Question

How do you prove $\cos 2A={\cos }^{2}A-{\sin }^{2}A$?

Answer 1

Proof of given relation:

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

Here we have $LHS=\cos 2A$

Therefore,

$\cos 2A=\cos \left(A+A\right)$ $[\because cos\left(A+B\right)=co\mathrm{sAc}o\mathrm{sB}-si\mathrm{nAs}i\mathrm{nB}]$

$=\cos A\cos A-\sin A\sin A$

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

$=RHS$

Hence, $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{A}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}$ is proved.