Question

A+B+C=270 then sin2A-sin2B+sin2C​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

sin2A-sin2B+sin2C​ = 2sin(A+C){cos(A-C) - cos(A+C)}

Step-by-step explanation:

A + B + C = 270° = 3π/2 = 2π - π/2 (GIVEN)

B+C = 2π - π/2 - A

sin(B+C) = sin(2π - π/2 - A ) = sin(- π/2 - A ) = - sin( π/2 + A ) =  -cos(A)

sin(B+C) = - cos(A)

cos(B+C) = cos(2π - π/2 - A ) = cos(- π/2 - A) = cos(π/2 + A) = -sin(A)

cos(B+C) = -sin(A)

sin2A-sin2B+sin2C​ = sin2A + sin2C - sin2B

                                  = 2sin(A+C)cos(A-C) - 2sinBcosB

                                  = 2sin(A+C) cos(A-C) - 2sin(A+C) cos(A+C)

                                  = 2sin(A+C){cos(A-C) - cos(A+C)}

sin2A-sin2B+sin2C​ = 2sin(A+C){cos(A-C) - cos(A+C)}

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