Question

If A+B+C=180 prove that sin2A +sin2B-sin2C = 4cosA cosB cosC

Answer 1

hope it helps you.....

Answer 2

Given:

A+B+C = 180°

=> A+B = 180° - C

=> sin(A+B) = sin(180-C)

= sinC -----(1)

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We know that,

i))SinC - SinD = 2cos[(C+D)/2]sin[(C-D)/2]

ii) SinC + sinD = 2sin[(C+D)/2]cos[(C+D)/2]

iii) Sin2B = 2sinBcosB

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Here,

LHS = sin2A+sin2B-sin2C

= sin2A-sin2C+sin2B

=2cos[(2A+2C)/2]sin[(2A-2C)/2]+sin2B

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

= -2cosBsin(A-C)+2sinBcosB

= (2cosB)[-sin(A-C)-sinB]

= 2cosB[-sin(A-C)+sin(A+C)]

= 2cosB[sin(A+C)-sin(A-C)]

= 2cosB{2cos[(A+C+A-C)/2]sin[(A+C-A+C)/2]}

= 4cosBcosAsinC

= 4cosAcosBsinC

= RHS

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