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.
 A: 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  $2\sin A \sin B = \cos (A-B) - \cos (A+B)$.
$$E=\sin^2(x+\alpha)-\frac{1}{2}\left[\cos(2\alpha-2\beta)-\cos(2x+2\alpha)\right]$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\color{green}{\cos(2x+2\alpha)}$$
$$E=\sin^2(x+\alpha)-\frac{1}{2}\cos(2\alpha-2\beta)+\frac{1}{2}\left[1-2\sin^2(x+\alpha)\right]$$
$$E=\frac{1}{2}-\frac{1}{2}\cos(2\alpha-2\beta)$$
A: HINT
Use the formulas $\sin^2t=\frac {1-\cos2t} 2$  and $\cos2p+\cos2q=2\cos(p+q)\cos(p-q)$.
A: $$F=\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$
$$=1-\{\underbrace{\cos^2(x+\alpha)-\sin^2(x+\beta)}\}-\cos(\alpha-\beta)\{\underbrace{2\sin(x+\alpha)\sin(x+\beta)}\}$$
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 $x$
A: $$\text {Let }\quad L=\sin^2 (x+a)+\sin^2(x+b).$$ $$\text {Let }\quad R=-2\cos (a-b)\sin (x+a) \sin (x+b).$$ $$\text {We have }\quad  L=\frac {1}{2}(1-\cos (2x+2a) +\frac {1}{2}(1-\cos (2x+2b).$$ Now with $y=2x+a+b$ and $z=a-b$ we have $$\cos (2x+2a)=\cos (y+z)=\cos y \cos z-\sin y \sin z. $$ $$\cos (2x+2b)=\cos (y-z)=\cos y \cos z +\sin y \sin z .$$  $$\text {Thus, }\quad\cos (2x+2a) + \cos (2x+2b )=2 \cos y \cos z.$$   $$\text {Therefore  }\quad L=1-\cos y \cos z.$$  The motivation for this comes from examining R : We always have $\sin U \sin V= \frac {1}{2}(\cos (U-V)-\cos (U+V)). $ So we have  $$\sin (x+a) \sin (x+b)=\frac {1}{2}(\cos (a-b)-\cos (2x+a+b))=\frac {1}{2}(\cos z-\cos y).$$ $$\text {Therefore }\quad R= (-2\cos z)\cdot \frac {1}{2}(\cos z-\cos y)=-\cos^2 z+\cos z \cos y.$$  $$\text {In conclusion,  } \quad L+R=1-\cos^2z=1-\cos^2 (a-b)$$ which is independent of $x.$
A: 
Taking @SiongthyeGoh's comment, in the above picture, $AE=1$, $\angle BAE=x$, $\angle BAC=\beta$, and $\angle BAD=\alpha$.
It follows that $DE=\sin(x+\alpha)$ and $CE=\sin(x+\beta)$. Applying the cosine theorem to $\triangle DCE$, the given expression is equal to $DC^2$, hence is a constant.
