Question

How do you prove $\sin \left(180°-a\right)=\sin a$?

Answer 1

Proof of given relation:

$\sin \left(180°-a\right)=\sin a$

Here we have $LHS=\sin \left(180°-a\right)$

Therefore,

$\sin \left(180°-a\right)=\sin 180°\cos a-\cos 180°\sin a$ $\left[\because \sin \left(a-b\right)=\mathrm{sina}\mathrm{cosb}-\mathrm{cosa}\mathrm{sinb}\right]$

$=\left(0\right)\cos a-\left(-1\right)\sin a$ $\left[\because \cos 180°=-1;\sin 180°=0\right]$

$=\sin a$

$=RHS$

Hence, $\mathit{s}\mathit{i}\mathit{n}\left(180°-a\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{a}$ is proved.