## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Test - Page 647: 12

#### Answer

$\sqrt 3$

#### Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta$ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ Thus, the reference angle of $\dfrac{ 7 \pi}{3}-\dfrac{6\pi}{3}=\dfrac{\pi}{3}$ So, $\tan \dfrac{ \pi}{3}=\sqrt 3$ Adding $n\pi$ to the angle will not change the value of the tangent function, since it repeats every $\pi$ or $180^{\circ}$.

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