Math, asked by doitll, 11 months ago

Prove :

sin2A + sin 2B + sin2C = 4sinAsinBsinC .

Answers

Answered by oOBADGIRLOo

Hmm, Good question :

See the attachment.

Attachments:

Answered by Anonymous

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$\large\mathcal\red{solution}$

=>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]

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$

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