Question

Given A,B,C are the angles of an triangle.
Shows that
sin (A)+sin(B)+sin(C)=4 cos (A/2) cos (B/2) cos (C/2)

Answer 1

A+B+C= Pi then taking B+C=D so sinA=Π−D=cosD=cos(B+C) .... cos(B+C) + sinB+ sinC =cosBcosC-sinBsinC + sinB + sinC

Answer 2

Erm if it includes angle
Please show it in degree... Radian a bit confusing me... Tq

Answer 3

Let
\[
x + y + z =\frac \pi 2\\
\sin (2 x)+\sin (2 y)+\sin (2 z)=2 \sin (x+y) \cos (x-y)+2 \sin
(z) \cos (z)=\\2 \sin (x+y) \cos (x-y)+2 \sin \left(\frac{\pi
}{2}-(x+y)\right) \cos \left(\frac{\pi }{2}-(x+y)\right)\\=2
\sin (x+y) \cos (x-y)+2 \sin (x+y) \cos (x+y)\\=2 \sin (x+y)
(\cos (x-y)+\cos (x+y))=\\2 \sin (x+y) (2 \cos (x) \cos (y))=\\4
\cos (x) \cos (y) \sin \left(\frac{\pi }{2}-z\right)=\\4 \cos
(x) \cos (y) \cos (z)
\]

This is the solution of your problem.

Answer 4

@eliassaab but u have to prove as 4 cos (A/2) cos (B/2) cos (C/2)

Answer 5

A=2x, x =A/2
B= 2y, y=B/2
C= 2y, y=C/2
Got it.