trigonometry

Asked by  | 21st Sep, 2008, 09:29: PM

Expert Answer:

let us denote α = A and β = B

2tanA = 3tanB   ==> tanA =3/2 tanB

tan(A-B) = (tanA - tanB) / ( 1+tanAtanB) = 1/2 tanB / ( 1+3/2 tan2B)

put tanB = sinB / cosB and simplify

= sinBcosB / (2cos2B + 3sin2B )  = 2sinBcosB / (4cos2B + 6sin2B) = sin2B / (4 +2sin2B)

=sin2B / (4 - (1-2sin2B)+1)

=sin2B / (5 - cos2B)

Answered by  | 21st Sep, 2008, 10:59: PM

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