Question

Prove this identity using the product-to-sum identity for sine:
$sin^2x=\frac{1-cos(2x)}{2}$
umm pls

Answer 1

It follows from

$\cos2x=\cos(x+x)=\cos x\cos x-\sin x\sin x=\cos^2x-\sin^2x=1-2\sin^2x$

$\implies\cos 2x-1=-2\sin^2x$

$\implies\sin^2x=\dfrac{1-\cos2x}2$