《2025年金版新学案高三总复习数学北师大版》

4. (多选题)(2025·安徽马鞍山模拟)若角$A,B,C$是$\triangle ABC$的三个内角，则下列结论中一定成立的是(AD)

A.$\cos(A + B)=-\cos C$
B.$\tan(B + C)=\tan A$
C.$\cos\frac{A + C}{2}=\sin B$
D.$\sin\frac{B + C}{2}=\cos\frac{A}{2}$

答案： 4.AD 对于$A,\cos(A + B)=\cos(\pi - C)=-\cos C，$故A正确；对于B，$\tan(B + C)=\tan(\pi - A)=-\tan A，$故B错误；对于$C,\cos\frac{A + C}{2}=\cos\frac{\pi - B}{2}=\cos(\frac{\pi}{2}-\frac{B}{2})=\sin\frac{B}{2}，$故C错误；对于D，$\sin\frac{B + C}{2}=\sin\frac{\pi - A}{2}=\sin(\frac{\pi}{2}-\frac{A}{2})=\cos\frac{A}{2}，$故D正确. 故选AD.

典例1 (1)(2025·山东泰安模拟)已知$\sin(\frac{3\pi}{2}+\alpha)=\frac{\sqrt{3}}{2}$且$\frac{\pi}{2}\lt\alpha\lt\pi$，则$\tan\alpha=$(B)

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

答案： 典例1
(1)B 由诱导公式得$\sin(\frac{3\pi}{2}+\alpha)=\sin(\pi+\frac{\pi}{2}+\alpha)=-\sin(\frac{\pi}{2}+\alpha)=-\cos\alpha=\frac{\sqrt{3}}{2}，$所以$\cos\alpha=-\frac{\sqrt{3}}{2}，$又因为$\alpha\in(\frac{\pi}{2},\pi)，$所以$\sin\alpha=\frac{1}{2}，$所以$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{\sqrt{3}}{3}. $故选B.

(2)(2023·全国乙卷)若$\theta\in(0,\frac{\pi}{2}),\tan\theta=\frac{1}{2}$，则$\sin\theta-\cos\theta=$$-\frac{\sqrt{5}}{5}$。
角度2 $\sin\alpha\pm\cos\alpha,\sin\alpha\cos\alpha$之间的关系

答案： $(2)-\frac{\sqrt{5}}{5} $因为$\theta\in(0,\frac{\pi}{2})，$则$\sin\theta＞0,\cos\theta＞0，$又因为$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1}{2}$则$\cos\theta = 2\sin\theta，$且$\cos^{2}\theta+\sin^{2}\theta = 4\sin^{2}\theta+\sin^{2}\theta=5\sin^{2}\theta = 1，$解得$\sin\theta=\frac{\sqrt{5}}{5}$或$\sin\theta=-\frac{\sqrt{5}}{5}（$舍去），所以$\sin\theta-\cos\theta=\sin\theta - 2\sin\theta=-\sin\theta=-\frac{\sqrt{5}}{5}.$

典例2 已知$\sin\alpha-\cos\alpha=\frac{1}{5},\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$，则$\frac{\sin\alpha\cos\alpha}{\sin\alpha+\cos\alpha}=$(D)

A.$-\frac{12}{5}$
B.$\frac{12}{5}$
C.$-\frac{12}{35}$
D.$\frac{12}{35}$

答案： 典例2 D 由题意可得，$(\sin\alpha-\cos\alpha)^{2}=1 - 2\sin\alpha\cos\alpha=\frac{1}{25}，$整理得$\sin\alpha\cos\alpha=\frac{12}{25}＞0，$且$\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})，$可得$\alpha\in(0,\frac{\pi}{2})，$即$\sin\alpha＞0,\cos\alpha＞0，$可得$\sin\alpha+\cos\alpha＞0，$因为$(\sin\alpha+\cos\alpha)^{2}=1 + 2\sin\alpha\cos\alpha=\frac{49}{25}，$可得$\sin\alpha+\cos\alpha=\frac{7}{5}，$所以$\frac{\sin\alpha\cos\alpha}{\sin\alpha+\cos\alpha}=\frac{\frac{12}{25}}{\frac{7}{5}}=\frac{12}{35}. $故选D.

典例3 若$\tan\theta = 2$，则$\frac{(\sin\theta+\cos\theta)^{2}(\cos\theta-\sin\theta)}{\sin\theta}=$(B)

A.$-\frac{2}{5}$
B.$-\frac{9}{10}$
C.$\frac{2}{5}$
D.$\frac{9}{10}$

答案： 典例$3 B \frac{(\sin\theta+\cos\theta)^{2}(\cos\theta-\sin\theta)}{\sin\theta}=\frac{\sin\theta+\cos\theta}{\sin\theta}·\frac{\cos^{2}\theta-\sin^{2}\theta}{\cos^{2}\theta+\sin^{2}\theta}=\frac{\tan\theta + 1}{1+\tan^{2}\theta}·\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\frac{9}{10}. $故选B.

(1)已知$\alpha$是第三象限角，$\tan\alpha=\frac{5}{12}$，则$\cos\alpha$的值是(D)
A. $\frac{5}{13}$
B. $-\frac{5}{13}$
C. $\frac{12}{13}$
D. $-\frac{12}{13}$
(2)(多选题)(2025·辽宁葫芦岛期末)设$\alpha\in(0,\pi)$，已知$\sin\alpha,\cos\alpha$是方程$3x^{2}-x - m = 0$的两根，则下列等式正确的是(BD)
A. $m = -\frac{4}{3}$
B. $\sin\alpha-\cos\alpha=\frac{\sqrt{17}}{3}$
C. $\tan\alpha=\frac{7}{13}$
D. $\cos^{2}\alpha-\sin^{2}\alpha=-\frac{\sqrt{17}}{9}$
(3)(2025·陕西安康模拟)若$\tan(2025\pi-\alpha)=\frac{2}{3}$，则$\frac{\sin\alpha}{2\cos\alpha}-\frac{\cos^{2}\alpha}{\cos2\alpha}=$$-\frac{32}{15}$。

答案： 对点练1.
(1)D
(1)因为$\tan\alpha=\frac{5}{12}，$所以$\frac{\sin\alpha}{\cos\alpha}=\frac{5}{12}\cos\alpha，$又因为$\sin^{2}\alpha+\cos^{2}\alpha = 1，$所以$(\frac{5}{12}\cos\alpha)^{2}+\cos^{2}\alpha = 1，$解得$\cos\alpha=\pm\frac{12}{13}，$因为$\alpha$是第三象限角，所以$\cos\alpha=-\frac{12}{13}. $故选$D. (2)BD \sin\alpha,\cos\alpha$是方程$3x^{2}-x - m = 0$的两根，则有$\begin{cases}\sin\alpha+\cos\alpha=\frac{1}{3}\\\sin\alpha·\cos\alpha=-\frac{m}{3}\end{cases}$由$(\sin\alpha+\cos\alpha)^{2}=\sin^{2}\alpha+2\sin\alpha·\cos\alpha+\cos^{2}\alpha，$得$\frac{1}{9}=1-\frac{2m}{3}，$解得$m=\frac{4}{3}，$故A错误；$\alpha\in(0,\pi)，$有$\sin\alpha＞0，$由$\sin\alpha·\cos\alpha=-\frac{m}{3}=-\frac{4}{9}$＜0，有\cos\alpha＜0，(\sin\alpha-\cos\alpha)^{2}=\sin^{2}\alpha-2\sin\alpha·\cos\alpha+\cos^{2}\alpha=1+\frac{8}{9}=\frac{17}{9}，由\sin\alpha-\cos\alpha＞0，所以$\sin\alpha-\cos\alpha=\frac{\sqrt{17}}{3}，$故B正确；由$\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=\frac{1}{3}$得$\frac{\sin\alpha}{\cos\alpha}=\frac{1+\frac{\sqrt{17}}{3}}{1-\frac{\sqrt{17}}{3}}=\frac{9+\sqrt{17}}{8}，$故C错误；$\cos^{2}\alpha-\sin^{2}\alpha=(\cos\alpha+\sin\alpha)(\cos\alpha-\sin\alpha)=\frac{1}{3}×(-\frac{\sqrt{17}}{3})=-\frac{\sqrt{17}}{9}，$故D正确. 故选$BD. (3)-\frac{32}{15} $因为$\tan(2025\pi-\alpha)=\frac{2}{3}，$所以$\tan\alpha=-\frac{2}{3}，$所以$\frac{\sin\alpha}{2\cos\alpha}-\frac{\cos^{2}\alpha}{\cos2\alpha}=\frac{\tan\alpha}{2}·\frac{1}{1-\tan^{2}\alpha}=\frac{-\frac{2}{3}}{1 - \frac{4}{9}}=-\frac{32}{15}.$

已知$\alpha\in(\frac{3\pi}{2},\frac{7\pi}{4})$，且。
(1)求$\tan\alpha$的值；
(2)求$\frac{\sin(\frac{3\pi}{2}-\alpha)\cos(\pi+\alpha)\tan(-\alpha-\pi)}{\cos(2\pi-\alpha)\sin(\pi-\alpha)\tan(-\alpha)}$的值。

答案： 典例4 解：
(1)若选①，因为$\sin\alpha\cos\alpha=-\frac{2}{5}$所以$\frac{\sin\alpha\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=-\frac{2}{5}，$则$-\frac{\tan\alpha}{\tan^{2}\alpha + 1}=-\frac{2}{5}，$解得：$\tan\alpha=-2$或$\tan\alpha=-\frac{1}{2}，$因为$\alpha\in(\frac{3\pi}{2},\frac{7\pi}{4})，$所以$\tan\alpha=-2. $若选②，因为$\sin\alpha=-\frac{2\sqrt{5}}{5},\alpha\in(\frac{3\pi}{2},\frac{7\pi}{4})，$所以$\cos\alpha=\sqrt{1-\sin^{2}\alpha}=\frac{\sqrt{5}}{5}，$所以$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-2. (2)$由
(1)可知，$\tan\alpha=-2，$$\frac{\sin(\frac{3\pi}{2}-\alpha)\cos(\pi+\alpha)\tan(-\alpha-\pi)}{\cos(2\pi-\alpha)\sin(\pi-\alpha)\tan(-\alpha)}=\frac{(-\cos\alpha)(-\cos\alpha)(-\tan\alpha)}{\cos\alpha\sin\alpha\tan\alpha}=\frac{1}{\sin\alpha\tan\alpha}=\frac{1}{2}.$