When I was in high school, I had proved that $$\sin^2(x)-\sin^2(y)=\sin(x-y)\sin(x+y) $$
I think it is beautiful since it resembles the identity $a^2-b^2=(a+b)(a-b)$.
But I can not find it in sources,I think it should be known.(of course,it is not something remarkable just beatuful)
If anyone finds it in somewhere,I would be thankful also.
Here is the proof:
$$\sin(x+y)\sin(x-y)=(\sin(x)\cos(y)+\cos(x)\sin(y))(\sin(x)\cos(y)-\cos(x)\sin(y))$$ $$=\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)$$ $$=\sin^2(x)\cos^2(y)-\sin^2(y)(1-\sin^2(x))$$ $$=\sin^2(x)\cos^2(y)-\sin^2(y)+\sin^2(x)sin^2(y)$$ $$=\sin^2(x)(\cos^2(y)+\sin^2(y))-\cos^2(y)) $$ $$=\sin^2(x)-\sin^2(y)$$ If it is already known, it will not be surprise for me. I hope you would like this.
This identity appears in my CRC Standard Mathematical Tables in the Trigonometry chapter under the heading Angle-sum and angle-difference relations. Actually several similar ones appear:
$$\sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta = \cos^2 \beta - \cos^2 \alpha$$ $$\cos(\alpha + \beta) \cos(\alpha - \beta) = \cos^2 \alpha - \sin^2 \beta = \cos^2 \beta - \sin^2 \alpha$$