《2025年创新设计高考总复习数学浙江专版》

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.$-\frac{\sin\alpha}{\cos\alpha}$ $-\frac{\sin\alpha}{\cos\alpha}$ $\frac{\sin\alpha}{\cos\alpha}$ $\frac{\cos\alpha}{\sin\alpha}$ $-\frac{\sin\alpha}{\cos\alpha}$ $\frac{\cos\alpha}{\sin\alpha}$ $-\frac{\cos\alpha}{\sin\alpha}$ $\frac{1}{\tan\alpha}$
$-\frac{\tan\alpha}{\ }$ $-\frac{\tan\alpha}{\ }$

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$\alpha\in R$,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.$\frac{\sqrt{2}}{2}[\sin\frac{5\pi}{6}\cos(-\frac{\pi}{4})+\sin\frac{11\pi}{6}\cos\frac{5\pi}{4}$
$=\sin(-\frac{\pi}{6}+\pi)\cos\frac{\pi}{4}+\sin(-\frac{\pi}{6}+2\pi)\cdot$
$\cos(\frac{\pi}{4}+\pi)$
$=\sin\frac{\pi}{6}\cos\frac{\pi}{4}+(-\sin\frac{\pi}{6})(-\cos\frac{\pi}{4})$
$=2×\frac{1}{2}×\frac{\sqrt{2}}{2}×\frac{\sqrt{2}}{2}$.]

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 [
(1)$\because\frac{2\sin\theta - \cos\theta}{\sin\theta + 2\cos\theta}$
$=\frac{2\tan\theta - 1}{\tan\theta + 2}=\frac{1}{2}$,
$\therefore\tan\theta = \frac{4}{3}$.
则$\frac{\cos\theta(1 - 2\sin^{2}\theta)}{1 - 2\sin^{2}\theta÷\tan\theta + 1}$
$=\frac{\sin^{2}\theta}{\tan^{2}\theta + 1}=\frac{1 - 2×\frac{\tan^{2}\theta}{\tan^{2}\theta + 1}}{1 - \tan^{2}\theta}$
$=\frac{(\frac{4}{3} + 1)×(\frac{16}{9} + 1)}{1 - \frac{16}{9}}$
$=\frac{(\frac{4}{3} + 1)×(\frac{16}{9} + 1)}{(1 - \frac{16}{9})(\frac{16}{9} + 1)}=-\frac{3}{25}$.]

(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$\sin\theta - \cos\theta=-\frac{\sqrt{5}}{5}$.]