1. ## Trigonometry Identity

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

2. ## I revise

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

3. Hello wizard654zzz
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 $\displaystyle (x^2 -y^2)^2 = 16xy$