已知 $f\left(\theta \right)=\dfrac {\cos \left(\theta -\dfrac {3{\mathrm \pi}} 2 \right)\cdot \sin \left(\dfrac {7{\mathrm \pi}} 2 +\theta \right)} {\sin \left(-\theta -{\mathrm \pi}\right)} $．
【难度】
【出处】
无
【标注】
-
化简 $f\left(\theta \right)$；标注答案$\cos\theta$解析$f\left(\theta \right)=\dfrac {\cos\left(\dfrac {3{\mathrm \pi}} 2 -\theta \right)\cdot \sin \left(\dfrac {3{\mathrm \pi}} 2 +\theta \right)} {-\sin \left({\mathrm \pi}+\theta \right)} =\dfrac {\left(-\sin\theta \right)\cdot \left(-\cos\theta \right)} {\sin\theta } =\cos\theta $．
-
若 $f\left(\theta \right)=\dfrac 1 3 $，求 $\tan\theta $ 的值；标注答案$2\sqrt 2$ 或 $-2\sqrt 2$解析由题意得 $f\left(\theta \right)=\cos\theta =\dfrac 1 3 >0$，
故 $\theta $ 为第一或第四象限角．
当 $\theta $ 为第一象限角时，$\sin\theta =\sqrt {1-\cos^2\theta }=\dfrac {2\sqrt 2} 3 $，$\tan\theta =\dfrac {\sin\theta } {\cos\theta } =2\sqrt 2$；
当 $\theta $ 为第四象限角时，$\sin\theta =-\sqrt {1-\cos^2\theta }=-\dfrac {2\sqrt 2} 3 $，$\tan\theta =\dfrac {\sin\theta } {\cos\theta } =-2\sqrt 2$． -
若 $f\left(\dfrac {\mathrm \pi} 6 -\theta \right)=\dfrac 1 3 $，求 $f\left(\dfrac {5{\mathrm \pi}} 6 +\theta \right)$ 的值．标注答案$-\dfrac 13$解析由题意得 $f\left(\dfrac {\mathrm \pi} 6 -\theta \right)=\cos \left(\dfrac {\mathrm \pi} 6 -\theta \right)=\dfrac 1 3 $，所以$$f\left(\dfrac {5{\mathrm \pi}} 6 +\theta \right)=\cos \left(\dfrac {5{\mathrm \pi}} 6 +\theta \right)=\cos \left[{\mathrm \pi}-\left(\dfrac {\mathrm \pi} 6 -\theta \right)\right]=-\cos \left(\dfrac {\mathrm \pi} 6 -\theta \right)=-\dfrac 1 3.$$
题目
问题1
答案1
解析1
备注1
问题2
答案2
解析2
备注2
问题3
答案3
解析3
备注3
