# Chapter 2 - Review Exercises - Page 94: 15

$\displaystyle \sin(-1470^{o})= -\frac{1}{2}$ $\displaystyle \cos(-1470^{o})= \frac{\sqrt{3}}{2}$ $\displaystyle \tan(-1470^{o})= -\frac{\sqrt{3}}{3}$ $\cot(-1470^{o}) =-\sqrt{3}$ $\displaystyle \sec(-1470^{o})= \frac{2\sqrt{3}}{3}$ $\csc(-1470^{o}) =-2$

#### Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quadrant: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ $-1470^{o}$ is coterminal with $-1470^{o}+5\cdot 360^{o}$ $=-1470^{o}+1800^{o}=330^{o}$ (quadrant IV. ) $330^{o}$ lies in quadrant IV. The reference angle is $360^{o}-330^{o}=30^{o}$. Since $-1470^{o}$ is in quadrant IV, the cosine and secant are positive, the sine, tangent, cotangent, and cosecant are negative. From the Function Values of Special Angles, for $30^{\mathrm{o}}$: $\displaystyle \sin(-1470^{o})=-\sin 30^{\mathrm{o}}=-\frac{1}{2}$ $\displaystyle \cos(-1470^{o})=\cos 30^{\mathrm{o}}=\frac{\sqrt{3}}{2}$ $\displaystyle \tan(-1470^{o})=-\tan 30^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$ $\cot(-1470^{o})=-\cot 30^{\mathrm{o}}=-\sqrt{3}$ $\displaystyle \sec(-1470^{o})=\sec 30^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$ $\csc(-1470^{o})=-\csc 30^{\mathrm{o}}=-2$

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