Evalute this

dejalisco5 · May 14, 2008, 01:38 PM

Arctan(-squareroot of3)

galactus · May 14, 2008, 05:13 PM

The unit circle is a handy thing to have on hand for these.

Take a look at $tan(\frac{2\pi}{3}), \;\ tan(\frac{5\pi}{3})$

Notice the coordinates on the circle. For instance, look at 120 degrees or $\frac{2\pi}{3}$

See the $(-\frac{1}{2},\frac{\sqrt{3}}{2})?.$

They come from the triangle since $x=cos({\theta}), \;\ y=sin({\theta})$

Therefore, $tan^{-1}(\frac{-\sqrt{3}}{1})=\frac{2\pi}{3}=\frac{-\pi}{3}$

Same for $\frac{5\pi}{3}=\frac{-\pi}{3}$

Hence, $\fbox{tan(\frac{(3n-1){\pi}}{3})=-\sqrt{3}}$