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Estonian Math Competitions

Geometry Difficulty 4.8 AIME Prove it Estonia

Angles α\alpha and β\beta are such that tan⁡αtan⁡β=k≠1\frac{\tan \alpha}{\tan \beta} = k \neq 1. Express sin⁡(α+β)sin⁡(α−β)\frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)} in terms of kk.

Solution

We have k=tan⁡αtan⁡β=sin⁡α⋅cos⁡βcos⁡α⋅sin⁡βk = \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \cdot \sin \beta}, or sin⁡αcos⁡β=k⋅cos⁡αsin⁡β\sin \alpha \cos \beta = k \cdot \cos \alpha \sin \beta. Therefore
sin⁡(α+β)sin⁡(α−β)=sin⁡αcos⁡β+cos⁡αsin⁡βsin⁡αcos⁡β−cos⁡αsin⁡β=(k+1)⋅cos⁡αsin⁡β(k−1)⋅cos⁡αsin⁡β=k+1k−1. \frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \frac{(k+1) \cdot \cos \alpha \sin \beta}{(k-1) \cdot \cos \alpha \sin \beta} = \frac{k+1}{k-1}.

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