# Trigonometry Practise [closed]

Prove: $\sin^2 3α - \sin^2α = \sin4α\sin2α$

Also, simplify: $\sqrt{\dfrac {1+\cos6α} {2}}$

Any help would be appreciated.

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## closed as off-topic by Thomas, martini, Trevor Wilson, azimut, user1337Oct 9 '13 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, martini, Trevor Wilson, azimut, user1337
If this question can be reworded to fit the rules in the help center, please edit the question.

For the first use $$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$$ (Proof)

For the second use $$\cos2A=2\cos^2A-1$$

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You were right...now its fixed. –  Bak1139 Oct 9 '13 at 18:49
@Bak1139, so the answer is adjusted:) –  lab bhattacharjee Oct 9 '13 at 18:51
many Thanks mate –  Bak1139 Oct 10 '13 at 14:14
@Bak1139, my pleasure –  lab bhattacharjee Oct 10 '13 at 16:43