《2025年世纪金榜高中全程复习方略高中数学A版》

4. (记错公式)下列等式恒成立的是 ( )
A. $\cos(-\alpha)=-\cos\alpha$
B. $\sin(360^{\circ}-\alpha)=\sin\alpha$
C. $\tan(2\pi-\alpha)=\tan(\pi+\alpha)$
D. $\cos(\pi+\alpha)=\cos(\pi-\alpha)$

答案： 4. D 因为$\cos(-\alpha)=\cos\alpha$;$\sin(360^{\circ}-\alpha)=-\sin\alpha$;$\tan(2\pi-\alpha)=-\tan\alpha$,$\tan(\pi+\alpha)=\tan\alpha$;$\cos(\pi+\alpha)=-\cos\alpha$,$\cos(\pi-\alpha)=-\cos\alpha$.

[例1] (1) (2023·惠州模拟)已知$\tan\alpha = 2$，$\pi＜\alpha＜\frac{3\pi}{2}$，则$\cos\alpha-\sin\alpha=$ ( )
A. $\frac{\sqrt{5}}{5}$
B. $-\frac{\sqrt{5}}{5}$
C. $\frac{3\sqrt{5}}{5}$
D. $-\frac{3\sqrt{5}}{5}$
(2) 已知$\cos\alpha=-\frac{5}{13}$，则$13\sin\alpha + 5\tan\alpha=$________.

答案：
(1)A 因为$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=2$,且$\sin^{2}\alpha+\cos^{2}\alpha = 1$,$\pi＜\alpha＜\frac{3\pi}{2}$,所以$\sin\alpha=-\frac{2\sqrt{5}}{5}$,$\cos\alpha=-\frac{\sqrt{5}}{5}$,所以$\cos\alpha-\sin\alpha=-\frac{\sqrt{5}}{5}-(-\frac{2\sqrt{5}}{5})=\frac{\sqrt{5}}{5}$.
(2)【解析】因为$\cos\alpha=-\frac{5}{13}＜0$且$\cos\alpha\neq - 1$,所以$\alpha$是第二或第三象限角.
①若$\alpha$是第二象限角,则$\sin\alpha=\sqrt{1-\cos^{2}\alpha}=\sqrt{1 - (-\frac{5}{13})^{2}}=\frac{12}{13}$,所以$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5}$,此时$13\sin\alpha+5\tan\alpha=13\times\frac{12}{13}+5\times(-\frac{12}{5})=0$.
②若$\alpha$是第三象限角,则$\sin\alpha=-\sqrt{1-\cos^{2}\alpha}=-\sqrt{1 - (-\frac{5}{13})^{2}}=-\frac{12}{13}$,所以$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$,此时,$13\sin\alpha+5\tan\alpha=13\times(-\frac{12}{13})+5\times\frac{12}{5}=0$.
综上,$13\sin\alpha+5\tan\alpha = 0$.
答案:0

[例2] (1) 已知$\frac{\sin\alpha+3\cos\alpha}{3\cos\alpha-\sin\alpha}=5$，则$\cos^{2}\alpha+\frac{1}{2}\sin2\alpha=$ ( )
A. $\frac{3}{5}$
B. $-\frac{3}{5}$
C. $-3$
D. $3$
(2) (2023·黄冈模拟)已知$\alpha\in\mathbf{R}$，$\sin^{2}\alpha + 4\sin\alpha\cos\alpha+4\cos^{2}\alpha=\frac{5}{2}$，则$\tan\alpha=$________.

答案：
(1)A 因为$\frac{\sin\alpha+3\cos\alpha}{3\cos\alpha-\sin\alpha}=5$,所以$\frac{\tan\alpha + 3}{3-\tan\alpha}=5$,解得$\tan\alpha=2$,故$\cos^{2}\alpha+\frac{1}{2}\sin2\alpha=\frac{\cos^{2}\alpha+\sin\alpha\cos\alpha}{\cos^{2}\alpha+\sin^{2}\alpha}=\frac{1+\tan\alpha}{1+\tan^{2}\alpha}=\frac{3}{5}$.
(2)【解析】因为$\sin^{2}\alpha+4\sin\alpha\cos\alpha+4\cos^{2}\alpha=\frac{\sin^{2}\alpha+4\sin\alpha\cos\alpha+4\cos^{2}\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{\tan^{2}\alpha+4\tan\alpha + 4}{\tan^{2}\alpha+1}=\frac{5}{2}$,所以$3\tan^{2}\alpha-8\tan\alpha-3 = 0$,解得$\tan\alpha=3$或$-\frac{1}{3}$.
答案:$3$或$-\frac{1}{3}$

[例3] (1) 已知$\sin\alpha+\cos\alpha=-\frac{7}{13}$，$\alpha\in(\frac{\pi}{2},\pi)$，则$\sin\alpha-\cos\alpha=$ ( )
A. $\frac{12}{13}$
B. $-\frac{12}{13}$
C. $\frac{17}{13}$
D. $-\frac{17}{13}$
(2) 已知$\tan\theta+\frac{1}{\tan\theta}=4$，则$\sin^{4}\theta+\cos^{4}\theta=$ ( )
A. $\frac{3}{8}$
B. $\frac{1}{2}$
C. $\frac{3}{4}$
D. $\frac{7}{8}$

答案：
(1)C 因为$\sin\alpha+\cos\alpha=-\frac{7}{13}$,所以$(\sin\alpha+\cos\alpha)^{2}=(-\frac{7}{13})^{2}$,即$\sin^{2}\alpha+\cos^{2}\alpha+2\sin\alpha\cos\alpha=(-\frac{7}{13})^{2}$,$2\sin\alpha\cos\alpha=-\frac{120}{169}$,所以$\sin^{2}\alpha+\cos^{2}\alpha-2\sin\alpha\cos\alpha=\frac{289}{169}$,即$(\sin\alpha-\cos\alpha)^{2}=\frac{289}{169}$,因为$\alpha\in(\frac{\pi}{2},\pi)$,所以$\sin\alpha-\cos\alpha＞0$,$\sin\alpha-\cos\alpha=\frac{17}{13}$.
(2)D 由题意得$\tan\theta+\frac{1}{\tan\theta}=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}=4$,则$\sin\theta\cos\theta=\frac{1}{4}$,故$\sin^{4}\theta+\cos^{4}\theta=(\sin^{2}\theta+\cos^{2}\theta)^{2}-2\sin^{2}\theta\cos^{2}\theta=1 - 2\times\frac{1}{16}=\frac{7}{8}$.

[1. (2023·安康模拟)已知$\tan\theta=\frac{1}{2}$，则$\frac{\sin^{2}\theta+\sin\theta}{\cos^{3}\theta+\sin\theta\cos^{2}\theta}=$ ( )
A. $\frac{1}{2}$
B. 2
C. $\frac{1}{6}$
D. 6

答案： A 因为$\tan\theta=\frac{1}{2}$,所以$\frac{\sin^{3}\theta+\sin\theta}{\cos^{3}\theta+\sin\theta\cos^{2}\theta}=\frac{\sin^{3}\theta+\sin\theta(\sin^{2}\theta+\cos^{2}\theta)}{\cos^{3}\theta+\sin\theta\cos^{2}\theta}=\frac{2\sin^{3}\theta+\sin\theta\cos^{2}\theta}{\cos^{3}\theta+\sin\theta\cos^{2}\theta}=\frac{2\tan^{3}\theta+\tan\theta}{1+\tan\theta}=\frac{2\times(\frac{1}{2})^{3}+\frac{1}{2}}{1+\frac{1}{2}}=\frac{\frac{3}{4}}{\frac{3}{2}}=\frac{1}{2}$.

2. (2023·梅州模拟)已知$\cos\alpha=\frac{1}{3}$，且$\alpha$为第四象限角，则$\tan\alpha=$________.

答案： 【解析】因为$\alpha$为第四象限角,所以$\sin\alpha＜0$,所以$\sin\alpha=-\sqrt{1-\cos^{2}\alpha}=-\frac{2\sqrt{2}}{3}$,所以$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-2\sqrt{2}$.
答案:$-2\sqrt{2}$

3. (2023·聊城模拟)已知$\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$，且$\sin\alpha+\cos\alpha=\frac{\sqrt{5}}{5}$，则$\tan\alpha$的值为________.

答案： 【解析】因为$\sin\alpha+\cos\alpha=\frac{\sqrt{5}}{5}$,所以$\sin^{2}\alpha+\cos^{2}\alpha+2\sin\alpha\cos\alpha=\frac{1}{5}$,所以$\sin\alpha\cos\alpha=-\frac{2}{5}$,所以$\sin^{2}\alpha+\cos^{2}\alpha-2\sin\alpha\cos\alpha=\frac{9}{5}=(\sin\alpha-\cos\alpha)^{2}$.又$\sin\alpha\cos\alpha＜0$,$\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$,所以$\alpha\in(-\frac{\pi}{2},0)$,所以$\sin\alpha＜0$,$\cos\alpha＞0$,所以$\cos\alpha-\sin\alpha=\frac{3\sqrt{5}}{5}$,所以$\sin\alpha=-\frac{\sqrt{5}}{5}$,$\cos\alpha=\frac{2\sqrt{5}}{5}$,所以$\tan\alpha=-\frac{1}{2}$.
答案:$-\frac{1}{2}$