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Question

In triangle ABC, prove the following:
a cos A+bcos B+c cos C=2b sin A sin C=2 c sin A sin B

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Solution

LetasinA=bsinB=csinC=k ...1Consider the LHS of the equation a cos A+bcos B+c cos C.acosA+bcosB+ccosC=ksinAcosA+sinBcosB+sinCcosC =k22sinAcosA+2sinAcosA+2sinCcosC =k2sin2A+sin2B+sin2C =k22sinA+BcosA-B+2sinCcosC =k22sinπ-CcosA-B+2sinCcosC =k22sinCcosA-B+2sinCcosC =2ksinC2cosA-B+cosC
=ksinCcosA-B+cosπ-A+B=ksinCcosA-B-cosA+B=ksinC2sinAsinB=2ksinAsinBsinC ...(1)



Now,on putting ksinC=C in equation (1), we get:2csinAsinBand on putting ksinB=b in equation (1), we get:2bsinAsinC

So, from (1), we have
a cos A+bcos B+c cos C=2b sin A sin C=2 c sin A sin B .

Hence proved.

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