If #A+B+C=pi# then prove that #sin(B+2C)+sin(C+2A)+sin(A+2B)=4sin((B-C)/2)*sin((C-A)/2)*sin((A-B)/2)#?

1 Answer
Ratnaker Mehta
Nov 8, 2017

Kindly refer to a Proof in the Explanation.

Explanation:

#A+B+C=pi rArr A+B=pi-C.#

#:. A+2B=(A+B)+B=(pi-C)+B=pi-(C-B).#

#:. sin(A+2B)=sin(pi-(C-B))=sin(C-B).#

Similarly,

# sin(B+2C)=sin(A-C), and, sin(C+2A)=sin(B-A).#

#:. sin(A+2B)+sin(B+2C)+sin(C+2A),#

#=sin(C-B)+sin(A-C)+sin(B-A).#

Let, #C-B=x, A-C=y and B-A=z.#

Then, #x+y+z=0.#

#:. x+y=-z......(1) and z=-(x+y)......(2).#

#"The L.H.S.="sinx+siny+sinz,#

#=2sin((x+y)/2)cos((x-y)/2)+sinz,#

#=2sin(-z/2)cos((x-y)/2)+sinz,............[because (1)],#

#=-2sin(z/2)cos((x-y)/2)+2sin(z/2)cos(z/2),...........................................................................[because sin2a=2sinacosa],#

#=2sin(z/2){cos(z/2)-cos((x-y)/2)},#

#=2sin(z/2){cos(-(x+y)/2)-cos((x-y)/2)},.......[because (2)],#

#=2sin(z/2){cos((x+y)/2)-cos((x-y)/2)},#

#=2sin(z/2){-2sin(x/2)sin(y/2)},#

#=-4sin(x/2)sin(y/2)sin(z/2),#

#=-4sin((C-B)/2)sin((A-C)/2)sin((B-A)/2),#

#=-4sin(-(B-C)/2)sin(-(C-A)/2)sin(-(A-B)/2),#

#=+4sin((B-C)/2)sin((C-A)/2)sin((A-B)/2),#

#"=The R.H.S."#

Enjoy Maths.!

Related questions
Impact of this question
24513 views around the world
You can reuse this answer
Creative Commons License