Question

If cos (alpha+beta) = 4 / 5, sin (alpha+beta) = 5 /13 and (alpha) and (beta) lies between 0 and pi / 4, show that tan 2(alpha) = 56 / 33.

Answer 1

Wehave, cos(α+β)= 4 5 ⇒sin(α+β)= 1−( 4 5 )2 − − − − − − − − √ = 3 5 And sin(α−β)= 5 13 ⇒cos(α−β)= 1−( 5 13 )2 − − − − − − − − √ = 12 13 Now, sin2α=sin[(α+β)+(α−β)] =sin(α+β)cos(α−β) +sin(α−β)cos(α+β) = 3 5 × 12 13 + 5 13 × 4 5 = 36 65 + 20 65 = 56 65 ∴cos2α= 1−sin22α − − − − − − − − − √ = 1−( 56 65 )2 − − − − − − − − √ = 1− 3136 4225 − − − − − − − √ = 1089 4225 − − − − √ = 33 65 Hence,tan2α= sin2α cos2α = 56 65 33 65 = 56 33

Answer 2

find sin(2 alpha) and then cos(2 alpha)..

$cos(\alpha+\beta)=\frac{4}{5},\ \ \ sin(\alpha-\beta)=\frac{5}{13}\\\\sin(\alpha+\beta)=\sqrt{1-(\frac{4}{5})^2}=\frac{3}{5}\\\\cos(\alpha-\beta)=\frac{12}{13},\ similarly.\\\\sin(\alpha+\beta+\alpha-\beta)=sin(\alpha+\beta)cos(\alpha-\beta)+sin(\alpha-\beta)cos(\alpha+\beta)\\\\=\frac{3*12}{5*13}+\frac{5*4}{5*13}=\frac{56}{65}\\\\cos(\alpha+\beta+\alpha-\beta)=cos(\alpha+\beta)cos(\alpha-\beta)+sin(\alpha-\beta)sin(\alpha-\beta)\\\\=\frac{4*12}{5*13}-\frac{5*3}{13*5}=\frac{33}{65}\\\\tan(2\alpha)=56/33$