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Question

Prove that acosA+bcosB+ccosC=4RsinAsinBsinC

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Solution

L.H.S =acosA+bcosB+ccosC

By sine rule

asinA=bsinB=csinC=R

using trigonometric identities and simplifying

=2R(sinAcosA+sinBcosB+sinCcosC)

=R(sin2A+sin2B+sin2C)

=2R(sin(A+B)cos(AB)sinCcosC))

=2RsinC(cos(AB)cos(A+B))

=4RsinAsinBsinC

= R.H.S

Hence proved.

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