《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

4. (多选)若角$A,B,C$是$\triangle ABC$的三个内角,则下列等式中一定成立的是
(BCD)

A.$\cos(A+B)=\cos C$
B.$\sin(A+B)=\sin C$
C.$\cos\frac{A+C}{2}=\sin\frac{B}{2}$
D.$\sin\frac{B+C}{2}=\cos\frac{A}{2}$

答案： 4.BCD $\cos (A+B)=\cos (\pi-C)=-\cos C$,A错误;$\sin (A+B)=$
$\sin (\pi-C)=\sin C$,B正确;$\cos \frac{A+C}{2}=\cos \frac{\pi-B}{2}=$
$\cos \left(\frac{\pi}{2}-\frac{B}{2}\right)=\sin \frac{B}{2}$,C正确;$\sin \frac{B+C}{2}=\sin \frac{\pi-A}{2}=$
$\sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\cos \frac{A}{2}$,D正确. 故选BCD.

5. 已知$\sin(\frac{\pi}{4}-\alpha)=\frac{1}{3}$,则$\sin(\frac{3\pi}{4}+\alpha)=$$\frac{1}{3}$

答案： 5.$\frac{1}{3}\sin \left(\frac{3\pi}{4}+\alpha\right)=\sin \left[\pi-\left(\frac{\pi}{4}-\alpha\right)\right]=\sin \left(\frac{\pi}{4}-\alpha\right)=\frac{1}{3}$.

6. 化简$\frac{\cos(\frac{5}{2}\pi-\alpha)\cos(-\alpha)}{\sin(\frac{3}{2}\pi+\alpha)\cos(\frac{21}{2}\pi-\alpha)}=$

答案： 6.-1 原式$=\frac{\cos \left[2\pi+\left(\frac{\pi}{2}-\alpha\right)\right]\cos \alpha}{\sin \left[\pi+\left(\frac{\pi}{2}+\alpha\right)\right]\cos \left[10\pi+\left(\frac{\pi}{2}-\alpha\right)\right]}$
$=\frac{\cos \left(\frac{\pi}{2}-\alpha\right)\cos \alpha}{-\sin \left(\frac{\pi}{2}+\alpha\right)\cos \left(\frac{\pi}{2}-\alpha\right)}=\frac{\sin \alpha\cos \alpha}{-\cos \alpha · \sin \alpha}=-1$.

7. 已知$\sin(\alpha-3\pi)=2\cos(\alpha-4\pi)$,
求$\frac{\sin(\pi-\alpha)+5\cos(2\pi-\alpha)}{2\sin(\frac{3\pi}{2}-\alpha)-\sin(-\alpha)}$的值.

答案： 7.由$\sin (\alpha-3\pi)=2\cos (\alpha-4\pi)$,
得$\sin (\alpha-\pi)=2\cos \alpha$,
即$\sin \alpha=-2\cos \alpha$.
$\therefore \frac{\sin (\pi-\alpha)+5\cos (2\pi-\alpha)}{2\sin \left(\frac{3\pi}{2}-\alpha\right)-\sin (-\alpha)}$
$=\frac{\sin \alpha+5\cos \alpha}{-2\cos \alpha+\sin \alpha}$
$=\frac{-2\cos \alpha+5\cos \alpha}{-2\cos \alpha-2\cos \alpha}=-\frac{3}{4}$.

8. 已知$f(\alpha)=$
$\frac{\sin(\pi-\alpha)\cos(2\pi-\alpha)\sin(\frac{3\pi}{2}-\alpha)}{\cos(\pi+\alpha)\cos(\frac{7\pi}{2}-\alpha)}$.
(1)化简$f(\alpha)$;
(2)若$\theta$是第三象限角,且$f(\theta+\frac{\pi}{4})=\frac{3}{5}$,求$f(\theta-\frac{\pi}{4})$的值.

答案： 8.
(1)$f(\alpha)=\frac{\sin \alpha\cos \alpha(-\cos \alpha)}{(-\cos \alpha)(-\sin \alpha)}=-\cos \alpha$.
(2)因为$f(\alpha)=-\cos \alpha,f\left(\theta+\frac{\pi}{4}\right)=\frac{3}{5}$,
所以$\cos \left(\theta+\frac{\pi}{4}\right)=-\frac{3}{5}$,
又因为$\theta$是第三象限角,所以$\theta+\frac{\pi}{4}$为第三象限角,
所以$\sin \left(\theta+\frac{\pi}{4}\right)=-\sqrt{1-\cos ^{2}\left(\theta+\frac{\pi}{4}\right)}=-\frac{4}{5}$,
故$f\left(\theta-\frac{\pi}{4}\right)=-\cos \left(\theta-\frac{\pi}{4}\right)=-\cos \left[\frac{\pi}{2}-\left(\theta+\frac{\pi}{4}\right)\right]=-\sin \left(\theta+\frac{\pi}{4}\right)=\frac{4}{5}$.