<s>[INST] Deriving exact value of $\sin \pi/12$ using double angle identity 
Deriving the exact value of $\sin \pi/12$ using double angle identity

Double angle identity - $\sin 2A = 2\sin A \cos A$
So, $\sin (2 \frac{\pi}{12}) = 2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}  $
$\sin^2 A + \cos^2 A = 1 $ so, $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin (2 \frac{\pi}{12}) =2 \sin \frac{\pi}{12}\sqrt{1- \sin^2 \frac{\pi}{12}} $
I square both sides of the equation to get:
$(1/2)^2 = 4 \sin^2 \frac{\pi}{12} (1- \sin^2 \frac{\pi}{12})$
With this, I am moving away from finding the exact value of $\sin \pi/12$ which is $\frac{\sqrt{3}-1}{2\sqrt{2}}$
 [/INST] A double identity which gets you there more directly (perhaps not the one you are using) is $$\cos2x=1-2\sin^2x.$$  When $x=\pi/12$, we have \begin{align}\cos\frac\pi6&=1-2\sin^2\frac\pi{12}\\
2\sin^2\frac\pi{12}&=1-\frac{\sqrt3}2\\
\sin^2\frac\pi{12}&=\frac12-\frac{\sqrt3}4=\frac{2-\sqrt3}4\\
\sin\frac\pi{12}&=\pm\sqrt{\frac{2-\sqrt3}4}=\pm\frac{\sqrt{2-\sqrt3}}2=\pm\frac{\sqrt3-1}{2\sqrt2}.
\end{align}
Since $\frac\pi{12}\in(0,\frac\pi2)$, we can reject the negative answer, leaving us with $$\sin\frac\pi{12}=\frac{\sqrt3-1}{2\sqrt2}$$ as desired.
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