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
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.!