Question

How do you prove $\cos \left(\mathrm{\pi }-\mathrm{x}\right)=-\cos x$?

Answer 1

Proof of given relation:

$\cos \left(\mathrm{\pi }-\mathrm{x}\right)=-\cos x$

Here we have $LHS=\cos \left(\mathrm{\pi }-\mathrm{x}\right)$

Therefore,

$\cos \left(\mathrm{\pi }-\mathrm{x}\right)=\cos \pi \cos x-\sin \mathrm{\pi }\sin \mathrm{x}$ $\left[\because cos\left(a-b\right)=cosacosb-sinasinb\right]$

$=\left(-1\right)\cos x-\left(0\right)\sin x$ $\left[\because cos\pi =-1;sin\pi =0\right]$

$=-\cos x$

$=RHS$

Hence, $\mathit{c}\mathit{o}\mathit{s}\left(\pi -x\right)\mathbf{=}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}$ is proved.