<s>[INST] 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.
 [/INST] $cos(2A)$ = $1 - 2Sin^2(A)$
$\Rightarrow$ $cos(30)$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ - 1 = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{2}$   = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{4}$   = $-Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt{2 - \sqrt3}}{2}$   = $Sin(15)$
$\Rightarrow$ $\sqrt{\frac{4 - 2\sqrt3}{8}}$   = $Sin(15)$
$\Rightarrow$ $\sqrt{\frac{(\sqrt3 - 1)^2}{8}}$   = $Sin(15)$
$\Rightarrow$ $\frac{(\sqrt3 - 1)}{2\sqrt2}$   = $Sin(15)$
$\Rightarrow$ $\frac{(\sqrt3 - 1)}{2\sqrt2}$   = $Sin(15)$
</s>