I need a hint to solve prove $\sin^2(6x)-\sin^2(4x) = \sin(2x)\sin(10x)$
I tried several solutions, including taking $(\sin(6x)+\sin(4x))(\sin(6x)-\sin(4x))$ but every time I ended up with a ridiculously large equation, I'm sure there is a more elegant way to solve this.
Note: I'm looking to go from the L.H.S to the R.H.S.
By the sum to product formulas, $$ \sin6x+\sin4x=2\sin5x\cos x, \qquad \sin6x-\sin4x=2\sin x\cos5x, $$ so the left-hand side becomes $$ 4\sin5x\cos x\sin x\cos5x $$ Can you finish?
Notice, $$\sin^2(6x)-\sin^{2}(4x)=\sin(2x)\sin(10x)$$ $$(\sin(6x)-\sin(4x))(\sin(6x)+\sin(4x))=\sin(2x)\sin(10x)$$ $$(2\cos(5x)\cos(x))(2\sin(5x)\cos(x))=\sin(2x)\sin(10x)$$ $$2(2\sin(5x)\cos(5x))\cos^2(x)=\sin(2x)\sin(10x)$$ $$2\sin(10x)\cos^2(x)=\sin(2x)\sin(10x)$$ $$\sin(10x)(2\cos^2(x)-\sin(2x))=0$$ $$\sin(10x)(2\cos^2(x)-2\sin(x)\cos(x))=0$$ $$2\sin(10x)\cos(x)(\cos(x)-\sin(x))=0$$
Using $z=e^{i2x}$ and the definition of the sine and De Moivre's formula, rewrite the equation as
$$(z^3-z^{-3})^2-(z^2-z^{-2})^2=(z-z^{-1})(z^5-z^{-5}).$$ (The denominators $(2i)^2$ simplify.)
It is an easy matter to check that both members are
$$z^6-z^4-z^{-4}+z^{-6}.$$