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Help with higher maths trigonometric identities.

I have a few questions to ask about trigonometric identies, I have tried these questions several times and I can't get any of them for the life of me. Could someone please help me as my exam is on Thursday and I want to practice hundreds of questions.

The questions are in this image : http://img29.imageshack.us/img29/6159/mathsquestions.png

I have also uploaded them as an attachment as well.

Thanks in advanced.
Use the fact that sin2x=2sinxcosx\sin 2x = 2\sin x\cos x
and cos2x=2cos2x1\cos 2x = 2 \cos ^2 x - 1

for the first identity.

By simplifying you should get 2sinx2cosx\frac{2\sin x}{2\cos x}, and it should be relativley simple form here to show that this equals tanx.
Reply 2
I nearly had that one, for some reason I was thinking I couldn't simplify 2sinxcosx2cos2x\frac{2sinxcosx}{2cos^2x}. I then realised I just had to divide throughout by cosx to get 2sinx2cosx\frac{2sinx}{2cosx} which simplifies to become tanx.

Thanks for your help on that one, does anyone have an idea what to do for the last one.
again you can use the expansion cos2x=2cos2x1.\cos 2x = 2\cos ^2 x - 1. You should be able to fairly simply get the LHS to equal 2(1cos2x)2cos2x\frac{2(1 - \cos ^2 x)}{2\cos ^2 x}

To simplify the numerator you need to use the rule sin2x+cos2x=1\sin ^2 x + \cos ^2 x = 1 and from here it should be fairly simple to prove that this equals tanx\tan x