Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 420: 131

Answer

Simplifying the left side using the identity $\sin^{theta} +\sin^2{\theta}=1$ results to $1$, which is equal to the RHS. THus, the given statement is an identity.

Work Step by Step

\[ \begin{aligned} L H S &: \quad(\sin \theta \cos \phi)^{2}+(\sin \theta \sin \phi)^{2}+\cos^2( \theta )\\ &=\sin ^{2} \theta \cos ^{2} \phi+\sin ^{2} \theta \sin ^{2} \phi+\cos ^{2} \theta \end{aligned} \] Factoring out the common factor $\sin ^{2} \theta \quad$, then using the identity $\sin^{\theta}+\cos^2{\theta}=1$ yields \[ \begin{array}{l} =\sin ^{2} \theta\left(\cos^{2} \phi+\sin ^{2} \phi\right)+\cos ^{2} \theta \\ =(\sin ^{2} \theta)(1)+\cos^{2} \theta \\ =\sin ^{2} \theta+\cos^{2} \theta \\ =1 =R H S \end{array} \] Therefore LHS = RHS Hence the identity is established
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.