simplifying using half angle identity

**satis** · March 10th 2010, 02:11 PM

I think I got this right, but may be missing some steps or may be wrong entirely. The idea is to use the double angle identity to simplify

$sin4x$

$= 2sin2xcos2x$
$= 2(2sinxcosx)(1-2sin^2x)$
$= 4sinxcosx(1-2sin^2x)$
$= 4sinxcosx-8sin^3xcosx$

Is this right?

**bigwave** · March 10th 2010, 03:49 PM

Originally Posted by satis

I think I got this right, but may be missing some steps or may be wrong entirely. The idea is to use the half angle identity to simplify

$sin4x$

$= 2sin2xcos2x$
$= 2(2sinxcosx)(1-2sin^2x)$
$= 4sinxcosx(1-2sin^2x)$
$= 4sinxcosx-8sin^3xcosx$

Is this right?

not sure how you can really simplify just $\sin{4x}$

also, the half angle formula is:

$\sin\left({\frac{\theta}{2}}\right) = \pm\sqrt{\frac{1-\cos{\theta}}{2}}$

**satis** · March 10th 2010, 04:52 PM

that would be my mistake. I meant simplifying using the double angle formula. My apologies. Let's see if I can alter the thread title...yup!

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