## Trigonometry (10th Edition)

Published by Pearson

# Chapter 3 - Review Exercises - Page 129: 33

#### Answer

$\sqrt3$

#### Work Step by Step

RECALL: $\tan{x} = \dfrac{\sin{x}}{\cos{x}}$ The angle $\frac{\pi}{3}$ is a special angle and the following values are known: $\sin{\frac{\pi}{3}}=\frac{\sqrt3}{2}$ $\cos{\frac{\pi}{3}}=\frac{1}{2}$ Thus, $\tan{\frac{\pi}{3}} = \dfrac{\sin{\frac{\pi}{3}}}{\cos{\frac{\pi}{3}}}=\dfrac{\frac{\sqrt3}{2}}{\frac{1}{2}}=\frac{\sqrt3}{2} \cdot \frac{2}{1}=\sqrt3$

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