已知函数 $f\left(x\right)=\sqrt 2\cos \left(x-\dfrac {\mathrm \pi} {12} \right) $，$ x\in \mathbb R$．若 $\cos\theta =\dfrac 3 5 $，$ \theta \in \left(\dfrac {3{\mathrm \pi}} 2 ,2{\mathrm \pi}\right)$，则 $f\left(\theta -\dfrac {5{\mathrm \pi}} {12} \right)=$ ．
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【答案】
$ -\dfrac {4\sqrt 2} 5 $
【解析】
因为$$\begin{split} f\left(\theta -\dfrac {5{\mathrm \pi}} {12} \right)=&\sqrt 2\cos \left(\theta -\dfrac {5{\mathrm \pi}} {12} -\dfrac {\mathrm \pi} {12} \right)\\=&\sqrt 2\cos \left(\theta -\dfrac {\mathrm \pi} 2 \right)\\=&\sqrt 2\sin\theta ,\end{split} $$由已知可得 $\theta $ 为第四象限角，所以 $\sin \theta <0$，
故 $\sin\theta =-\sqrt {1-\cos^2\theta }=-\dfrac 4 5 $，从而有$$f\left(\theta -\dfrac {5{\mathrm \pi}} {12} \right)=\sqrt 2\sin\theta =\sqrt 2\times \left(-\dfrac 4 5 \right)=-\dfrac {4\sqrt 2} 5 .$$
故 $\sin\theta =-\sqrt {1-\cos^2\theta }=-\dfrac 4 5 $，从而有$$f\left(\theta -\dfrac {5{\mathrm \pi}} {12} \right)=\sqrt 2\sin\theta =\sqrt 2\times \left(-\dfrac 4 5 \right)=-\dfrac {4\sqrt 2} 5 .$$
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