《2025年创新设计高考总复习高三数学人教版A版》

1. 同角三角函数的基本关系
(1)平方关系：\sin^{2}\alpha+\cos^{2}\alpha = 1。
(2)商数关系：$\frac{\sin\alpha}{\cos\alpha}=\tan\alpha\left(\alpha\neq\frac{\pi}{2}+k\pi,k\in\mathbf{Z}\right)$。

答案： $1.(1)\sin^{2}\alpha+\cos^{2}\alpha = 1$

2. 三角函数的诱导公式

答案： $2.-\sin\alpha -\sin\alpha \sin\alpha \cos\alpha \cos\alpha -\cos\alpha$
$-\cos\alpha -\cos\alpha \sin\alpha -\sin\alpha \tan\alpha$
$-\tan\alpha -\tan\alpha$

1. 思考辨析(在括号内打“√”或“×”)
(1)若$\alpha$，$\beta$为锐角，则$\sin^{2}\alpha+\cos^{2}\beta = 1$。(×)
(2)$\sin(\pi+\alpha)=-\sin\alpha$成立的条件是$\alpha$为锐角。(×)
(3)若$\alpha\in\mathbf{R}$，则$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$恒成立。(×)
(4)若$\sin(k\pi-\alpha)=\frac{1}{3}(k\in\mathbf{Z})$，则$\sin\alpha=\frac{1}{3}$。(×)

答案： 1.
(1)×
(2)×
(3)×
(4)×
[
(1)对任意的角$\alpha,\sin^{2}\alpha+\cos^{2}\alpha = 1.$
(2)中对于任意$\alpha\in\mathbf{R},$恒有$\sin(\pi+\alpha)=-\sin\alpha.$
(3)中当$\alpha$的终边落在y轴上时，商数关系不成立.
(4)当k为奇数时，$\sin\alpha=\frac{1}{3}，$
当k为偶数时，$\sin\alpha = -\frac{1}{3}.]$

2. (湘教必修一P168例5改编)已知$\alpha$是第三象限角，$\sin\alpha=-\frac{3}{5}$，则$\tan\alpha=$(B)

A.$-\frac{3}{4}$
B.$\frac{3}{4}$
C.$-\frac{4}{3}$
D.$\frac{4}{3}$

答案： 2.B[由题意得$\cos\alpha = -\frac{4}{5}，$
故$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{3}{4}.]$

3. (人教A必修一P195T5改编)已知$\sin\left(\frac{7\pi}{2}+\alpha\right)=\frac{3}{5}$，那么$\cos\alpha=$(B)

A.$-\frac{4}{5}$
B.$-\frac{3}{5}$
C.$\frac{3}{5}$
D.$\frac{4}{5}$

答案： 3.B[因为$\sin(\frac{7\pi}{2}+\alpha)=-\cos\alpha = -\frac{3}{5}，$
所以$\cos\alpha = -\frac{3}{5}.$

4. (北师大必修二P24例8(3)改编)求值：$\sin\frac{5\pi}{6}\cos\left(-\frac{\pi}{4}\right)+\sin\frac{11\pi}{6}\cos\frac{5\pi}{4}=$\frac{\sqrt{2}}{2}。

答案： $4.\frac{\sqrt{2}}{2}[\sin\frac{5\pi}{6}\cos(-\frac{\pi}{4})+\sin\frac{11\pi}{6}\cos\frac{5\pi}{4}$
$=\sin(-\frac{\pi}{6}+\pi)\cos\frac{\pi}{4}+\sin(-\frac{\pi}{6}+2\pi).$
$\cos(\frac{\pi}{4}+\pi)$
$=\sin\frac{\pi}{6}\cos\frac{\pi}{4}+(-\sin\frac{\pi}{6})(-\cos\frac{\pi}{4})$
$=2×\frac{1}{2}×\frac{\sqrt{2}}{2}×\frac{\sqrt{2}}{2}.]$

例1 (1)(2025·济南质检)若$\frac{2\sin\theta-\cos\theta}{\sin\theta+2\cos\theta}=\frac{1}{2}$，则$\frac{\cos\theta(1 - 2\sin^{2}\theta)}{\sin\theta+\cos\theta}=$(C)

A.$-\frac{4}{25}$
B.$\frac{4}{25}$
C.$-\frac{3}{25}$
D.$\frac{3}{25}$

答案： 例$1(1)C(2)-\frac{\sqrt{5}}{5}[(1)\because\frac{2\sin\theta - \cos\theta}{\sin\theta + 2\cos\theta}=\frac{1}{2}，$
$\therefore\frac{2\tan\theta - 1}{\tan\theta + 2}=\frac{1}{2}，$
$\therefore\tan\theta=\frac{4}{3}，$
则$\frac{\cos\theta(1 - 2\sin^{2}\theta)}{\sin\theta+\cos\theta}=\frac{1 - 2\sin^{2}\theta}{\tan\theta + 1}$
$=\frac{1 - 2×\frac{\sin^{2}\theta}{\sin^{2}\theta+\cos^{2}\theta}}{\tan\theta + 1}=\frac{1 - 2×\frac{\tan^{2}\theta}{\tan^{2}\theta + 1}}{\tan\theta + 1}$
$=\frac{(1 - \tan^{2}\theta)(1+\tan^{2}\theta)}{1 - \tan^{2}\theta}$
$=(\frac{\frac{4}{3}+1}{\frac{16}{9}+1})×\frac{3}{25}.$
(2)由$\begin{cases}\tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac{1}{2}\\\sin^{2}\theta+\cos^{2}\theta = 1,\end{cases}$且$\theta\in(0,\frac{\pi}{2})，$
解得$\begin{cases}\sin\theta=\frac{\sqrt{5}}{5}\\\cos\theta=\frac{2\sqrt{5}}{5}\end{cases}，$故$\sin\theta - \cos\theta=-\frac{\sqrt{5}}{5}.]$

(2)(2023·全国乙卷)若$\theta\in\left(0,\frac{\pi}{2}\right)$，$\tan\theta=\frac{1}{2}$，则$\sin\theta-\cos\theta=$-\frac{\sqrt{5}}{5}。

答案： $(2)-\frac{\sqrt{5}}{5}[(2)$由$\begin{cases}\tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac{1}{2}\\\sin^{2}\theta+\cos^{2}\theta = 1,\end{cases}$且$\theta\in(0,\frac{\pi}{2})，$解得$\begin{cases}\sin\theta=\frac{\sqrt{5}}{5}\\\cos\theta=\frac{2\sqrt{5}}{5}\end{cases}，$故$\sin\theta - \cos\theta=-\frac{\sqrt{5}}{5}.]$