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 $\frac{\sqrt{3}-1}{2\sqrt{2}}$
 A: Let $x := \sin^2\frac{\pi}{12}$. Then your last equality is
$$\frac{1}{4} = 4x(1-x) = 4x - 4x^2.$$
The solutions of this quadratic equation are $x_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$. Clearly, $x$ is smaller than $\frac{1}{2}$, so we must have $x = \frac{1}{2} - \frac{\sqrt{3}}{4}$. Taking the square root gives
$$\sin\frac{\pi}{12} = \frac{\sqrt{2 - \sqrt{3}}}{2}.$$
Now since $(\sqrt{3}-1)^2 = 4 - 2\sqrt{3} = 2(2 - \sqrt{3})$, we get
$$\sin\frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$
as expected.
A: 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.
