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In any triangle A B C , prove that: ac...

In any triangle `A B C` , prove that: `acosA+bcosB+c cosC=2asinBsinC`

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From sine law, we have,
`a/sinA = b/sinb = c/sinC = k`
`=> a = ksinA, b = ksinB, c = ksinC`
Now,
`L.H.S. = acosA+bcosB+c cosC`
`=ksinAcosA+ksinBcosB+ksinCcosC`
`=k/2[2sinAcosA+2sinBcosB+2sinCcosC]`
`=k/2[sin2A+sin2B+sin2C]`
...
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