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# Composition of Inverse Reciprocal Trig Functions

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

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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 sec1\begin{align*}sec^{-1}\end{align*}, csc1\begin{align*}csc^{-1}\end{align*}, and cot1\begin{align*}cot^{-1}\end{align*}.

Consider the following problem:

csc(cot13)\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.

1. Without a calculator, find cos(cot13)\begin{align*}\cos \left ( \cot^{-1} \sqrt{3} \right )\end{align*}.

First, find cot13\begin{align*}\cot^{-1} \sqrt{3}\end{align*}, which is also tan133\begin{align*}\tan^{-1}\frac{\sqrt{3}}{3}\end{align*}. This is π6\begin{align*}\frac{\pi}{6}\end{align*}. Now, find cosπ6\begin{align*}\cos \frac{\pi}{6}\end{align*}, which is 32\begin{align*}\frac{\sqrt{3}}{2}\end{align*}. So, our answer is 32\begin{align*}\frac{\sqrt{3}}{2}\end{align*}.

2. Without a calculator, find sec1(cscπ3)\begin{align*}\sec^{-1} \left ( \csc \frac{\pi}{3} \right )\end{align*}.

First, cscπ3=1sinπ3=132=23=233\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 sec1233=cos132=π6\begin{align*}\sec^{-1} \frac{2\sqrt{3}}{3} = \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}\end{align*}.

3. Evaluate cos(sin135)\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 θ=sin135\begin{align*}\theta = \sin^{-1} \frac{3}{5} \end{align*} or sinθ=35\begin{align*}\sin \theta = \frac{3}{5}\end{align*}, which means θ\begin{align*}\theta\end{align*} is in the Quadrant 1 (from our restricted domain, it cannot also be in Quadrant II). Substituting in θ\begin{align*}\theta\end{align*} we get cos(sin135)=cosθ\begin{align*}\cos \left ( \sin^{-1} \frac{3}{5} \right ) = \cos \theta\end{align*} and cosθ=45\begin{align*}\cos \theta = \frac{4}{5}\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to solve  csc(cot13)\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 2\begin{align*}\sqrt{2}\end{align*}?"

Since values for "x" and "y" around the unit circle are all fractions, and cotangent is equal to xy\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 2\begin{align*}\sqrt{2}\end{align*} as the answer.

When looking around the unit circle, you can see that cot30=3212=3\begin{align*}\cot 30^\circ = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\end{align*}

Therefore, cot13=30\begin{align*}\cot^{-1} \sqrt{3} = 30^\circ\end{align*}

Then you can apply the next function:

csc30=hypotenuseopposite=112=2\begin{align*}\csc 30^\circ = \frac{hypotenuse}{opposite} = \frac{1}{\frac{1}{2}} = 2\end{align*}

And so

csc(cot13)=2\begin{align*}\csc (\cot^{-1} \sqrt{3}) = 2\end{align*}

#### Example 2

Find the exact value of csc(cos132)\begin{align*}\csc \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right )\end{align*} without a calculator, over its restricted domains.

csc(cos132)=cscπ6=2\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 sec1(tan(cot11))\begin{align*}\sec^{-1} ( \tan (\cot^{-1} 1))\end{align*} without a calculator, over its restricted domains.

sec1(tan(cot11))=sec1(tanπ4)=sec11=0\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 tan1(cosπ2)\begin{align*}\tan^{-1} \left ( \cos \frac{\pi}{2} \right )\end{align*} without a calculator, over its restricted domains.

tan1(cosπ2)=tan10=0\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. sin(sec12)\begin{align*}\sin \left ( \sec^{-1} \sqrt{2} \right )\end{align*}
2. cos(csc11)\begin{align*}\cos \left ( \csc^{-1} 1 \right )\end{align*}
3. tan(cot13)\begin{align*}\tan \left ( \cot^{-1} \sqrt{3} \right )\end{align*}
4. cos(csc12)\begin{align*}\cos \left ( \csc^{-1} 2 \right )\end{align*}
5. cot(cos11)\begin{align*}\cot \left ( \cos^{-1} 1 \right )\end{align*}
6. csc(sin122)\begin{align*}\csc \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )\end{align*}
7. sec1(cosπ)\begin{align*}\sec^{-1} \left ( \cos \pi \right )\end{align*}
8. cot1(tanπ4)\begin{align*}\cot^{-1} \left ( \tan \frac{\pi}{4} \right )\end{align*}
9. sec1(cscπ4)\begin{align*}\sec^{-1} \left ( \csc \frac{\pi}{4} \right )\end{align*}
10. csc1(secπ3)\begin{align*}\csc^{-1} \left ( \sec \frac{\pi}{3} \right )\end{align*}
11. cos1(cotπ4)\begin{align*}\cos^{-1} \left ( \cot -\frac{\pi}{4} \right )\end{align*}
12. tan(cot10)\begin{align*}\tan \left ( \cot^{-1} 0 \right )\end{align*}
13. sin(csc1233)\begin{align*}\sin \left ( \csc^{-1} \frac{2\sqrt{3}}{3} \right )\end{align*}
14. cot1(sinπ2)\begin{align*}\cot^{-1} \left ( \sin \frac{\pi}{2} \right )\end{align*}
15. cos(sec1233)\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.

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### 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 $y=\frac{1}{x}$.