## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Test - Page 97: 6

#### Answer

$\displaystyle \sin(-135^{o})=-\frac{\sqrt{2}}{2}$ $\displaystyle \cos(-135^{o})=-\frac{\sqrt{2}}{2}$ $\tan(-135^{o})=1$ $\csc(-135^{o})=-\sqrt{2}$ $\sec(-135^{o})=-\sqrt{2}$ $\cot(-135^{o})=1$

#### Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quadrant: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta \end{array}\right]$ $-135^{o}$ is between $-90^{o}$ and $-180^{o}$ ,in quadrant III, and has the same terminal side as the angle $-135^{o}+360^{o}=225^{o}$ for which the reference angle is $\theta^{\prime}=\theta-180^{o}=225^{o}-180^{o}=45^{o}$ Signs: In quadrant III, tan and cot are positive, all the others are negative We use the table Function Values of Special Angles with the corresponding signs: $\displaystyle \sin(-135^{o})=-\sin 45^{o}=-\frac{\sqrt{2}}{2}$ $\displaystyle \cos(-135^{o})=-\cos 45^{o}=-\frac{\sqrt{2}}{2}$ $\tan(-135^{o})=\tan 45^{o}=1$ $\csc(-135^{o})=-\csc 45^{o}=-\sqrt{2}$ $\sec(-135^{o})=-\sec 45^{o}=-\sqrt{2}$ $\cot(-135^{o})=\cot 45^{o}=1$

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