<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

Composition of Inverse Reciprocal Trig Functions

Application of one of six trigonometric functions or inverses and then another.

Estimated5 minsto complete
%
Progress
Practice Composition of Inverse Reciprocal Trig Functions
Progress
Estimated5 minsto complete
%
Composition of Inverse Reciprocal Trig Functions

Composing functions involves applying one function and then applying another function afterward. In the case of inverse reciprocal functions, you could create compositions of functions such as \begin{align*}sec^{-1}\end{align*}, \begin{align*}csc^{-1}\end{align*}, and \begin{align*}cot^{-1}\end{align*}.

Consider the following problem:

\begin{align*}\csc (\cot^{-1} \sqrt{3})\end{align*}

Can you solve this problem?

Composition of Inverse Reciprocal Trig Functions

Just as you can apply one function and then another whenever you'd like, you can do the same with inverse reciprocal trig functions. This process is called composition.

Here we'll explore some examples of composition for these inverse reciprocal trig functions by doing some problems.

Without a calculator, find \begin{align*}\cos \left ( \cot^{-1} \sqrt{3} \right )\end{align*}.

First, find \begin{align*}\cot^{-1} \sqrt{3}\end{align*}, which is also \begin{align*}\tan^{-1}\frac{\sqrt{3}}{3}\end{align*}. This is \begin{align*}\frac{\pi}{6}\end{align*}. Now, find \begin{align*}\cos \frac{\pi}{6}\end{align*}, which is \begin{align*}\frac{\sqrt{3}}{2}\end{align*}. So, our answer is \begin{align*}\frac{\sqrt{3}}{2}\end{align*}.

Without a calculator, find \begin{align*}\sec^{-1} \left ( \csc \frac{\pi}{3} \right )\end{align*}.

First, \begin{align*}\csc \frac{\pi}{3} = \frac{1}{\sin \frac{\pi}{3}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\end{align*}. Then \begin{align*}\sec^{-1} \frac{2\sqrt{3}}{3} = \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}\end{align*}.

Evaluate \begin{align*}\cos \left ( \sin^{-1} \frac{3}{5} \right )\end{align*}.

Even though this problem is not a critical value, it can still be done without a calculator. Recall that sine is the opposite side over the hypotenuse of a triangle. So, 3 is the opposite side and 5 is the hypotenuse. This is a Pythagorean Triple, and thus, the adjacent side is 4. To continue, let \begin{align*}\theta = \sin^{-1} \frac{3}{5} \end{align*} or \begin{align*}\sin \theta = \frac{3}{5}\end{align*}, which means \begin{align*}\theta\end{align*} is in the Quadrant 1 (from our restricted domain, it cannot also be in Quadrant II). Substituting in \begin{align*}\theta\end{align*} we get \begin{align*}\cos \left ( \sin^{-1} \frac{3}{5} \right ) = \cos \theta\end{align*} and \begin{align*}\cos \theta = \frac{4}{5}\end{align*}.

Examples

Example 1

Earlier, you were asked to solve  \begin{align*}\csc (\cot^{-1} \sqrt{3})\end{align*}.

The first step in this problem is to ask yourself  "What angle would produce a cotangent of \begin{align*}\sqrt{2}\end{align*}?"

Since values for "x" and "y" around the unit circle are all fractions, and cotangent is equal to \begin{align*}\frac{x}{y}\end{align*}, you need to find a pair of equations on the unit circle which, when divided by each other, give \begin{align*}\sqrt{2}\end{align*} as the answer.

When looking around the unit circle, you can see that \begin{align*}\cot 30^\circ = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\end{align*}

Therefore, \begin{align*}\cot^{-1} \sqrt{3} = 30^\circ\end{align*}

Then you can apply the next function:

\begin{align*}\csc 30^\circ = \frac{hypotenuse}{opposite} = \frac{1}{\frac{1}{2}} = 2\end{align*}

And so

\begin{align*}\csc (\cot^{-1} \sqrt{3}) = 2\end{align*}

Example 2

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

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

Example 3

Find the exact value of \begin{align*}\sec^{-1} ( \tan (\cot^{-1} 1))\end{align*} without a calculator, over its restricted domains.

\begin{align*}\sec^{-1} (\tan (\cot^{-1} 1)) = \sec^{-1} \left ( \tan \frac{\pi}{4} \right ) = \sec^{-1} 1 = 0\end{align*}

Example 4

Find the exact value of \begin{align*}\tan^{-1} \left ( \cos \frac{\pi}{2} \right )\end{align*} without a calculator, over its restricted domains.

\begin{align*}\tan^{-1} \left ( \cos \frac{\pi}{2} \right ) = \tan^{-1} 0 = 0\end{align*}

Review

Without using technology, find the exact value of each of the following. Use the restricted domain for each function.

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

To see the Review answers, open this PDF file and look for section 4.8.

Vocabulary Language: English

inverse function

Inverse functions are functions that 'undo' each other. Formally $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$.

Reciprocal Function

A reciprocal function is a function with the parent function $y=\frac{1}{x}$.

Explore More

Sign in to explore more, including practice questions and solutions for Composition of Inverse Reciprocal Trig Functions.