# How do you prove sin x = 1 - 2 cos^2x?

For $x = \frac{\pi}{6}$:
$\sin \left(x\right) = \frac{1}{2} , \cos \left(x\right) = \frac{\sqrt{3}}{2}$
$1 - 2 \cdot {\left(\frac{\sqrt{3}}{2}\right)}^{2} = 1 - 2 \cdot \frac{3}{4} = 1 - \frac{3}{2} = - \frac{1}{2} \ne \frac{1}{2}$//