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 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 496: 12



Work Step by Step

Recall: $(a-b)(a+b)=a^2-b^2$ Use the rule above to obtain: \begin{align*} \dfrac{\sin{\theta}}{1+\cos{\theta}} \times \dfrac{1-\cos{\theta}}{1-\cos{\theta}} &= \dfrac{\sin{\theta}(1-\cos{\theta})}{(1+\cos{\theta})(1-\cos{\theta})}\\\\ &= \dfrac{\sin{\theta}(1-\cos{\theta})}{1-\cos^2{\theta}} \end{align*} Since $1-\cos^2{\theta} = \sin^2{\theta}$, then: $\dfrac{\sin{\theta}(1-\cos{\theta})}{1-\cos^2{\theta}} = \dfrac{\sin{\theta}(1-\cos{\theta})}{\sin^2{\theta}} = \boxed{\dfrac{1-\cos{\theta}}{\sin{\theta}}}$
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