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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

share|cite|improve this question

$\sin^26x-\sin^24x=\sin(6x+4x)\sin(6x-4x)=\sin2x\sin10x$ 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 $\cos C - \cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$

share|cite|improve this answer

Your Answer


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.