#a+b+c=180^@#, #cos(b+c-a)+cos(c+a-b)+cos(a+b-c) =1+4cos(a)cos(b)cos(c)#?? As plz

1 Answer
Feb 11, 2018

#a+b+c=180^@#,

#LHS=cos(b+c-a)+cos(c+a-b)+cos(a+b-c)#

#=cos(180-2a)+cos(180-2b)+cos(180-2c)#

#=-cos(2a)-cos(2b)-cos(2c)#

#=-(2cos^2(a)-1)-(cos(2b)+cos(2c))#

#=1-2cos^2(a)-(2cos(b+c)cos(b-c))#
#=1-2cos^2(a)-2cos(180-a)cos(b-c)#

#=1-2cos^2(a)+2cos(a)cos(b-c)#

#=1-2cos(a)(cos(a)-cos(b-c))#

#=1-2cos(a)(cos(180-(b+c))-cos(b-c))#

#=1-2cos(a)(-cos(b+c)-cos(b-c))#

#=1+2cos(a)(cos(b+c)+cos(b-c))#

# =1+4cos(a)cos(b)cos(c)#