## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Review Exercises - Page 94: 14

#### Answer

$\displaystyle \sin 120^{\mathrm{o}}=\frac{\sqrt{3}}{2}$ $\displaystyle \cot 120^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$ $\displaystyle \cos 120^{\mathrm{o}}=-\frac{1}{2}$ $\tan 120^{\mathrm{o}}=-\sqrt{3}$ $\sec 120^{\mathrm{o}}=-2$ $\displaystyle \csc 120^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$

#### Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quadrant: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta \end{array}\right]$ $120^{0}$ is in quadrant II, so the reference angle is $180^{o}-120^{0}=60^{\mathrm{o}}$. Since $120^{0}$ is in quadrant II, the sine and cosecant are positive, the cosine, tangent, cotangent and secant are negative. From the Function Values of Special Angles, for $60^{\mathrm{o}}$: $\displaystyle \sin 120^{\mathrm{o}}=\sin 60^{\mathrm{o}}=\frac{\sqrt{3}}{2}$ $\displaystyle \cot 120^{\mathrm{o}}=-\cot 60^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$ $\displaystyle \cos 120^{\mathrm{o}}=-\cos 60^{\mathrm{o}}=-\frac{1}{2}$ $\tan 120^{\mathrm{o}}=-\tan 60^{\mathrm{o}}=-\sqrt{3}$ $\sec 120^{\mathrm{o}}=-\sec 60^{\mathrm{o}}=-2$ $\displaystyle \csc 120^{\mathrm{o}}=\csc 60^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$

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