Double angle formulas

**bobred** · February 18th 2009, 04:02 AM

Hi

I'm having some problems seeing how they've come to an answer in the course book, here we go.

$sin^2x \ cos^2x = \frac{1}{4}\ sin^2(2x)$

and then through several stages, some skipped, we get

$=\frac{1}{8}\ (1-cos(4x))$

Any help would be appreciated.

Thanks

**red_dog** · February 18th 2009, 07:32 AM

We have $\sin 2x=2\sin x\cos x$ and $\cos 4x=1-2\sin^22x$

Now $\sin^2x\cos^2x=(\sin x\cos x)^2=\left(\frac{\sin 2x}{2}\right)^2=\frac{1}{4}\sin^22x=$

$=\frac{1}{8}\cdot 2\sin^22x=\frac{1}{8}(1-1+2\sin^22x)=\frac{1}{8}[1-(1-2\sin^22x)]=\frac{1}{8}(1-\cos 4x)$

**darence** · February 18th 2009, 07:33 AM

look here
Some formulae may be helpfull

**theindian** · February 18th 2009, 09:17 PM

use cos2x formula

**metlx** · February 19th 2009, 06:53 AM

$ cos(4x) = cos(2x + 2x) = .... = 8cos^4 x - 8cos^2 x + 1 $

$ cos^4 x = (1-sin^2 x)^2 = 1 - 2sin^2 x + sin^4 x $
------------------------------------------
$ sin^2 xcos^2 x = \frac{1}{8} (1 - cos(4x)) $

$ sin^2 xcos^2 x = \frac {1}{8} (1 - 8cos^4 x + 8cos^2 x - 1) $

$ sin^2 xcos^2 x = -cos^4 x + cos^2 x $

$ sin^2 xcos^2 x = -1 + 2sin^2x - sin^4 x + (1- sin^2 x) $

$ sin^2 xcos^2 x = sin^2 x - sin^4 x $

$ sin^2 xcos^2 x = sin^2 x (1 - sin^2 x) $

$ cos^2 x = 1 - sin^2 x $

$ cos^2 x = cos^2 x $

sorry for the long explanation :P

EDIT: just realized this wasn't really about proving the equalities.. meh I'll leave it unedited anyways

)

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