# 5.1: What’s your Inverse?

**At Grade**Created by: CK-12

*This activity is intended to supplement Trigonometry, Chapter 4, Lesson 3.*

**Time Required: 30 minutes**

## Activity Overview

This activity is intended to be teacher-led, where students will use their graphing calculators to graph the six inverse trig functions. The inverse reciprocal properties will be derived as a class.

**Topics Covered**

- Graphing inverses
- Finding the inverse of a function

**Teacher Preparation and Notes**

- Make sure students have cleared \begin{align*}Y=\end{align*}
Y= menu before starting. - Students will need to know how to find the inverse of a function algebraically. It might be helpful to have a warm-up covering this topic as a quick review.
- Go over the definition of a restricted domain and make sure students understand why they are necessary to find the inverse of trigonometric functions.
- Make sure students’ calculators are in
**Radians**.

**Associated Materials**

- Student Worksheet: What's your Inverse http://www.ck12.org/flexr/chapter/9702

## Problem 1

For the domain and range of \begin{align*}y = \sin^{-1}x\end{align*}

## Problem 2

Again, review the domain and range of \begin{align*}y = \cos x\end{align*}

## Problem 3

The domain of \begin{align*}y = \tan x\end{align*}

**For secant, cosecant and cotangent guide students through how to derive the equation that is needed to plug into the calculator.**

## Problem 4

Prove \begin{align*}\cos^{-1}x = \sec^{-1} \left(\frac{1}{x} \right)\end{align*}*Walk students through these steps*.

\begin{align*}y &= \cos^{-1}x\\
\cos y &= x\\
\frac{1}{\cos y} &= \frac{1}{x}\\
\sec y &= \frac{1}{x}\\
y &= \sec^{-1} \left(\frac{1}{x} \right)\end{align*}

This means that \begin{align*}\cos^{-1}x = \sec^{-1}\left(\frac{1}{x}\right) \end{align*}

In \begin{align*}Y1\end{align*}

## Problem 5

Prove \begin{align*}\sin^{-1}x = \csc^{-1}\left(\frac{1}{x}\right)\end{align*}*Walk students through these steps*.

\begin{align*}y &= \sin^{-1}x\\
\sin y &= x\\
\frac{1}{\sin y} &= \frac{1}{x}\\
\csc y &= \frac{1}{x}\\
y &= \csc^{-1} \left(\frac{1}{x}\right)\end{align*}

This means that \begin{align*}\sin^{-1}x = \csc^{-1}\left(\frac{1}{x} \right)\end{align*}

In \begin{align*}Y1\end{align*}

## Problem 6

Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs from tangent by a reflection over the \begin{align*}y-\end{align*}

\begin{align*}y &= - \tan \left(x - \frac{\pi}{2}\right)\\
x &= - \tan \left(y - \frac{\pi}{2}\right)\\
-x &= \tan \left(y - \frac{\pi}{2}\right)\\
\tan^{-1}(-x) &= y - \frac{\pi}{2}\\
\frac{\pi}{2} + \tan^{-1}(-x) &= y\\
\frac{\pi}{2} - \tan^{-1}x &= y\end{align*}

This means that \begin{align*}\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x\end{align*}

Because tangent is an odd function, or \begin{align*}\tan (-x) = - \tan x\end{align*}, then its inverse is also odd. In \begin{align*}Y1\end{align*} students should input \begin{align*}\frac{\pi}{2} - \tan^{-1}x\end{align*} in order to graph \begin{align*}y = \cot^{-1} x\end{align*}

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