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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, user1337 Oct 9 '13 at 19:53

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1 Answer 1

up vote 1 down vote accepted

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

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