Question #8f7b6

Question

2668 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-24T01:07:18

Given
#A+B+C=pi#

#=>A/2+B/2+C/2=pi/2#

#=>C/2=pi/2-(A+B)/2#

#LHS=sin(A/2)+sin(B/2)+sin(C/2)#

#=sin(A/2)+sin(B/2)+sin(pi/2-(A+B)/2)#

#=sin(A/2)+sin(B/2)+cos((A+B)/2)#

#=sin(A/2)+sin(B/2)+1-sin^2((A+B)/4)#

#=1+2sin((A+B)/4)cos((A-B)/4)-sin^2((A+B)/4)#

#=1+2sin((A+B)/4)(cos((A-B)/4)-sin((A+B)/4)#

#=1+2sin(pi/4-C/4)(cos((A-B)/4)-cos(pi/2-(A+B)/4))#

#=1+4sin(pi/4-C/4)sin(pi/4-B/4)sin(pi/4-A/4)#

#=1+4sin(pi/4-A/4)sin(pi/4-B/4)sin(pi/4-C/4)#

#=RHS#

Proved