If #A+B+C+D=2pi# then prove that #cos^2(A/2)+cos^2(B/2)+cos^2(C/2)+cos^2(D/2)=2-2cos((A+B)/2)*cos((A+C)/2)*cos((A+D)/2)#?

1 Answer
Nov 8, 2017

#LHS=cos^2(A/2)+cos^2(B/2)^2+cos^2(C/2)+cos^2(D/2)#

#=1/2(2cos^2(A/2)+2cos^2(B/2)+2cos^2(C/2)+2cos^2(D/2))#

#=1/2(1+cosA+1+cosB+1+cosC+1+cosD)#

#=2+1/2(cosA+cosB+cosC+cosD)#

#=2+1/2(2cos((A+B)/2)cos((A-B)/2)+2cos((C+D)/2)cos((C-D)/2))#

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

#=2+1/2(2cos((A+B)/2)cos((A-B)/2)-2cos((A+B)/2)cos((C-D)/2))#

#=2+cos((A+B)/2)(cos((A-B)/2)-cos((C-D)/2))#

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

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

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

#=2-2cos((A+B)/2)*cos((A+C)/2)*cos((A+D)/2)#