How to derive: $\sin(x-a)\sin(x+a) = \sin^2(x) - \sin^2(a)?$

Question

I was looking at the solution of

$$\int\frac{\sqrt{\sin(x−a)}}{\sqrt{\sin(x+a)}}dx.$$

Where the formula, $$\sin(x-a)\sin(x+a) = \sin^2x - \sin^2a = \cos^2 a - \cos^2 x$$ is used, I want to know how it was derived as I couldn't find it on the net.

Answer 1

$\begin{align}\sin(x-a)\sin(x+a) &=\frac{\cos(2a)-\cos(2x)}2=\frac{1-2\sin^2a-(1-2\sin^2x)}2= \sin^2x - \sin^2a\end{align}$

Answer 2

\begin{align} \sin(x-a)\sin(x+a)&=(\sin x\cos a-\cos x\sin a)(\sin x\cos a+\cos x\sin a) \\ &=\sin^2x\cos^2 a-\cos^2 x\sin^2 a \\ &=\sin^2x(1-\sin^2a)-(1-\sin^2 x)\sin^2 a \\ &=\sin^2 x-\sin^2 a \end{align}