In a triangle ABC, the value of sin A + sin B | vedclass

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Question

(b) In $\Delta ABC,A + B + C = 180^\circ $

$ \Rightarrow \sin A + \sin B + \sin C $

$= 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} + 2\sin \fr

Vedclass · Accepted Answer

$4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

Vedclass · Answer

(b) In $\Delta ABC,A + B + C = 180^\circ $

$ \Rightarrow \sin A + \sin B + \sin C $

$= 2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}$ 

$ = 2\sin \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\cos \frac{{A - B}}{2} + 2\cos \frac{C}{2}\sin \left( {\frac{\pi }{2} - \frac{{\overline {A + B} }}{2}} \right)$

$ = 2\cos \frac{C}{2}\cos \frac{{A - B}}{2} + 2\cos \frac{C}{2}\cos \frac{{A + B}}{2}$ 

$ = 2\cos \frac{C}{2}\left[ {\cos \frac{{A - B}}{2} + \cos \frac{{A + B}}{2}} \right]$

$ = 2\cos \frac{C}{2}\left( {2\cos \frac{A}{2}\cos \frac{B}{2}} \right) $

$= 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$ .

In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is

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