《2025年世纪金榜高中全程复习方略高中数学A版》

[例4] (1) (2023·黑龙江模拟)$\sin495^{\circ}=$ ( )
A. 1
B. $-\frac{1}{2}$
C. $\frac{\sqrt{3}}{2}$
D. $\frac{\sqrt{2}}{2}$
(2) 已知$x\in\mathbf{R}$，则下列等式恒成立的是 ( )
A. $\sin(3\pi - x)=-\sin x$
B. $\sin\frac{\pi - x}{2}=-\cos\frac{x}{2}$
C. $\cos(\frac{5\pi}{2}+3x)=\sin3x$
D. $\cos(\frac{3\pi}{2}-2x)=-\sin2x$
(3) 已知$\sin(\alpha+\frac{\pi}{12})=\frac{1}{3}$，则$\cos(\alpha+\frac{7}{12}\pi)$的值为________；$\sin(\frac{11}{12}\pi-\alpha)$的值为________.

答案：
(1)D $\sin495^{\circ}=\sin(360^{\circ}+135^{\circ})=\sin135^{\circ}=\sin(180^{\circ}-45^{\circ})=\sin45^{\circ}=\frac{\sqrt{2}}{2}$.
(2)D $\sin(3\pi - x)=\sin(\pi - x)=\sin x$,$\sin\frac{\pi - x}{2}=\sin(\frac{\pi}{2}-\frac{x}{2})=\cos\frac{x}{2}$,$\cos(\frac{5\pi}{2}+3x)=\cos(\frac{\pi}{2}+3x)=-\sin3x$,$\cos(\frac{3\pi}{2}-2x)=-\sin2x$.
(3)【解析】$\cos(\alpha+\frac{7\pi}{12})=\cos(\frac{\pi}{2}+\alpha+\frac{\pi}{12})=-\sin(\alpha+\frac{\pi}{12})=-\frac{1}{3}$.$\sin(\frac{11\pi}{12}-\alpha)=\sin[\pi-(\alpha+\frac{\pi}{12})]=\sin(\alpha+\frac{\pi}{12})=\frac{1}{3}$.
答案:$-\frac{1}{3}$ $\frac{1}{3}$

1. $\frac{\tan(\pi-\alpha)\cos(2\pi-\alpha)\sin(-\alpha+\frac{3\pi}{2})}{\cos(-\alpha-\pi)\sin(-\pi-\alpha)}$的值为 ( )
A. $-2$
B. $-1$
C. 1
D. 2

答案： B 原式$=\frac{-\tan\alpha\cdot\cos\alpha\cdot(-\cos\alpha)}{\cos(\pi+\alpha)\cdot[-\sin(\pi+\alpha)]}=\frac{\tan\alpha\cdot\cos^{2}\alpha}{-\cos\alpha\cdot\sin\alpha}=\frac{-\sin\alpha}{\cos\alpha}\cdot\frac{\cos\alpha}{\sin\alpha}=-1$.

2. (2023·茂名模拟)已知$\sin(\theta-\frac{\pi}{6})=\frac{1}{2}$，则$\cos(\theta+\frac{\pi}{3})=$ ( )
A. $-\frac{\sqrt{3}}{2}$
B. $-\frac{1}{2}$
C. $\frac{1}{2}$
D. $\frac{\sqrt{3}}{2}$

答案： B $\cos(\theta+\frac{\pi}{3})=\cos(\theta-\frac{\pi}{6}+\frac{\pi}{2})=-\sin(\theta-\frac{\pi}{6})=-\frac{1}{2}$.

[例5] (1) 已知$\sin(\pi-\alpha)+\sin(\alpha-\frac{\pi}{2})=\frac{1}{2}$，则$\frac{\cos(\frac{3}{2}\pi+\alpha)}{1+\tan(-\alpha)}$的值为 ( )
A. $-\frac{3}{4}$
B. $\frac{3}{4}$
C. $-\frac{3}{16}$
D. $\frac{3}{16}$
(2) (2023·阳泉模拟)已知$\sin(\alpha+\frac{\pi}{6})=\frac{\sqrt{3}}{3}$，且$\alpha\in(-\frac{\pi}{4},\frac{\pi}{4})$，则$\sin(\frac{\pi}{3}-\alpha)=$________.

答案：
(1)A 由已知得$\sin\alpha-\cos\alpha=\frac{1}{2}$,两边平方得$1 - 2\sin\alpha\cos\alpha=\frac{1}{4}$,解得$\sin\alpha\cos\alpha=\frac{3}{8}$,则原式$=\frac{\sin\alpha}{1-\tan\alpha}=\frac{\sin\alpha}{1-\frac{\sin\alpha}{\cos\alpha}}=\frac{\sin\alpha\cos\alpha}{\cos\alpha-\sin\alpha}=-\frac{3}{4}$.
(2)【解析】因为$\alpha\in(-\frac{\pi}{4},\frac{\pi}{4})$,所以$\alpha+\frac{\pi}{6}\in(-\frac{\pi}{12},\frac{5\pi}{12})$,故$\cos(\alpha+\frac{\pi}{6})＞0$,所以$\cos(\alpha+\frac{\pi}{6})=\sqrt{1 - (\frac{\sqrt{3}}{3})^{2}}=\frac{\sqrt{6}}{3}$.$\sin(\frac{\pi}{3}-\alpha)=\sin[\frac{\pi}{2}-(\alpha+\frac{\pi}{6})]=\cos(\alpha+\frac{\pi}{6})=\frac{\sqrt{6}}{3}$.
答案:$\frac{\sqrt{6}}{3}$

1. 若$\alpha\in(0,\pi)$，$\sin(\pi-\alpha)+\cos\alpha=\frac{\sqrt{2}}{3}$，则$\sin\alpha-\cos\alpha$的值为 ( )
A. $\frac{\sqrt{2}}{3}$
B. $-\frac{\sqrt{2}}{3}$
C. $\frac{4}{3}$
D. $-\frac{4}{3}$

答案： C 由诱导公式得,$\sin(\pi-\alpha)+\cos\alpha=\sin\alpha+\cos\alpha=\frac{\sqrt{2}}{3}$,所以$(\sin\alpha+\cos\alpha)^{2}=1 + 2\sin\alpha\cos\alpha=\frac{2}{9}$,则$2\sin\alpha\cos\alpha=-\frac{7}{9}＜0$,因为$\alpha\in(0,\pi)$,所以$\sin\alpha＞0$,所以$\cos\alpha＜0$,所以$\sin\alpha-\cos\alpha＞0$,因为$(\sin\alpha-\cos\alpha)^{2}=1 - 2\sin\alpha\cos\alpha=\frac{16}{9}$,所以$\sin\alpha-\cos\alpha=\frac{4}{3}$.

2. (2023·成都模拟)已知$\sin\alpha = 2\cos\alpha$，则$\frac{\sin\alpha-\sin^{3}\alpha}{\sin(\alpha+\frac{\pi}{2})}=$ ( )
A. $\frac{3}{5}$
B. $\frac{2}{5}$
C. $-\frac{2}{5}$
D. $-\frac{3}{5}$

答案： B 由$\sin\alpha=2\cos\alpha$,显然$\cos\alpha\neq0$,可得$\tan\alpha=2$.因为$\frac{\sin\alpha-\sin^{3}\alpha}{\sin(\alpha+\frac{\pi}{2})}=\frac{\sin\alpha(1 - \sin^{2}\alpha)}{\cos\alpha}=\frac{\sin\alpha\cdot\cos^{2}\alpha}{\cos\alpha}=\frac{\sin\alpha\cdot\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{\tan\alpha}{\tan^{2}\alpha+1}=\frac{2}{5}$.