Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Review Exercises - Page 94: 16


$\displaystyle \sin(-225^{\mathrm{o}})= =\frac{\sqrt{2}}{2}$ $\displaystyle \cos(-225^{\mathrm{o}})= -\frac{\sqrt{2}}{2}$ $\tan(-225^{\mathrm{o}})= -1$ $\cot(-225^{\mathrm{o}})= -1$ $\sec(-225^{\mathrm{o}})= -\sqrt{2}$ $\csc(-225^{\mathrm{o}}) =\sqrt{2}$

Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quadr.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ $-225^{0}$ is coterminal with $-225^{0}+360^{0}=135^{\mathrm{o}}$. $135^{\mathrm{o}}$ lies in quadrant II. The reference angle is$ 180^{o}-135^{\mathrm{o}}=45^{\mathrm{o}}$. Since $-225^{\mathrm{o}}$ 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 $45^{\mathrm{o}}$: $\displaystyle \sin(-225^{\mathrm{o}})=\sin 45^{\mathrm{o}}=\frac{\sqrt{2}}{2}$ $\displaystyle \cos(-225^{\mathrm{o}})=-\cos 45^{\mathrm{o}}=-\frac{\sqrt{2}}{2}$ $\tan(-225^{\mathrm{o}})=-\tan 45^{\mathrm{o}}=-1$ $\cot(-225^{\mathrm{o}})=-\cot 45^{\mathrm{o}}=-1$ $\sec(-225^{\mathrm{o}})=-\sec 45^{\mathrm{o}}=-\sqrt{2}$ $\csc(-225^{\mathrm{o}})=\csc 45^{\mathrm{o}}=\sqrt{2}$
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