Find the exact value of: (a) $\cos \dfrac{17\pi }{6}$

(b) $\tan\left( -\dfrac{\pi }{3}\right)$

Solution  (a) Refer to Figure 58(a) on page A-31. The angle $\dfrac{17\pi }{6}$ is in quadrant II, where the cosine function is negative. The reference angle for $\dfrac{17\pi }{6}$ is $\dfrac{\pi }{6}.$ Since $\cos \dfrac{\pi }{6}=\dfrac{\displaystyle\sqrt{3}}{2}$, $$\cos \dfrac{17\pi }{6}=-\cos \dfrac{\pi }{6}=-\dfrac{\displaystyle\sqrt{3}}{2}$$

(b) Refer to Figure 58(b). The angle $-\dfrac{\pi }{3}$ is in quadrant IV, where the tangent function is negative. The reference angle for $-\dfrac{\pi }{3}$ is $\dfrac{\pi }{3}.$ Since $\tan \dfrac{\pi }{3}=\displaystyle\sqrt{3}$, $$\tan \left( -\dfrac{\pi }{3}\right) =-\tan \dfrac{\pi }{3}=-\displaystyle\sqrt{3}$$

A-31

Figure 58