函数 $f(x)=\cos x-\sin x-\cos 2x$ 的最小值是_______．
答案 $-\dfrac 14\sqrt{\dfrac{71+17\sqrt{17}}{2}}$．
解析 题中函数即\[f(x)=-\sqrt 2\sin\left(x-\dfrac{\pi}4\right)+\sin2\left(x-\dfrac{\pi}4\right),\]于是所求最小值即\[y=-\sqrt 2\sin x+2\sin x\cos x\]也即\[y=-\sqrt 2\sin x\left(1-\sqrt 2\cos x\right)\]的最小值．令 $\sqrt 2\cos x=t$，则\[y^2=(2-t^2)(t-1)^2=-t^4+2t^3+t^2-4t+2,\]而\[\left(-t^4+2t^3+t^2-4t+2\right)'_t=-2(t-1)(2t^2-t-2),\]于是\[\max\{-t^4+2t^3+t^2-4t+2\}=\left(-t^4+2t^3+t^2-4t+2\right)\Big|_{t=\frac{1-\sqrt{17}}4}=\dfrac{71+17\sqrt{17}}{32},\]进而可得所求最小值为 $-\dfrac 14\sqrt{\dfrac{71+17\sqrt{17}}{2}}$．