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Question

In any ABC< prove that
acosA+bcosB+ccosC=2asinBsinC.

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Solution

LHSksinAcosA+k=sinBcosB+ksinCcosC
=k2[sin2A+sin2B+sin2C]
=k2[2sin(A+B)cos(AB)+2sinCcosC]
=k[sin(πC)cos(AB)+sinCcos(π(A+B))]
=k[sinCcos(AB)sinCcos(A+B)]
=ksinC[2sinA.sin(B)]
=2ksinCsinAsinB
=2asinBsinC
=RHS

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