# If A+B+C=pi then prove that sinA+sinB+sinC=4cos(A/2)*cos(B/2)*cos(C/2)?

Nov 4, 2017

#### Explanation:

We have,

$\underline{\sin A + \sin B} + \sin C = 2 \sin \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right) + \sin C \ldots . \left(\star\right) .$

But, $A + B + C = \pi \therefore A + B = \pi - C \therefore \left(\frac{A + B}{2}\right) = \frac{\pi - C}{2.}$

$\therefore \left(\frac{A + B}{2}\right) = \frac{\pi}{2} - \frac{C}{2.}$

$\therefore \sin \left(\frac{A + B}{2}\right) = \sin \left(\frac{\pi}{2} - \frac{C}{2}\right) = \cos \left(\frac{C}{2}\right) \ldots . \left({\star}^{1}\right) .$

Also, $\sin C = 2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{C}{2}\right) \ldots \ldots \ldots \left({\star}^{2}\right) .$

Utilising $\left({\star}^{1}\right) \mathmr{and} \left({\star}^{2}\right)$ in $\left(\star\right) ,$ we get,

$\sin A + \sin B + \sin C ,$

$= 2 \cos \left(\frac{C}{2}\right) \cos \left(\frac{A - B}{2}\right) + 2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{C}{2}\right) ,$

$= 2 \cos \left(\frac{C}{2}\right) \left\{\cos \left(\frac{A - B}{2}\right) + \sin \left(\frac{C}{2}\right)\right\} ,$

=2cos(C/2){cos((A-B)/2)+sin((pi-(A+B))/2)...[as, A+B+C=pi],

$= 2 \cos \left(\frac{C}{2}\right) \left\{\cos \left(\frac{A - B}{2}\right) + \sin \left(\frac{\pi}{2} - \frac{A + B}{2}\right)\right\} ,$

$= 2 \cos \left(\frac{C}{2}\right) \left\{\cos \left(\frac{A - B}{2}\right) + \cos \left(\frac{A + B}{2}\right)\right\} ,$

=2cos(C/2){2cos(((A-B)/2+(A+B)/2)/2)cos(((A-B)/2-(A+B)/2)/2),

$= 2 \cos \left(\frac{C}{2}\right) \left\{2 \cos \left(\frac{A}{2}\right) \cos \left(- \frac{B}{2}\right)\right\} ,$

$= 4 \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right) \ldots . \left[\because , \cos \left(- x\right) = \cos x\right] .$

Hence, the Proof.

Enjoy Maths.!