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

Question

52964 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-04T17:36:51

Please refer to the 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.!

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