[SOLVED] prove equality using double angle identity

**satis** · October 23rd 2009, 03:25 AM

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

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
$8(sin^2xcos^2x)$
which is
$2(2sinxcosx)(2sinxcosx)$
since 2sinxcosx is the double angle identity for sine I subbed that in
$2(sin2x)(sin2x)$
which gets turned into
$2sin^2x$

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?

**mathaddict** · October 23rd 2009, 03:29 AM

Originally Posted by satis

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

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
$8(sin^2xcos^2x)$
which is
$2(2sinxcosx)(2sinxcosx)$
since 2sinxcosx is the double angle identity for sine I subbed that in
$2(sin2x)(sin2x)$
which gets turned into
$2sin^2x$

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 .

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

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

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

$=2\sin^2 2x$

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

try continuing from here .

**satis** · October 23rd 2009, 03:44 AM

Got it, thank you very much.

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