Prove :
sin2A + sin 2B + sin2C = 4sinAsinBsinC .
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Hmm, Good question :
See the attachment.
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=>A+B+C=π
Sin2A+Sin2B+Sin2C
=>2sin(A+B)cos(A-B)+Sin2C
=>2sin(π-c)cos(A-B)+2sinCcosC
=>2sinCcos(A-B+2sinCCosC
=>2sinC(cos(A-B)+cosC)
=>2sinC[cos(A-B)+cos(π-(A+B))
=>2sinC[cos(A-B)-cos(A+B)]
=>2sinC(2sinA sinB)
=>4sinAsinBsinC=[proved]
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