# What is the exact value of tan -330?

May 10, 2015

Assuming you mean degrees, $\tan \left(- 330\right) = \tan \left(30\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

$\sin \left(30\right) = \frac{1}{2}$, $\cos \left(30\right) = \frac{\sqrt{3}}{2}$ and $\tan \left(30\right) = \sin \frac{30}{\cos} \left(30\right)$.

To see this for yourself, consider an equilateral triangle with sides of length 1, then cut it in half to produce 2 right angled triangles.

These smaller triangles will have internal angles of $30$, $60$ and $90$ degrees.

The shortest side is of length $\frac{1}{2}$, so

$\sin \left(30\right) = \frac{\frac{1}{2}}{1} = \frac{1}{2}$.

The hypotenuse is of length 1, so the other side will have length
$\sqrt{{1}^{2} - {\left(\frac{1}{2}\right)}^{2}}$ = $\sqrt{\frac{3}{4}}$ = $\frac{\sqrt{3}}{2}$.