Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\sin^26x-\sin^24x=\sin2x\sin10x$$ Hi, Please help me to solve the above trigonometric function as i am trying to solve this question since last an hour.

share|improve this question
    
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
4  
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
show 1 more comment

1 Answer

$\sin^26x-\sin^24x=\sin(6x+4x)\sin(6x-4x)=\sin2x\sin10x$ applying $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$

Alternatively,

$\sin^26x-\sin^24x=\frac{1}{2}(2\sin^26x-2\sin^24x)$ $=\frac{1}{2}(1-\cos12x-(1-\cos8x))$ as $\cos2A=1-2sin^2A$

So, $\sin^26x-\sin^24x=\frac{\cos8x-\cos12x}{2}=\sin10x\sin2x$ applying $\cos C - \cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.