write down the expression for $\sin (15°)$ using the double angle formula. show that $\sin 15^\circ=\frac {\sqrt3 -1}{2\sqrt2}$ using $\cos2A=1-2\sin^2A$
However I got $\sin 15^\circ= \sqrt{\frac {2-\sqrt 3}{4}}$ instead.

Your form, $\sqrt{\frac{2-\sqrt3}{4}}$, is already close:
$$\begin{align*}
\sin15^\circ &=\sqrt{\frac{2-\sqrt3}{4}}\\
&= \sqrt{\frac{4-2\sqrt3}{8}}\\
&= \sqrt{\frac{1-2\sqrt3+3}{8}}\\
&= \sqrt{\frac{(1-\sqrt3)^2}{8}}\\
&= \frac{\sqrt3 -1 }{2\sqrt2}
\end{align*}$$
