# Thread: [SOLVED] prove equality using double angle identity

1. ## [SOLVED] prove equality using double angle identity

The equation is
$\displaystyle 8sin^2xcos^2X = 1-cos4x$

I wasn't sure what to do with the right side, so I did all my work on the left. I factored out the 8
$\displaystyle 8(sin^2xcos^2x)$
which is
$\displaystyle 2(2sinxcosx)(2sinxcosx)$
since 2sinxcosx is the double angle identity for sine I subbed that in
$\displaystyle 2(sin2x)(sin2x)$
which gets turned into
$\displaystyle 2sin^2x$

I may have just gone down the wrong path as I'm not sure how that's supposed to turn into 1-cos4x. Any pointers?

2. Originally Posted by satis
The equation is
$\displaystyle 8sin^2xcos^2X = 1-cos4x$

I wasn't sure what to do with the right side, so I did all my work on the left. I factored out the 8
$\displaystyle 8(sin^2xcos^2x)$
which is
$\displaystyle 2(2sinxcosx)(2sinxcosx)$
since 2sinxcosx is the double angle identity for sine I subbed that in
$\displaystyle 2(sin2x)(sin2x)$
which gets turned into
$\displaystyle 2sin^2x$

I may have just gone down the wrong path as I'm not sure how that's supposed to turn into 1-cos4x. Any pointers?
HI

lets start from RHS .

$\displaystyle 1-\cos 4x=1-\cos 2(2x)$

$\displaystyle =1-[\cos^2 2x-\sin^2 2x]$

$\displaystyle =\cos^2 2x+ \sin^2 2x-\cos^2 2x+\sin^2 2x$

$\displaystyle =2\sin^2 2x$

$\displaystyle =2(2\sin x\cos x)^2$

try continuing from here .

3. Got it, thank you very much.