Trigonometry Identity

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

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

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

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

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

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

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

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

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

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