Question

Prove that $\frac{{\sin }^{2}3A}{{\sin }^{2}A}-\frac{{\cos }^{2}3A}{{\cos }^{2}A}=4\cos 2A.$.

Answer 1

$\frac{{\sin }^{2}3A}{{\sin }^{2}A}-\frac{{\cos }^{2}3A}{{\cos }^{2}A}\Rightarrow \frac{{\cos }^{2}A{\sin }^{2}3A-{\cos }^{2}3A{\sin }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$
$\Rightarrow \frac{{\cos }^{2}A(1-{\cos }^{2}3A)-{\cos }^{2}3A(1-{\cos }^{2}A)}{{\sin }^{2}A{\cos }^{2}A}$
$\Rightarrow \frac{{\cos }^{2}A-\left({\cos }^{2}A{\cos }^{2}3A\right)-{\cos }^{2}3A+\left({\cos }^{2}3A{\cos }^{2}A\right)}{{\sin }^{2}A{\cos }^{2}A}$
$\Rightarrow \frac{{\cos }^{2}A-{\cos }^{2}3A}{{\sin }^{2}A{\cos }^{2}A}\Rightarrow -\frac{\sin (A-3A)\sin (A+3A)}{{\sin }^{2}A{\cos }^{2}A}$
$\Rightarrow \frac{4\sin 2A\sin 4A}{{\sin }^{2}2A}\Rightarrow 4\frac{\sin 4A}{\sin 2A}=8\frac{\sin 2A\cos 2A}{\sin 2A}=8\cos 2A$