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 $x$ 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.

Let 
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$ 
Now use  $2\cos A \sin B = \sin (A+B) - \sin (A-B)$.
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(-\beta -x\right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)+\sin\left(x+\beta \right)\right]\sin(x+\beta)$$
$$E=\sin^2(x+\alpha)-\color {red}{\sin\left(\alpha-\beta +x+\alpha\right)\sin(x+\beta)}$$
Now use  $2\sin A \sin B = \cos (A-B) - \cos (A+B)$.
$$E=\sin^2(x+\alpha)-\frac{1}{2}\left[\cos(2\alpha-2\beta)-\cos(2x+2\alpha)\right]$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\color{green}{\cos(2x+2\alpha)}$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\left[1-2\sin^2(x+\alpha)\right]$$
$$E=\frac{1}{2}-\frac{1}{2}\cos(2\alpha-2\beta)$$
