<meta http-equiv="refresh" content="1; url=/nojavascript/"> What’s your Inverse? | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Trigonometry Student Edition Go to the latest version.

Created by: CK-12
0  0  0

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 $Y=$ 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

## Problem 1

For the domain and range of $y = \sin^{-1}x$, review what the domain and range are of $y = \sin x$. The domain is all real numbers and the range is between -1 and 1. For the inverse, they will be switched, so the domain is between -1 and 1 and the range should be all real numbers. However, from looking at the graph, we know this is not true. Hence, there is a restricted domain on $y = \sin x$ so that it can have an inverse (recall that $y = \sin x$ is periodic). The domain is restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and the range of $y = \sin^{-1}x$ would be between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

## Problem 2

Again, review the domain and range of $y = \cos x$, which are all real numbers for the domain and between -1 and 1 for the range. By looking at the graph, domain of the inverse is also between -1 and 1 and the range is between 0 and $\pi$.

## Problem 3

The domain of $y = \tan x$ is all real numbers, except for every odd multiple of $\frac{\pi}{2}$ $\left(x \ne \frac{\pi}{2},\frac{3 \pi}{2}, \frac{5 \pi}{2},\ldots \right)$. The range is all real numbers. So, the range of $y = \tan^{-1}x$ is between two of these asymptotes $\left (-\frac{\pi}{2}, \frac{\pi}{2} \right )$ and there are horizontal asymptotes. The domain is all real numbers.

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

## Problem 4

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

$y &= \cos^{-1}x\\ \cos y &= x\\ \frac{1}{\cos y} &= \frac{1}{x}\\ \sec y &= \frac{1}{x}\\ y &= \sec^{-1} \left(\frac{1}{x} \right)$

This means that $\cos^{-1}x = \sec^{-1}\left(\frac{1}{x}\right)$ and $\sec^{-1}x = \cos^{-1}\left(\frac{1}{x}\right)$.

In $Y1$, students should input $\cos^{-1}\left(\frac{1}{x} \right)$ in order to graph $y = \sec^{-1}x$.

## Problem 5

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

$y &= \sin^{-1}x\\\sin y &= x\\\frac{1}{\sin y} &= \frac{1}{x}\\\csc y &= \frac{1}{x}\\y &= \csc^{-1} \left(\frac{1}{x}\right)$

This means that $\sin^{-1}x = \csc^{-1}\left(\frac{1}{x} \right)$ and $\csc^{-1}x = \sin^{-1}\left(\frac{1}{x} \right)$.

In $Y1$, students should input $\sin^{-1}\left(\frac{1}{x} \right)$ in order to graph $y = \csc^{-1}x$.

## Problem 6

Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs from tangent by a reflection over the $y-$axis and a shift of $\frac{\pi}{2}$. As an equation, the relationship would be $\cot x = - \tan \left(x -\frac{\pi}{2}\right )$. Students will need to take the inverse of $y = - \tan \left(x - \frac{\pi}{2}\right )$ to find how to graph $y = \cot^{-1}x$ in their calculators.

$y &= - \tan \left(x - \frac{\pi}{2}\right)\\x &= - \tan \left(y - \frac{\pi}{2}\right)\\-x &= \tan \left(y - \frac{\pi}{2}\right)\\\tan^{-1}(-x) &= y - \frac{\pi}{2}\\\frac{\pi}{2} + \tan^{-1}(-x) &= y\\\frac{\pi}{2} - \tan^{-1}x &= y$

This means that $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$ and $\tan^{-1} x = \frac{\pi}{2} - \cot^{-1}x$.

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

Feb 23, 2012

Aug 19, 2014