Question

In any triangle ABC prove that a.cosA+b.cosB+c.cosC=2a.sinBsinC

Answer 1

A/sinA=b/sinB=c/sinC=k (say)
∴, a=ksinA, b=ksinB, c=ksinC
∴, acosA+bcosB+ccosC
=ksinAcosA+ksinBcosB+csinCcosC
=k/2(2sinAcosA+2sinBcosB+2sinCcosC)
=k/2(sin2A+sin2B+sin2C)
=k/2[{2sin(2A+2B)/2cos(2A-2B)/2}+sin2C]
[∵, sinC+sinD=2sin(C+D)/2cos(C-D)/2]
=k/2[2sin(A+B)cos(A-B)+2sinCcosC]
=k[sin(π-C)cos(A-B)+sinCcos{π-(A+B)}] [∵, A+B+C=π]
=k[sinCcos(A-B)+sinC{-cos(A+B)}]
=ksinC[cos(A-B)-cos(A+B)]
=ksinC[2sin(A-B+A+B)/2sin(A+B-A+B)/2]
[∵, cosC-cosD=2sin(C+D)/2sin(D-C)/2]
=ksinC(2sinAsinB)
=2(ksinA)sinBsinC
=2asinBsinC (Proved)