Q. 1

If A + B + C = π, prove that

sin 2A + sin 2B – sin 2C = 4cos A cos B sin C

Answer :

= sin 2A + sin 2B – sin 2C


= 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C
using formula,
sin (A + B) = sin A cos B + cos A sin B
= sin 2A + sin 2B - sin 2C


Using formula


sin2A = 2sinAcosA
= 2sinAcosA + 2sinBcosB - 2sinCcosC


since A + B + C = π



And sin(π – A) = sinA
= 2sin(B + C)cos A + 2sin(A + C)cosB - 2sin(A + B)cosC
= 2 ( sin B cos C + cos B sin C ) cos A + 2(sinAcosC + cosAsinC)cosB - 2(sinAcosB + cosAsinB )cosC
= 2cosAsinBcosC + 2cosAcosBsinC + 2sinAcosBcosC + 2cosAcosBsinC– 2sinAcosBcosC – 2cosAsinBcosC
= 2cosAcosBsinC + 2cosAcosBsinC
= 4cosAcosBsinC


= R.H.S


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RELATED QUESTIONS :

If A + B + C = π, prove that


RS Aggarwal - Mathematics

If A + B + C = π, prove that


RS Aggarwal - Mathematics