# simplifying using half angle identity

• Mar 10th 2010, 02:11 PM
satis
simplifying using double angle identity
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

$\displaystyle sin4x$

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

Is this right?
• Mar 10th 2010, 03:49 PM
bigwave
Quote:

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

$\displaystyle sin4x$

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

Is this right?

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

also, the half angle formula is:

$\displaystyle \sin\left({\frac{\theta}{2}}\right) = \pm\sqrt{\frac{1-\cos{\theta}}{2}}$
• Mar 10th 2010, 04:52 PM
satis
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! :)