Answer
$$\sin195^\circ=-\frac{\sqrt{2-\sqrt 3}}{2}$$
Work Step by Step
$$A=\sin195^\circ$$
We can separate $195^\circ$ into $180^\circ$ and $15^\circ$. Therefore, we can rewrite A as follows $$A=\sin(180^\circ+15^\circ)$$ $$A=\sin180^\circ\cos15^\circ+\cos180^\circ\sin15^\circ$$ $$A=0\times\cos15^\circ+(-1)\sin15^\circ$$ $$A=-\sin15^\circ$$
Since $15^\circ$ is half of $30^\circ$, we can rewrite $$A=-\sin\Bigg(\frac{1}{2}30^\circ\Bigg)$$ $$A=-\Bigg[\pm\sqrt{\frac{1-\cos30^\circ}{2}}\Bigg]$$
However, since $15^\circ$ is in quadrant I, in which $\sin X\gt0$, so $$A=-\Bigg[\sqrt{\frac{1-\cos30^\circ}{2}}\Bigg]$$ $$A=-\sqrt{\frac{1-\frac{\sqrt 3}{2}}{2}}$$ $$A=-\sqrt\frac{\frac{2-\sqrt 3}{2}}{2}$$ $$A=-\sqrt{\frac{2-\sqrt 3}{4}}$$ $$A=-\frac{\sqrt{2-\sqrt 3}}{2}$$
