Question

How do you prove ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$?

Answer 1

Proof of given relation:

${\cos }^{4}x-{\sin }^{4}x=\cos 2x$

Here we have $LHS={\cos }^{4}x-{\sin }^{4}x$

Therefore,

${\cos }^{4}x-{\sin }^{4}x={\left({\cos }^{2}x\right)}^2-{\left({\sin }^{2}x\right)}^2$

$=\left({\cos }^{2}x+{\sin }^{2}x\right)\left({\cos }^{2}x-{\sin }^{2}x\right)$ $\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]$

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

$=\cos 2x$ $\left[\because co{s}^{2}x-si{n}^{2}x=cos2x\right]$

$=RHS$

Hence, $\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{4}}\mathit{x}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{4}}\mathit{x}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}$ is proved.