《2025年中职一年级假期作业语文数学英语》

3. 已知$tanα= 3$,则$tan(α+\frac {π}{4})$的值是(D)
A.1
B.$\frac {1}{2}$
C.2
D.-2

答案： 3. D 【解析】$\tan \left(\alpha+\frac{\pi}{4}\right)=\frac{\tan \alpha+1}{1-\tan \alpha}=-2$.

4. 已知$sin(\frac {π}{6}-α)= cos(\frac {π}{6}+α)$,则$tanα$的值是(A)
A.-1
B.0
C.$\frac {1}{2}$
D.1

答案： 4. A 【解析】由题意可得 $\frac{1}{2} \cos \alpha-\frac{\sqrt{3}}{2} \sin \alpha=\frac{\sqrt{3}}{2} \cos \alpha-\frac{1}{2} \sin \alpha$，即 $\sin \alpha+\cos \alpha=0$，则 $\tan \alpha=-1$，故选 A.

5. 若α,β为锐角,且满足$cosα= \frac {4}{5}$,$cos(α+β)= \frac {5}{13}$,则$sinβ$的值为(B)
A.$-\frac {16}{65}$
B.$\frac {33}{65}$
C.$\frac {56}{65}$
D.$\frac {63}{65}$

答案： 5. B 【解析】因 $\alpha, \beta$ 为锐角，$\cos \alpha=\frac{4}{5}, \cos (\alpha+\beta)=\frac{5}{13}$，故 $\sin \alpha=\frac{3}{5}$，$\sin (\alpha+\beta)=\frac{12}{13}$，故 $\sin \beta=\sin [(\alpha+\beta)-\alpha]=\frac{12}{13} × \frac{4}{5}-\frac{5}{13} × \frac{3}{5}=\frac{33}{65}$，故选 B.

6. 已知$cosα= -\frac {4}{5}$,且$α∈(\frac {π}{2},π)$,则$tan(α+\frac {π}{4})=$(B)
A.$-\frac {1}{7}$
B.$\frac {1}{7}$
C.-7
D.7

答案： 6. B 【解析】因为 $\cos \alpha=-\frac{4}{5}, \alpha \in\left(\frac{\pi}{2}, \pi\right)$，所以 $\sin \alpha=\frac{3}{5}, \tan \alpha=-\frac{3}{4}$，$\tan \left(\alpha+\frac{\pi}{4}\right)=\frac{1}{7}$，故选 B.

7. 下列各式中,值为$\frac {\sqrt {3}}{2}$的是(B).
A.$sin15^{\circ }cos15^{\circ }$
B.$cos^{2}\frac {π}{12}-sin^{2}\frac {π}{12}$
C.$\frac {1+tan15^{\circ }}{1-tan15^{\circ }}$
D.$\sqrt {\frac {1+cos30^{\circ }}{2}}$

答案： 7. B 【解析】选项 A，$\sin 15^{\circ} \cos 15^{\circ}=\frac{1}{2} \sin 30^{\circ}=\frac{1}{4}$；选项 B，$\cos ^{2} \frac{\pi}{12}-\sin ^{2} \frac{\pi}{12}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$；选项 C，$\frac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}=\tan 60^{\circ}=\sqrt{3}$；选项 D，$\sqrt{\frac{1+\cos 30^{\circ}}{2}}=\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$，故选 B.

8. 若α为第三象限角,则$\frac {cosα}{\sqrt {1-sin^{2}α}}+\frac {2sinα}{\sqrt {1-cos^{2}α}}$的值是(D).
A.1
B.-1
C.3
D.-3

答案： 8. D 【解析】因为角 $\alpha$ 为第三象限角，$\frac{\cos \alpha}{\sqrt{1-\sin ^{2} \alpha}}+\frac{2 \sin \alpha}{\sqrt{1-\cos ^{2} \alpha}}=\frac{\cos \alpha}{\sqrt{\cos ^{2} \alpha}}+\frac{2 \sin \alpha}{\sqrt{\sin ^{2} \alpha}}=\frac{\cos \alpha}{-\cos \alpha}+\frac{2 \sin \alpha}{-\sin \alpha}=-1+(-2)=-3$，故选 D.

9. 已知$tanα= \sqrt {3}$,$π＜α＜\frac {3π}{2}$,那么$cosα-sinα$的值是(C).
A.$\frac {1+\sqrt {3}}{2}$
B.$\frac {1-\sqrt {3}}{2}$
C.$\frac {-1+\sqrt {3}}{2}$
D.$-\frac {1+\sqrt {3}}{2}$

答案： 9. C 【解析】$\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\sqrt{3}, \sin \alpha=\sqrt{3} \cos \alpha, \sin ^{2} \alpha=3 \cos ^{2} \alpha, \sin ^{2} \alpha+\cos ^{2} \alpha=3 \cos ^{2} \alpha+\cos ^{2} \alpha=4 \cos ^{2} \alpha=1, \cos ^{2} \alpha=\frac{1}{4}$，因为 $\pi＜\alpha＜\frac{3 \pi}{2}, \cos \alpha=-\frac{1}{2}, \sin \alpha=\sqrt{3} \cos \alpha=\sqrt{3} ×\left(-\frac{1}{2}\right)=-\frac{\sqrt{3}}{2}$，则 $\cos \alpha-\sin \alpha=-\frac{1}{2}-\left(-\frac{\sqrt{3}}{2}\right)=\frac{-1+\sqrt{3}}{2}$，故选 C.

10. 设α为第三象限角,则$\frac {sinα}{cosα}\cdot \sqrt {\frac {1}{sin^{2}α}-1}=$(A).
A.1
B.$tan^{2}α$
C.$-tan^{2}α$
D.-1

答案： 10. A 【解析】因为 $\alpha$ 为第三象限角，$\frac{\sin \alpha}{\cos \alpha} \cdot \sqrt{\frac{1}{\sin ^{2} \alpha}-1}=\frac{\sin \alpha}{\cos \alpha} \cdot \sqrt{\frac{1-\sin ^{2} \alpha}{\sin ^{2} \alpha}}=\frac{\sin \alpha}{\cos \alpha} \cdot \sqrt{\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}}=\frac{\sin \alpha}{\cos \alpha} \cdot \frac{-\cos \alpha}{-\sin \alpha}=1$，故选 A.