《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》

1.$\sin2024°$的值为(B)

A.$\sin44°$
B.$-\sin44°$
C.$\sin48°$
D.$-\sin48°$

答案： 1.B $\sin 2024° = \sin(5 × 360° + 224°) = \sin(180° + 44°) = -\sin 44°$. 故选 B.

2.$\sin^2150°+\sin^2135°+2\sin210°+\cos^2225°$的值是(A)

A.$\frac{1}{4}$
B.$\frac{3}{4}$
C.$\frac{11}{4}$
D.$\frac{9}{4}$

答案： 2.A 原式$= \sin^2 30° + \sin^2 45° - 2 \sin 30° + \cos^2 45° = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 - 2 × \frac{1}{2} + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{4}$.

3.$\sin(\pi-2)+\cos(\pi-2)$的值为(B)

A.$\sin2+\cos2$
B.$\sin2 - \cos2$
C.$-\sin2+\cos2$
D.$-\sin2 - \cos2$

答案： 3.B

4.已知$\sin(\frac{\pi}{4}+\alpha)=\frac{\sqrt{3}}{2}$,则$\sin(\frac{3\pi}{4}-\alpha)$的值为(C)

A.$\frac{1}{2}$
B.$-\frac{1}{2}$
C.$\frac{\sqrt{3}}{2}$
D.$-\frac{\sqrt{3}}{2}$

答案： 4.C $\because \sin\left(\frac{\pi}{4} + \alpha\right) = \frac{\sqrt{3}}{2}$,
$\therefore \sin\left(\frac{3\pi}{4} - \alpha\right) = \sin\left[\pi - \left(\frac{\pi}{4} + \alpha\right)\right] = \sin\left(\frac{\pi}{4} + \alpha\right) = \frac{\sqrt{3}}{2}$.

5.已知角$\alpha$和$\beta$的终边关于$x$轴对称,则下列各式中正确的是(C)

A.$\sin\alpha=\sin\beta$
B.$\sin(\alpha - 2\pi)=\sin\beta$
C.$\cos\alpha=\cos\beta$
D.$\cos(2\pi-\alpha)=-\cos\beta$

答案： 5.C 由角$\alpha$和$\beta$的终边关于$x$轴对称,可知$\beta = -\alpha + 2k\pi (k \in \mathbf{Z})$,故$\cos \alpha = \cos \beta$.

6.已知函数$f(x)=\cos\frac{x}{2}$,则下列等式成立的是 (D)

A.$f(2\pi-x)=f(x)$
B.$f(2\pi+x)=f(x)$
C.$f(-x)=-f(x)$
D.$f(-x)=f(x)$

答案： 6.D 对于A,$f(2\pi - x) = \cos \frac{2\pi - x}{2} = \cos\left(\pi - \frac{x}{2}\right) = -\cos \frac{x}{2} \neq f(x)$, A 不成立; 对于 B,$f(2\pi + x) = \cos \frac{2\pi + x}{2} = \cos\left(\frac{x}{2} + \pi\right) = -\cos \frac{x}{2} \neq f(x)$, B 不成立; 对于 C,$f(-x) = \cos \frac{-x}{2} = \cos \frac{x}{2} = f(x) \neq -f(x)$, C 不成立, D 成立. 故选 D.

7.$\sin780° =$$\frac{\sqrt{3}}{2}$.

答案： 7.$\frac{\sqrt{3}}{2}$ $\sin 780° = \sin(2 × 360° + 60°) = \sin 60° = \frac{\sqrt{3}}{2}$.

8.已知$\cos(508°-\alpha)=\frac{12}{13}$,则$\cos(212°+\alpha)=$$\frac{12}{13}$.

答案： 8.$\frac{12}{13}$ 由于$\cos(508° - \alpha) = \cos(360° + 148° - \alpha) = \cos(148° - \alpha) = \frac{12}{13}$,所以$\cos(212° + \alpha) = \cos(360° + \alpha - 148°) = \cos(\alpha - 148°) = \cos(148° - \alpha) = \frac{12}{13}$.

9.设函数$f(x)=a\sin(\pi x+\alpha)+b\cos(\pi x+\beta)$,其中$a,b,\alpha,\beta$都是非零实数,且满足$f(2018)= -1$,则$f(2019)$的值为1.

答案： 9.1 因为$f(2018) = a \sin(2018\pi + \alpha) + b \cos(2018\pi + \beta) = -1$,
所以$f(2019) = a · \sin(2019\pi + \alpha) + b \cos(2019\pi + \beta) = a \sin[\pi + (2018\pi + \alpha)] + b \cos[\pi + (2018\pi + \beta)] = -[a \sin(2018\pi + \alpha) + b \cos(2018\pi + \beta)] = 1$.

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

答案： 10.$\because \frac{\sin(2\pi - \alpha) \cos(\pi + \alpha)}{\cos(\pi - \alpha) \sin(3\pi - \alpha) \sin(-\pi - \alpha)} = \frac{\sin(-\alpha) · (-\cos \alpha)}{-\cos \alpha · \sin(\pi - \alpha) · [-\sin(\pi + \alpha)]} = \frac{-\sin \alpha · (-\cos \alpha)}{-\cos \alpha · \sin \alpha · \sin \alpha} = 3$.
$\therefore \sin \alpha = -\frac{1}{3}$.
又$\sin(-\alpha) = -\sin \alpha, \therefore \sin(-\alpha) = \frac{1}{3}$.

1.下列各式错误的是 (B)

A.$\sin(\alpha+180°)=-\sin\alpha$
B.$\cos(-\alpha+\beta)=-\cos(\alpha-\beta)$
C.$\sin(-\alpha-360°)=-\sin\alpha$
D.$\cos(-\alpha-\beta)=\cos(\alpha+\beta)$

答案： 1.B 对于B,$\cos(-\alpha + \beta) = \cos[-(\alpha - \beta)] = \cos(\alpha - \beta)$, B错误,由诱导公式知 A、C、D 都正确, 故选 B.

2.已知$A=\frac{\sin(k\pi+\alpha)}{\sin\alpha}+\frac{\cos(k\pi+\alpha)}{\cos\alpha}(k\in\mathbf{Z})$,则$A$构成的集合是
(C)

A.$\{ -1,1,-2,2\}$
B.$\{1,-1\}$
C.$\{2,-2\}$
D.$\{ -2,-1,0,1,2\}$

答案： 2.C 当$k$为偶数时,$A = 2$;当$k$为奇数时,$A = -2$. 故$A$构成的集合为$\{ -2, 2 \}$.