## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Review Exercises - Page 291: 29

#### Answer

$\frac{\sqrt7}{4}$

#### Work Step by Step

RECALL: (1) $\theta = \arccos{x} \longrightarrow \cos{\theta}=x$ (2) $\sin^2{\theta} + \cos^2{\theta} = 1$ Let $\theta = \arccos{(\frac{3}{4})}$ This means that $\cos{\theta} = \frac{3}{4}$. Since arccosine is defined in Quadrant I and II only, then the angle is in Quadrant I since cosine is positive. Use the (2) in the recall part above to obtain: $\sin^2{\theta} + \cos^2{\theta}=1 \\\sin^2{\theta} + \left(\frac{3}{4}\right)^2=1 \\\sin^2{\theta}+\frac{9}{16}=1 \\\sin^2{\theta}=1-\frac{9}{16} \\\sin^2{\theta}=\frac{7}{16} \\\sin{\theta} = \pm \sqrt{\frac{7}{16}} \\\sin{\theta} = \pm \frac{\sqrt{7}}{4}$ Since the angle is in Quadrant I, then sine must be positive. Thus, $\sin{\theta} = \frac{\sqrt7}{4}$. Therefore, $\sin{(\arccos{(\frac{3}{4}})}=\frac{\sqrt7}{4}$

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