Given \(\displaystyle{2}{\sin{{\left(\alpha\right)}}}+{2}{\sin{{\left(\beta\right)}}}={3}{\sin{{\left(\alpha+\beta\right)}}}\) prove \(\displaystyle{\tan{{\left({\frac{{\alpha}}{{{2}}}}\right)}}}{\tan{{\left({\frac{{\beta}}{{{2}}}}\right)}}}={\frac{{{1}}}{{{5}}}}\)

Harold Hoover

Harold Hoover

Answered question

2022-03-22

Given 2sin(α)+2sin(β)=3sin(α+β) prove tan(α2)tan(β2)=15

Answer & Explanation

Jayda Burch

Jayda Burch

Beginner2022-03-23Added 10 answers

Applying sum to product rule on LHS and sine of the sum of 2 angles on RHS
2sin(α)+2sin(β)=3sin(α+β)
4sin(α+β2)cos(αβ2)=6sin(α+β2)cos(α+β2)
2cos(α2β2)=3cos(α2+β2)
2cos(α2)cos(β2)+2sin(α2)sin(β2)=3cos(α2)cos(β2)3sin(α2)sin(β2)
5sin(α2)sin(β2)=cos(α2)cos(β2)
tan(α2)tan(β2)=15

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