TITLE: APICS Mathematics Contest 1999: Prove $\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$ is a constant function of $x$
QUESTION [12 upvotes]: This is question 3 from the APICS Mathematics Competition paper of 1999:

Prove that $$\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ is a constant function of $x$.

Expanding it seems rather daunting, in particular the last term, and nothing I've tried has been useful towards cancelling terms out.
It was assigned in a pre-calculus course, so it should be possible to solve without using derivatives. However, showing that $f'(x)=0$ would obviously be a valid solution.
Any ideas are greatly welcome.

REPLY [4 votes]: $$F=\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$
$$=1-\{\underbrace{\cos^2(x+\alpha)-\sin^2(x+\beta)}\}-\cos(\alpha-\beta)\{\underbrace{2\sin(x+\alpha)\sin(x+\beta)}\}$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and Werner Formula$(2\sin A\sin B=\cdots),$
$$D=1-\cos(2x+\alpha+\beta)\cos(\alpha-\beta)-\cos(\alpha-\beta)[\cos(\alpha-\beta)-\cos(2x+\alpha+\beta)]$$
$$=\cdots=1-\cos^2(\alpha-\beta)$$ which is independent of $x$