
Double angle formulas
Hi
I'm having some problems seeing how they've come to an answer in the course book, here we go.
$\displaystyle sin^2x \ cos^2x = \frac{1}{4}\ sin^2(2x) $
and then through several stages, some skipped, we get
$\displaystyle =\frac{1}{8}\ (1cos(4x)) $
Any help would be appreciated.
Thanks

We have $\displaystyle \sin 2x=2\sin x\cos x$ and $\displaystyle \cos 4x=12\sin^22x$
Now $\displaystyle \sin^2x\cos^2x=(\sin x\cos x)^2=\left(\frac{\sin 2x}{2}\right)^2=\frac{1}{4}\sin^22x=$
$\displaystyle =\frac{1}{8}\cdot 2\sin^22x=\frac{1}{8}(11+2\sin^22x)=\frac{1}{8}[1(12\sin^22x)]=\frac{1}{8}(1\cos 4x)$

look here
Some formulae may be helpfull(Nod)


$\displaystyle
cos(4x) = cos(2x + 2x) = .... = 8cos^4 x  8cos^2 x + 1
$
$\displaystyle
cos^4 x = (1sin^2 x)^2 = 1  2sin^2 x + sin^4 x
$

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