函数 $f(x)=\sin 2 x+\sin \left(x+\frac{\pi}{2}\right)+\cos \left(x+\frac{\pi}{2}\right)$ 的最小值是
【答案】 $ -1-\sqrt{2}$

【解析】 $$ \begin{aligned}
& f(x)=\sin 2 x+\cos x-\sin x=2 \sin x \cos x+\cos x-\sin x \text {, 令 } t=\cos x-\sin x=\sqrt{2} \cos \left(x+\frac{\pi}{4}\right), \\
& \therefore t \in[-\sqrt{2}, \sqrt{2}] \text {, 且 } t^2=1-2 \sin x \cos x, \quad 2 \sin x \cos x=1-t^2, \therefore y=1-t^2+t, t \in[-\sqrt{2}, \sqrt{2}], \\
& \therefore y_{\min }=1-2-\sqrt{2}=-1-\sqrt{2} \text {. }
\end{aligned}
$$
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