Question

a cos A + b cos B + c cos C = 2a sin B. sin C​

Answer 1

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$\large \tt \green{✦Answer✦}$

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)

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Answer 2

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]

[∵, 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)

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