Answer
$-\frac{\sqrt{2-\sqrt{3}}}{2}$
Work Step by Step
Use the half-angle formula, $\sin\frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Note that $195^\circ$ is in Quadrant III, where sine is negative, so we take the positive square root.
$\sin 195^\circ$
$=\sin \frac{390^\circ}{2}$
$=-\sqrt{\frac{1-\cos 390^\circ}{2}}$
$=-\sqrt{\frac{1-\cos 30^\circ}{2}}$
$=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$
$=-\sqrt{\frac{(1-\frac{\sqrt{3}}{2})*2}{2*2}}$
$=-\sqrt{\frac{2-\sqrt{3}}{4}}$
$=-\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}$
$=-\frac{\sqrt{2-\sqrt{3}}}{2}$
