每日一题[734]三角恒等变换与最值

已知$\cos 2\alpha+\cos2\beta+\cos2\gamma =1$，$\sin\alpha+\sin\beta+\sin\gamma =0$，则$\tan\gamma$的最大值为_______．

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分析与解　$\sqrt 2$．

已知条件即

$$\begin{cases} \cos^2\gamma =\sin^2\alpha+\sin^2\beta,\\ \sin \gamma =-\sin\alpha-\sin\beta,\end{cases}$$ 于是 $$\tan^2\gamma =\dfrac{\left(\sin\alpha+\sin\beta\right)^2}{\sin^2\alpha+\sin^2\beta}\leqslant 2,$$ 等号当$\sin\alpha=\sin\beta$时取得．

又当$\sin\alpha=\sin\beta=\dfrac{\sqrt 6}6$，$\sin\gamma =-\dfrac{\sqrt 6}3$，$\cos\gamma =-\dfrac{\sqrt 3}3$时，$\tan\gamma =\sqrt 2$，于是所求的最大值为$\sqrt 2$．