## Trigonometry (10th Edition)

$\cot{(\frac{2\pi}{3})}=-\frac{\sqrt3}{3}$
Convert $\dfrac{2\pi}{3}$ to degrees to obtain: $\require{cancel} \\=\dfrac{2\pi}{3} \times \dfrac{180^o}{\pi} \\=\dfrac{2\cancel{\pi}}{\cancel{3}} \times \dfrac{\cancel{180^o}^{60^o}}{\cancel{\pi}} \\=120^o$ The reference angle of $120^o$ is $60^o$. Note that: $\sin{60^o}= \frac{\sqrt{3}}{2} \\\cos{60^o}=\frac{1}{2}$ However, since $\dfrac{2\pi}{3} = 120^o$ and this angle terminates in Quadrant II, then $\sin{120^o}= \frac{\sqrt{3}}{2} \\\cos{120^o}=-\frac{1}{2}$ RECALL: $\cot {\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$ Therefore, $\cot{(\frac{2\pi}{3})} \\=\cot{120^o} \\=\dfrac{\cos{120^o}}{\sin{120^o}} \\=\dfrac{-\frac{1}{2}}{\frac{\sqrt3}{2}} \\=-\frac{1}{2} \times \frac{2}{\sqrt3} \\=-\frac{1}{\sqrt3} \\=-\frac{1}{\sqrt3} \times \frac{\sqrt3}{\sqrt3} \\=-\frac{\sqrt3}{3}$