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

$$\sin^26x-\sin^24x=\sin2x\sin10x$$ Hi, 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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I tried to edit your question but the last term on the right isn't clear: did you mean $\,\sin^{10}x\,$ or $\,\sin(10x)\,$? Besides this, perhaps you should do something about your accept rate... –  DonAntonio Aug 29 '12 at 8:14
Its sin^2(6x)-sin^2(4x)=sin2xsin10x –  Drownpc Aug 29 '12 at 8:23
Ok. Please (1) Try to use LaTeX when writing maths in this site, (2) Try to show a little more effort, what ideas you have ,background, etc. You've asked quite a few questions in the last hours, and (3) Try to improve your accept rate in order to have mroe people interested in helping you out. –  DonAntonio Aug 29 '12 at 8:25
Thank you for your advice. –  Drownpc Aug 29 '12 at 8:27
... and (4) don't say "solve the trigonometric function". You solve equations, prove identitites, etc. You don't solve functions. –  Hans Lundmark Aug 29 '12 at 9:08
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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 $\cos C - \cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$