题目

已知 $ m=\dfrac{\tan\left( \alpha+\beta+\gamma \right)}{\tan\left( \alpha-\beta+\gamma \right)} $ ，若 $ \sin2\left( \alpha+\gamma \right)=3\sin2\beta $ ，求 $ m $.

解法一(角度变换)

设 $ \alpha+\gamma=\theta $ ,则有 $$ \sin2\theta=3\sin2\beta,$$ 令 $ x=\theta+\beta,y=\theta-\beta $ ,则 $$ 2\theta=x+y,2\beta=x-y,$$ 所以 $$ \sin\left( x+y \right)=3\sin\left( x-y \right)$$ 所以 $$ \sin x\cos y+\cos x\sin y=3\sin x\cos y-3\cos x\sin y,$$ 所以 $$ 4\cos x\sin y=2\sin x\cos y,$$ 所以 $ \tan x=2\tan y $ ,从而
$$ m=\dfrac{\tan\left( \theta+\beta \right)}{\tan\left( \theta-\beta \right)}=\dfrac{\tan x}{\tan y}=2.$$

解法二(名称变换)

设 $ \alpha+\gamma=\theta $ ,则有 $$ \sin2\theta=3\sin2\beta,$$ 所以 $$ \dfrac{2\sin\theta\cos\theta}{\cos^2\theta+\sin^2\theta}=\dfrac{6\sin\beta\cos\beta}{\cos^2\beta+\sin^2\beta},$$ 所以 $$ \dfrac{\tan\theta}{1+\tan^2\theta}=\dfrac{3\tan\beta}{1+\tan^2\beta},$$ 令 $ x=\tan\theta,y=\tan\beta $ ,上式可整理为 $$ x+xy^2=3y+3x^2y$$ 故 $$ \begin{array}{rl} m=\dfrac{\tan\left( \theta+\beta \right)}{\tan\left( \theta-\beta \right)}&=\dfrac{\tan\theta+\tan\beta}{1-\tan\theta\tan\beta}\times\dfrac{1+\tan\theta\tan\beta}{\tan\theta-\tan\beta}\\\ &=\dfrac{x+y}{1-xy}\times\dfrac{1+xy}{x-y}\\\ &=\dfrac{x+xy^2+y+x^2y}{x+xy^2-y-x^2y}\\\ &=\dfrac{3y+3x^2y+y+x^2y}{3y+3x^2y-y-x^2y}\\\ &=\dfrac{1+x^2+3+3x^2}{3+3x^2-1-x^2}=\dfrac{4x^2+4}{2x^2+2}=2. \end{array} $$