Math, asked by jaragalnikhil, 1 year ago

If A+B+C=π , prove that sin 2A + sin2B + sin2C =4sinA*sinB*sinC

Answers

Answered by akshaymourya2003
0

Answer:

LHS

=sin2A+sin2B+sin2C

=(sin2A+sin2B)+sin2C

=2sin (A+B)cos (A-B)+2sicCcosC

=2sin (pi-C)cos(A-B)+2sinCcosC

=2sinC [cos (A-B)+cosC]

=2sinC [cos(A-B)+cos{pi-(A+B)}]

=2sinC [cos(A-B)-cos (A+B)]

=2sinC [2sinAsinB]

=4sinAsinBsinC (proved)=RHS

Hope it help you ☝☝

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