Trigonometry (11th Edition) Clone

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

Chapter 2 - Test - Page 97: 8


$135^{o},\quad 225^{o}$

Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quad.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ Cosine is negative in quadrants II and III. Browsing through Function Values of Special Angles, we find that $\displaystyle \cos 45^{o}=\frac{\sqrt{2}}{2}$, so the reference angle is $45^{o}$ In quadrant II, $\theta^{\prime}=180^{o}-\theta$ so $\theta=180^{o}-\theta^{\prime}=180^{o}-45^{o}=135^{o}$ In quadrant III, $\theta^{\prime}=\theta-180^{o}$ so $\theta=180^{o}+\theta^{\prime}=180^{o}+45^{o}=225^{o}$
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