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 6 - Analytic Trigonometry - Section 6.5 Sum and Difference Formulas - 6.5 Assess Your Understanding - Page 509: 64


See proof below.

Work Step by Step

Apply the sum and difference formulas: $\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$ and $\cos(A-B)=\cos(A)\cos(B) +\sin(A) \ \sin(B)$ In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows: $\text{LHS}=\dfrac{\cos{(\alpha+\beta)}}{\cos (\alpha - \beta)}\\ =\dfrac{\cos \alpha \cos \beta-\sin \alpha \sin \beta }{\cos \alpha \cos \beta +\sin \alpha \sin \beta} $ Divide the numerator and denominator by $\cos \alpha \cos \beta$ to obtain: $\dfrac{1-\dfrac{\sin \alpha }{\cos \alpha}\dfrac{\sin \beta }{\cos \beta} }{ 1 +\dfrac{\sin \alpha }{\cos \alpha}\dfrac{\sin \beta }{\cos \beta}} \\=\dfrac{1-\tan \alpha \ \tan \beta}{1 + \tan \alpha \ \tan \beta} \\=\text{ RHS}$
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