Question

Prove that
(i) $\cos 2x={\cos }^{2}x-{\sin }^{2}x$
(ii) $\sin 2x=2\sin x\cos x$
(iii) $\tan 2x=\frac{2\tan x}{1-{\tan }^2}$

Answer 1

(i) $\cos 2x=\cos (x+x)$

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

$=\cos x\times \cos x-\sin x\times \sin x$

$={\cos }^{2}x-{\sin }^{2}x$

(ii) $\sin 2x=\sin (x+x)$

$=\sin x\cos x+\cos x\sin x$

$=2\sin x\cos x$

(iii) $\tan \left(2x\right)=\tan (x+x)$

$\left\{\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\right\}$

$=\frac{\tan x+\tan x}{1-\tan x\times \tan x}$

$=\frac{2\tan x}{1-{\tan }^{2}x}$