 0 \right )\end{align*}
13. \begin{align*}\sin \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right )\end{align*}
14. \begin{align*}\tan^{-1} \left ( \cos \frac{\pi}{2} \right )\end{align*}
15. \begin{align*}\cos \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )\end{align*}

To see the Review answers, open this PDF file and look for section 4.6.

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Color Highlighted Text Notes
composite function A composite function is a function $h(x)$ formed by using the output of one function $g(x)$ as the input of another function $f(x)$. Composite functions are written in the form $h(x)=f(g(x))$ or $h=f \circ g$.