# Prove tha $\sin^26x-\sin^24x=\sin2x\sin10x$

$$\sin^26x-\sin^24x=\sin2x\sin10x$$

Please help me to solve the above trigonometric function as I am trying to solve this question since last an hour.

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$\sin^26x-\sin^24x=\sin(6x+4x)\sin(6x-4x)=\sin2x\sin10x$ applying $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$
$\sin^26x-\sin^24x=\frac{1}{2}(2\sin^26x-2\sin^24x)$ $=\frac{1}{2}(1-\cos12x-(1-\cos8x))$ as $\cos2A=1-2sin^2A$
So, $\sin^26x-\sin^24x=\frac{\cos8x-\cos12x}{2}=\sin10x\sin2x$ applying $\cos C - \cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$