## Trigonometry (11th Edition) Clone

$-\sqrt3$
Convert the angle measure to degrees to obtain: $=\frac{5\pi}{3} \cdot \frac{180^o}{\pi} = 5(60^o)=300^0$ Thus, $\tan{(\frac{5\pi}{3})} \\= \tan{300^o}$ 300 degrees is in Quadrant IV so : (i) its reference angle is $360^o-300^o=60^o$ (ii) its tangent is negative From Section 2.1 (page 50) , we learned that: $\tan{60^o} = \sqrt3$ This means that: $\tan{300^o} = -\tan{60^o} = -\sqrt3$.