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

Published by Pearson

# Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 85

#### Answer

$-1$

#### Work Step by Step

Recall: Sine is an odd function therefore $\sin{(-\theta)} = - \sin{\theta}$. This means that: $\sin{\left(-\dfrac{\pi}{12} \right)} = - \sin{\left(\dfrac{\pi}{12} \right)}$ Also, $\sin{\left(-\dfrac{\pi}{12} \right)} \cdot \csc{\left(\dfrac{25 \pi }{12} \right)} = - \sin{\left(\dfrac{\pi}{12} \right)} \cdot \csc{\left(\dfrac{25 \pi }{12} \right)}$ Recal that: $\sin{(\theta+2 \pi)} = \sin{\theta}$ Thus, $\sin{\left(\dfrac{\pi}{12} \right)} = \sin{\left(\dfrac{\pi}{12}+2\pi \right)} = \sin{\left(\dfrac{25 \pi}{12} \right)}$ Therefore, $-\sin{\left(\dfrac{\pi}{12} \right)} \cdot \csc{\left(\dfrac{25 \pi }{12} \right)} =- \sin{\left(\dfrac{25 \pi}{12} \right)}\cdot \csc{\left(\dfrac{25 \pi }{12} \right)}$ Recall: $\csc{\theta} =\dfrac{1}{\sin{\theta}}$ Hence, $-\sin{\left(\dfrac{25 \pi}{12} \right)}\cdot \csc{\left(\dfrac{25 \pi }{12} \right)} =- \sin{\left(\dfrac{25 \pi}{12} \right)}\cdot \dfrac{1}{ \sin{\left(\dfrac{25 \pi}{12} \right)}} = \boxed{-1}$

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