Question

If $\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\pi }$ then prove that $\sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)=4\cos \left(\frac{A}{2}\right)\times \cos \left(\frac{B}{2}\right)\times \cos \left(\frac{C}{2}\right)$

Answer 1

Prove the given result

Given that ,

$A+B+C=\mathrm{\pi }$

And,$(\mathrm{A}+\mathrm{B})=\mathrm{\pi }-C$

We have to prove:

$\sin \left(A\right)+\sin \left(B\right)+\sin \left(C\right)=4\cos \left(\frac{A}{2}\right)\times \cos \left(\frac{B}{2}\right)\times \cos \left(\frac{C}{2}\right)$ .

LHS :-

$$\begin{gathered} \sin A+\sin B+\sin C=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\sin C[\mathrm{Standard}\mathrm{identity}] \\ =2\sin \left(\frac{\mathrm{\pi }-C}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right)\left[\because \sin 2x=2\sin x\cos x\right] \\ =2\sin \left(\frac{\mathrm{\pi }}{2}-\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\cos \left(\frac{\mathrm{\pi }-\mathrm{A}-\mathrm{B}}{2}\right)\cos \left(\frac{C}{2}\right) \\ =2\cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)+2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{C}{2}\right)\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cos \theta \right] \\ =2\cos \left(\frac{C}{2}\right)\times \left(\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)\right)\left[\mathrm{Standard}\mathrm{identity}\right] \\ =4\cos \left(\frac{C}{2}\right)\times \cos \left(\frac{B}{2}\right)\times \cos \left(\frac{A}{2}\right) \end{gathered}$$

$\therefore$ LHS=RHS

Hence, proved.