# Trigonometry Identity

• Nov 13th 2009, 11:36 PM
wizard654zzz
Trigonometry Identity
$tan2\theta- sin2\theta = x and tan2\theta + sin2\theta = y$ prove that $(x^2 -y^2)^2 = 16xy$
• Nov 13th 2009, 11:39 PM
wizard654zzz
I revise
I revise : $tan2\theta- sin2\theta = x$ and $tan2\theta + sin2\theta = y$ prove that $(x^2 -y^2)^2 = 16xy$
• Nov 14th 2009, 12:34 AM
Hello wizard654zzz
Quote:

Originally Posted by wizard654zzz
I revise : $tan2\theta- sin2\theta = x$ and $tan2\theta + sin2\theta = y$ prove that $(x^2 -y^2)^2 = 16xy$

Factorise the LHS using difference of two squares:
$(x^2-y^2)^2$
$=\Big((x+y)(x-y)\Big)^2$

$=\Big(2\tan2\theta(-2\sin2\theta)\Big)^2$

$= 16\tan^22\theta\sin^22\theta$

$=16\tan^22\theta(1-\cos^22\theta)$

$=16(\tan^22\theta-\tan^22\theta\cos^22\theta)$

$=16(\tan^22\theta-\sin^22\theta)$

$=16(\tan2\theta - \sin2\theta)(\tan2\theta +\sin2\theta)$

$=16xy$
I revise : $tan2\theta- sin2\theta = x$ and $tan2\theta + sin2\theta = y$ prove that $(x^2 -y^2)^2 = 16xy$