# Chapter 5 - Trigonometric Functions - Chapter Review - Review Exercises - Page 458: 14

$-1$

#### Work Step by Step

Since the trigonometric function $\sin$ is an odd function, we can write $f(-\theta)=-f(\theta) \implies \sin ({-40^\circ})=-\sin (40^\circ)$. Therefore, $\dfrac{\sin (-40^\circ)}{\sin (40^\circ) }=\dfrac{-\sin (40^\circ)}{\sin (40^\circ) } \\=-1$

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