《2025年暑假作业甘肃教育出版社高一数学》

5. 已知$\sin(\pi+\theta)= -\frac{1}{3}$,则$\cos2\theta$等于(D)
A.$-\frac{7}{9}$
B.$-\frac{5}{9}$
C.$\frac{5}{9}$
D.$\frac{7}{9}$

答案： D

6. 已知$\sin(\frac{\pi}{4}-\alpha)= \frac{5}{13}$,$0＜\alpha＜\frac{\pi}{4}$,则$\cos2\alpha$的值为(D)
A.$\frac{117}{169}$
B.$\frac{118}{169}$
C.$\frac{119}{169}$
D.$\frac{120}{169}$

答案： D

7. 已知$\tan\alpha=2$,$\tan(\alpha+\beta)= -1$,则$\tan\beta=$3.

答案： 3

8. 已知$\sin\alpha=-\frac{4}{5}$,$\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$,则$\sin2\alpha=$$-\frac{24}{25}$.

答案： $-\frac{24}{25}$

9. 函数$y= \sqrt{3}\cos4x+\sin4x$的最小正周期为$\frac{\pi}{2}$.

答案： $\frac{\pi}{2}$

10. 已知$\sin\theta+\cos\theta=\frac{1}{5}$,且$\frac{\pi}{2}\leq\theta\leq\frac{3\pi}{4}$,则$\cos2\theta=$$-\frac{7}{25}$.

答案： $-\frac{7}{25}$

11. 设$\alpha\in(0,\frac{\pi}{3})$,满足$\sqrt{6}\sin\alpha+\sqrt{2}\cos\alpha=\sqrt{3}$.
(1)求$\cos(\alpha+\frac{\pi}{6})$的值；
(2)求$\cos(2\alpha+\frac{\pi}{12})$的值.

答案： 解：
(1) $\because \sqrt{6} \sin \alpha+\sqrt{2} \cos \alpha=\sqrt{3}$，
$\therefore 2 \sqrt{2}\left(\frac{\sqrt{3}}{2} \sin \alpha+\frac{1}{2} \cos \alpha\right)=\sqrt{3}$，
$\therefore \sin \left(\alpha+\frac{\pi}{6}\right)=\frac{\sqrt{6}}{4}$，
$\because \alpha \in\left(0, \frac{\pi}{3}\right), \therefore \alpha+\frac{\pi}{6} \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$，
$\therefore \cos \left(\alpha+\frac{\pi}{6}\right)=\frac{\sqrt{10}}{4}$
(2) 由
(1) 可得 $\cos \left(2 \alpha+\frac{\pi}{3}\right)$
$=2 \cos ^2\left(\alpha+\frac{\pi}{6}\right)-1=2 ×\left(\frac{\sqrt{10}}{4}\right)^2-1=\frac{1}{4}$
$\because \alpha \in\left(0, \frac{\pi}{3}\right), \therefore 2 \alpha+\frac{\pi}{3} \in\left(\frac{\pi}{3}, \pi\right)$，
$\therefore \sin \left(2 \alpha+\frac{\pi}{3}\right)=\frac{\sqrt{15}}{4}$
$\therefore \cos \left(2 \alpha+\frac{\pi}{12}\right)=\cos \left[\left(2 \alpha+\frac{\pi}{3}\right)-\frac{\pi}{4}\right]$
$=\cos \left(2 \alpha+\frac{\pi}{3}\right) \cos \frac{\pi}{4}+\sin \left(2 \alpha+\frac{\pi}{3}\right) \sin \frac{\pi}{4}$
$=\frac{\sqrt{30}+\sqrt{2}}{8}$.