已知 $2\tan(\alpha+\beta)=3\tan \alpha$,则 $A=\sin^2\beta-\sin(2\alpha+\beta)$ 的最大值为_______.
答案 $\dfrac{26}{25}$.
解法一 记 $\tan(\alpha+\beta)=3m$,$\tan\alpha=2m$,则\[\tan\beta=\tan((\alpha+\beta)-\alpha)=\dfrac{m}{1+6m^2},\]而\[\tan(2\alpha+\beta)=\tan((\alpha+\beta)+\alpha)=\dfrac{5m}{1-6m^2},\]于是\[\begin{split} A&\leqslant \dfrac{m^2}{(1+6m^2)^2+m^2}+\dfrac{5|m|}{\sqrt{(1-6m^2)^2+(5m)^2}}\\ &=\dfrac{m^2}{36m^4+13m^2+1}+5\sqrt{\dfrac{m^2}{36m^4+13m^2+1}},\end{split}\]而\[\dfrac{m^2}{36m^2+13m^2+1}\leqslant \dfrac{m^2}{12m^2+13m^2}=\dfrac{1}{25},\]等号当 $m=-\dfrac 16$ 时取得,于是 $A$ 的最大值为 $\dfrac{26}{25}$.
解法二 根据题意,有\[\dfrac{\sin(\alpha+\beta)\cos\alpha}{\sin\alpha\cos(\alpha+\beta)}=\dfrac 32,\]于是\[\dfrac{\sin(\alpha+\beta)\cos\alpha+\sin\alpha\cos(\alpha+\beta)}{\sin(\alpha+\beta)\cos\alpha-\sin\alpha\cos(\alpha+\beta)}=\dfrac {3+2}{3-2},\]即\[\dfrac{\sin(2\alpha+\beta)}{\sin\beta}=5,\]于是\[A=\sin^2\beta-\sin\beta,\]而\[\tan\beta=\tan((\alpha+\beta)-\alpha)=\dfrac{\dfrac 32\tan\alpha-\tan\alpha}{1+\dfrac 32\tan\alpha\cdot \tan\alpha}=\dfrac{\tan\alpha}{2+3\tan^2\alpha},\]于是\[-\dfrac{1}{2\sqrt 6}\leqslant\tan\beta\leqslant \dfrac{1}{2\sqrt 6},\]进而\[-\dfrac 15\leqslant \sin\beta\leqslant\dfrac 15,\]因此\[A\leqslant \dfrac{26}{25},\]等号当 $\tan\alpha=-\sqrt{\dfrac 23}$ 时取得,因此所求最大值为 $\dfrac{26}{25}$.