Question

How do you prove $\cos 2x=2{\cos }^{2}x-1$?

Answer 1

Proof of given relation:

$\cos 2x=2{\cos }^{2}x-1$

Here we have $LHS=\cos 2x$

Therefore,

$\cos 2x=\cos \left(x+x\right)$ $\left[\because cos\left(A+B\right)=cosAcosB-sinAsinB\right]$

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

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

$={\cos }^{2}x-\left(1-{\cos }^{2}x\right)$ $\left[\because si{n}^{2}x+co{s}^{2}x=1\right]$

$=2{\cos }^{2}x-1$

$=RHS$

Hence, $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{x}\mathbf{-}\mathbf{1}$ is proved.