# 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

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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 $$\cos2A=2\cos^2A-1$$