## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Review Exercises - Page 249: 34

#### Answer

$sin~B = -\frac{\sqrt{7}}{4}$ $cos~B = \frac{3}{4}$

#### Work Step by Step

If $540^{\circ} \lt 2B \lt 720^{\circ},~~$ then $~~270^{\circ} \lt B \lt 360^{\circ},~~$ so the angle $B$ is in quadrant IV. Since $B$ is in quadrant IV, $sin~B$ is negative. We can find the value of $sin~B$: $cos~2B = 1-2~sin^2~B$ $sin^2~B = \frac{1-cos~2B}{2}$ $sin ~B = -\sqrt{\frac{1-cos~2B}{2}}$ $sin~B = -\sqrt{\frac{1-(\frac{1}{8})}{2}}$ $sin~B = -\sqrt{\frac{(\frac{7}{8})}{2}}$ $sin~B = -\sqrt{\frac{7}{16}}$ $sin~B = -\frac{\sqrt{7}}{4}$ Since $B$ is in quadrant IV, $cos~B$ is positive. We can find the value of $cos~B$: $cos~2B = 2~cos^2~B-1$ $cos^2~B = \frac{1+cos~2B}{2}$ $cos~B = \sqrt{\frac{1+cos~2B}{2}}$ $cos~B = \sqrt{\frac{1+(\frac{1}{8})}{2}}$ $cos~B = \sqrt{\frac{(\frac{9}{8})}{2}}$ $cos~B = \sqrt{\frac{9}{16}}$ $cos~B = \frac{3}{4}$

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