Answer
$$\sin165^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
Work Step by Step
$$\sin165^\circ$$
We would write $165^\circ$ as the difference of $180^\circ$ and $15^\circ$, as the angle $15^\circ$ can be applied the half-angle identities.
$$\sin165^\circ=\sin(180^\circ-15^\circ)$$
- Difference Identity for sine: $\sin(A+B)=\sin A\cos B+\cos A\sin B$ with $A=180^\circ$ and $B=15^\circ$
$$\sin165^\circ=\sin180^\circ\cos15^\circ-\cos180^\circ\sin15^\circ$$
$\cos180^\circ=-1$ and $\sin180^\circ=0$
$$\sin165^\circ=0\times\cos15^\circ-(-1)\times\sin15^\circ$$
$$\sin165^\circ=\sin15^\circ$$
*Calculate $\sin15^\circ$
$$\sin15^\circ=\sin\Big(\frac{30^\circ}{2}\Big)$$
Apply the identity $\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$ with $A=30^\circ$.
$$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$
$$\sin15^\circ=\pm\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}$$
$$\sin15^\circ=\pm\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}$$
$$\sin15^\circ=\pm\sqrt{\frac{2-\sqrt3}{4}}$$
$$\sin15^\circ=\frac{\pm\sqrt{2-\sqrt3}}{2}$$
As $15^\circ$ is in quadrant I, where $\sin\theta\gt0$, so $\sin15^\circ\gt0$, so we would select the positive square root.
$$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
Therefore, $$\sin165^\circ=\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
