## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Review Exercises - Page 249: 38

#### Answer

$sin~\frac{A}{2} = \sqrt{\frac{7}{8}}$

#### Work Step by Step

If $90^{\circ} \lt A \lt 180^{\circ}$, then the angle $A$ is in quadrant II. Then $45^{\circ} \lt \frac{A}{2} \lt 90^{\circ}$, so the angle $\frac{A}{2}$ is in quadrant I. We can find the value of $sin~\frac{A}{2}$: $sin~\frac{A}{2} = \sqrt{\frac{1-cos~A}{2}}$ $sin~\frac{A}{2} = \sqrt{\frac{1-(-\frac{3}{4})}{2}}$ $sin~\frac{A}{2} = \sqrt{\frac{(\frac{7}{4})}{2}}$ $sin~\frac{A}{2} = \sqrt{\frac{7}{8}}$

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