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4.2: Trig Proofs

Difficulty Level: At Grade Created by: CK-12
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This activity is intended to supplement Trigonometry, Chapter 3, Lesson 2.

ID: 9776

Time required: 30 minutes

Activity Overview

Students will perform trigonometric proofs and use the graphing capabilities of the calculator for verification.

Topic: Trigonometric Identities

  • Use fundamental trigonometric identities to prove more complex trigonometric identities.
  • Verify trigonometric identities by graphing.

Teacher Preparation and Notes

  • Students should already be familiar with the Pythagorean trigonometric identities as well the fact that \begin{align*}tangent = \frac{sine}{cosine}, cosecant =\frac{1}{sine}, secant = \frac{1}{cosine}\end{align*}tangent=sinecosine,cosecant=1sine,secant=1cosine, and \begin{align*}cotangent = \frac{1}{tangent}\end{align*}cotangent=1tangent.
  • This activity is intended to be teacher-led. You may use the following pages to present the material to the class and encourage discussion. Students will follow along using their calculators, although the majority of the ideas and concepts are only presented in this document; be sure to cover all the material necessary for students’ total comprehension.
  • The worksheet is intended to guide students through the main ideas of the activity, while providing more detailed instruction on how they are to perform specific actions using the tools of the calculator. It also serves as a place for students to record their answers. Alternatively, you may wish to have the class record their answers on separate sheets of paper, or just use the questions posed to engage a class discussion.

Associated Materials

Problem 1 – Using the Calculator for verification

Prove: \begin{align*}(1 + \cos x)(1 - \cos x) = \sin^2x\end{align*}(1+cosx)(1cosx)=sin2x.

\begin{align*}(1 + \cos(x))(1 - \cos(x)) &= 1 - \cos^2(x)\\ &= [\sin^2(x) + \cos^2(x)] - \cos^2(x)\\ &= \sin^2x\end{align*}(1+cos(x))(1cos(x))=1cos2(x)=[sin2(x)+cos2(x)]cos2(x)=sin2x

To verify this proof graphically, you will determine if the graph of the expression on the left side of the equation coincides with the graph of the expression on the right side of the equation.

Enter the left side of the equation \begin{align*}(1 + \cos x) (1 - \cos x)\end{align*}(1+cosx)(1cosx) in \begin{align*}Y_1\end{align*}Y1.

Enter the right side of the equation \begin{align*}(\sin x)^2\end{align*}(sinx)2 in \begin{align*}Y_2\end{align*}Y2.

Arrow over to the icon to the left of \begin{align*}Y_2\end{align*}Y2 and press ENTER to change it to a thick line. Now \begin{align*}Y_1\end{align*}Y1 will be graphed as a thin line and \begin{align*}Y_2\end{align*}Y2 will be graphed as a thick line. This will help you distinguish the two graphs (and make sure that they are equal).

Press ZOOM and select ZTrig to set the window size.

The two graphs coincide, so the two sides of the equation are equal, as we proved. Note that the calculator is only verifying what we have proven. The graph in no way constitutes a proof.

Problems 2-5

For problems 2 through 5, prove the equation given and then verify it graphically. For \begin{align*}\cot x\end{align*}cotx, type \begin{align*} \left(\frac{1}{\tan x} \right)\end{align*}(1tanx). For \begin{align*}\sec x\end{align*}secx, type \begin{align*}\left(\frac{1}{\cos x} \right)\end{align*}(1cosx).

2. \begin{align*}\sin x \cdot \cot x \cdot \sec x = 1\end{align*}sinxcotxsecx=1

3. \begin{align*}\frac{\sec^2 x - 1}{\sec^2 x}=\sin^2 x\end{align*}sec2x1sec2x=sin2x

4. \begin{align*}\tan x + \cot x = \sec x(\csc x) \end{align*}tanx+cotx=secx(cscx)

5. \begin{align*}\frac{\sin^2 x - 49}{\sin^2x+14\sin x+ 49}=\frac{\sin x - 7}{\sin x + 7}\end{align*}sin2x49sin2x+14sinx+49=sinx7sinx+7

Solutions

Problem 2

\begin{align*}\sin (x) \cdot \cot (x) \cdot \sec (x) &= \sin (x) \cdot \frac{\cos (x)}{\sin (x)} \cdot \frac{1}{\cos(x)}\\ &= 1\end{align*}sin(x)cot(x)sec(x)=sin(x)cos(x)sin(x)1cos(x)=1

Problem 3

\begin{align*}\frac{\sec^2(x) - 1}{\sec^2(x)} &= \frac{[1 + \tan^2 x]-1}{\sec^2 (x)}\\ &= \frac{\tan^2 (x)}{\sec^2 (x)}\\ &= \frac{\left(\frac{\sin (x)}{\cos (x)}\right)^2}{\frac{1}{\cos^2 (x)}}\\ &=\sin^2 (x)\end{align*}sec2(x)1sec2(x)=[1+tan2x]1sec2(x)=tan2(x)sec2(x)=(sin(x)cos(x))21cos2(x)=sin2(x)

Problem 4

\begin{align*}\tan (x) + \cot (x) &= \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\sin (x)}\\ &= \frac{\sin^2 (x)}{\cos (x) \cdot \sin(x)} + \frac{\cos^2 (x)}{\cos (x) \cdot \sin (x)}\\ &= \frac{\sin^2(x) + \cos^2 (x)} {\cos (x) \cdot \sin (x)}\\ &= \frac{1}{\cos(x) \cdot \sin (x)}\\ &= \sec (x) \cdot \csc (x)\end{align*}tan(x)+cot(x)=sin(x)cos(x)+cos(x)sin(x)=sin2(x)cos(x)sin(x)+cos2(x)cos(x)sin(x)=sin2(x)+cos2(x)cos(x)sin(x)=1cos(x)sin(x)=sec(x)csc(x)

Problem 5

\begin{align*}\frac{\sin^2 (x) - 49}{\sin^2(x) + 14 \sin(x) + 49} &= \frac{(\sin(x) - 7) \cdot (\sin (x) + 7)}{(\sin (x) + 7) \cdot (\sin(x) + 7)}\\ &= \frac{\sin(x) - 7}{\sin (x) + 7}\end{align*}sin2(x)49sin2(x)+14sin(x)+49=(sin(x)7)(sin(x)+7)(sin(x)+7)(sin(x)+7)=sin(x)7sin(x)+7

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Date Created:
Aug 19, 2014
Last Modified:
Nov 04, 2014
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