1668

已知 ${\displaystyle x,y\in (-\frac{\pi }{2},\frac{\pi }{2})}$, ${\displaystyle \sin x+\sqrt{2}\sin y=\frac{1}{2}}$, 则 ${\displaystyle \cos (x+y)}$ 的最大值为_____.

类似 1596, 采用主元的思想. 记 ${\displaystyle x+y=z}$, 则用 ${\displaystyle x=z-y}$ 代换条件式: $$\begin{array}{rl}\frac{1}{2}& =\sin (z-y)+\sqrt{2}\sin y=\sin y(\sqrt{2}-\cos z)+\cos y(\sin z)\\ & \le \sqrt{{\sin }^{2}z+(\sqrt{2}-\cos z{)}^2}=\sqrt{3-2\sqrt{2}\cos z},\end{array}$$ 所以 ${\displaystyle \cos z\le \frac{11\sqrt{2}}{16}}$.

记 ${\displaystyle t=\cos x-\sqrt{2}\cos y}$. 则 $$t^2+\frac{1}{4}=(\sin x+\sqrt{2}\sin y{)}^2+(\cos x-\sqrt{2}\cos y{)}^2=3-2\sqrt{2}\cos (x+y),$$ 所以 $$\cos (x+y)=\frac{\sqrt{2}}{4}(\frac{11}{4}-t^2)\le \frac{11\sqrt{2}}{16}.$$