Answer
$$\sin165^\circ=\frac{\sqrt6-\sqrt2}{4}$$
Work Step by Step
$$X=\sin165^\circ$$
There are several ways to rewrite $165^\circ$. Here, it would be rewritten as the sum of $135^\circ$ and $30^\circ$
$$X=\sin(135^\circ+30^\circ)$$
The identity of the sum of sines states that
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
Therefore, $$X=\sin135^\circ\cos30^\circ+\cos135^\circ\sin30^\circ$$
*About $\sin135^\circ$ and $\cos135^\circ$
We can rewrite $135^\circ$ as the sum of $90^\circ$ and $45^\circ$ and apply the identities of sines sum and cosines sum.
$$\sin135^\circ=\sin(90^\circ+45^\circ)=\sin90^\circ\cos45^\circ+\cos90^\circ\sin45^\circ$$
$$\sin135^\circ=1\times\frac{\sqrt2}{2}+0\times\sin45^\circ=\frac{\sqrt2}{2}$$
$$\cos135^\circ=\cos(90^\circ+45^\circ)=\cos90^\circ\cos45^\circ-\sin90^\circ\sin45^\circ$$
$$\cos135^\circ=0\times\cos45^\circ-1\times\frac{\sqrt2}{2}=-\frac{\sqrt2}{2}$$
Apply them back to $X$:
$$X=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\Big(-\frac{\sqrt2}{2}\Big)\frac{1}{2}$$
$$X=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$$
$$X=\frac{\sqrt6-\sqrt2}{4}$$
In conclusion, $$\sin165^\circ=\frac{\sqrt6-\sqrt2}{4}$$
