A geometry problem by Syed Hamza Khalid

Let $\sin(22.5^o)=y$

We use two trigonometric identities:

$\sin(2x)=2 \sin(x)\cos(x)$

$(\sin(x))^2+(\cos(x))^2=1$

$\dfrac{1}{\sqrt{2}} = 2 y \sqrt{1-y^2} \Rightarrow y=\dfrac{\sqrt{2-\sqrt{2}}}{2} \Rightarrow \sin(22.5^o)=\dfrac{\sqrt{2-\sqrt{2}}}{2} \approx 0.3826834323651$

Note: We ignore the other solution of $y=\dfrac{\sqrt{2+\sqrt{2}}}{2}$ as it is a value of $\cos(22.5^o)$ not $\sin$ .