# How do you evaluate sin((-17pi)/3)?

$\sin \left(- 17 \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3} - 3 \left(2 \pi\right)\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
How do I know that $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$?
Picture an equilateral triangle with sides of length 1. Now cut the triangle into two equal pieces with a line segment running from one vertex to the middle of the opposite side. This will split the triangle into two right-angled triangles with angles 30, 60 and 90 degrees i.e. $\frac{\pi}{6}$, $\frac{\pi}{3}$ and $\frac{\pi}{2}$ radians. The shortest side of one of these right-angled triangles has length $\frac{1}{2}$. Using Pythagoras theorem, the other side adjoining the right angle must have length:
$\sqrt{{1}^{2} - {\left(\frac{1}{2}\right)}^{2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
Divide this by the length of the hypotenuse (1) to get the sine of the opposite angle (i.e. $\sin \left(\frac{\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}$).