The purpose of this post is to show the interesting identity

\[ \sin(x)^2-\sin(y)^2 = \sin(x+y)\sin(x-y) . \]

In fact, this is a simple consequence of the sine addition identity

\[ \sin(x\pm y) = \sin x\cos y\pm\cos x\sin y \]

and the fundamental Pythagorean identity $\sin(\theta)^2+\cos(\theta)^2=1$.

The demonstration is a pretty straightforward usage of the above identities, but involves a little bit of trickery—on the third step, we add and subtract the quantity $\sin(x)^2\sin(y)^2$:

\begin{align*}
\sin(x+y)\sin(x-y) &= (\sin x\cos y+\cos x\sin y)(\sin x\cos y-\cos x\sin y) \\
&= \sin(x)^2\cos(y)^2-\cos(x)^2\sin(y)^2 \\
&\qquad+\cos x\sin y\sin x\cos y-\cos x\sin y\sin x\cos y \\
&= \sin(x)^2\cos(y)^2-\cos(x)^2\sin(y)^2 \\
&\qquad+\sin(x)^2\sin(y)^2-\sin(x)^2\sin(y)^2 \\
&= \sin(x)^2(\cos(y)^2+\sin(y)^2) \\
&\qquad-\sin(y)^2(\cos(x)^2+\sin(x)^2) \\
&= \sin(x)^2-\sin(y)^2
\end{align*}

This was not an identity I knew off the top of my head, but it came up on a problem I was working on. We’ll have reason to use it in a later post, which is why I wanted to single it out right now.