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

1 Answer
Nov 4, 2017

Please refer to the Explanation.

Explanation:

We have,

ul(sinA+sinB)+sinC=2sin((A+B)/2)cos((A-B)/2)+sinC....(star).

But, A+B+C=pi :. A+B=pi-C :. ((A+B)/2)=(pi-C)/2.

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

:. sin((A+B)/2)=sin(pi/2-C/2)=cos(C/2)....(star^1).

Also, sinC=2sin(C/2)cos(C/2).........(star^2).

Utilising (star^1) and (star^2) in (star), we get,

sinA+sinB+sinC,

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

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

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

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

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

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

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

=4cos(A/2)cos(B/2)cos(C/2)....[because, cos(-x)=cosx].

Hence, the Proof.

Enjoy Maths.!