Pic mai h sirji
Asked by nandrammeena2
20th February 2019,
10:08 PM
Answered by Expert
Answer:
3 sin2A + 2 sin2B = 0 , hence sin2B = (-3/2)sin2A ....................(1)
3 sin2A + 2 sin2B = 1 or 3 sin2A = 1 - 2 sin2B = cos 2B ......................(2)
cos(A+2B) = cosA cos2B + sinA sin2B ..........................(3)
let us substitute for sin2B and cos2B in eqn.(3), using eqn.(1) and eqn.(2)
cos(A+2B) = cosA (3sin2A ) +(-3/2) sinA sin2A = cosA (3sin2A ) +(-3/2) sinA (2 sinA cosA ) = 0
Hence (A+2B) = π/2
Answered by Expert
21st February 2019,
12:41 AM
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