# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 58

$\dfrac{\pi}{6}$

#### 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: $(2\pi - \theta)$ We can see that the angle $\dfrac{17\pi}{6}$ is close to $\dfrac{18\pi}{6}=3\pi$. Thus, the reference angle of $\dfrac{17\pi}{6}$ is: $= 3\pi - \dfrac{17\pi}{6}$ $\implies =\dfrac{18\pi}{6} - \dfrac{17\pi}{6}$ $=\dfrac{\pi}{6}$

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