Question

If a b c are angle of triangle then prove that sin 2A + sin2B + sin2c is equals to 4 sin a sin b sin c

Answer 1

Answer:

Step-by-step explanation:

A+B+C=180

LHS=sin2A+sin2B+sin2C

=2sin(A+B)cos(A−B)+2sinCcosC

=2sinCcos(A−B)+2sinCcosC

=2sinC(cos(A−B)+cosC)

=2sinC(cos(A−B)−cos(A+B))

=2sinC2sinAsinB

=4sinAsinBsinC

=RHS