Question

LONG ANSWER TYPE QUESTIONS 1. If A+B+C = 180°, prove that B i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
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Answer 1

Answer:

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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

Answer 2

Answer:

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Step-by-step explanation:

A+B+C=180

A+B+C=180LHS= sin2A+sin2B+sin2C

sin2A+sin2B+sin2C=2sin(A+B)cos(A−B)+2sinCcosC

sin2A+sin2B+sin2C=2sin(A+B)cos(A−B)+2sinCcosC=2sinCcos(A−B)+2sinCcosC

sin2A+sin2B+sin2C=2sin(A+B)cos(A−B)+2sinCcosC=2sinCcos(A−B)+2sinCcosC=2sinC(cos(A−B)+cosC)

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))

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

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

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