《2025年创新设计高考总复习数学浙江专版》

例2(多选)已知$\theta \in (0, \pi)$，$\sin \theta + \cos \theta = \frac{1}{5}$，则下列结论正确的是(ABD)

A.$\sin \theta = \frac{4}{5}$
B.$\cos \theta = -\frac{3}{5}$
C.$\tan \theta = -\frac{3}{4}$
D.$\sin \theta - \cos \theta = \frac{7}{5}$ABD

答案： 例2 ABD [由题意知$\sin\theta+\cos\theta=\frac{1}{5}$，
$\therefore(\sin\theta+\cos\theta)^{2}=1 + 2\sin\theta\cos\theta=\frac{1}{25}$，
$\therefore2\sin\theta\cos\theta=-\frac{24}{25}＜0$，
又$\because\theta\in(0,\pi)$，$\therefore\frac{\pi}{2}＜\theta＜\pi$，
$\therefore\sin\theta - \cos\theta＞0$，
$\therefore\sin\theta - \cos\theta=\sqrt{1 - 2\sin\theta\cos\theta}$
$=\sqrt{1 - (-\frac{24}{25})}=\sqrt{\frac{49}{25}}=\frac{7}{5}$，
$\therefore\sin\theta=\frac{4}{5},\cos\theta=-\frac{3}{5}$.
$\therefore\tan\theta=-\frac{4}{3}$,$\therefore$A,B,D正确.]

训练1(1)已知$x \in \left( -\frac{\pi}{2}, 0 \right)$，$\sin^{4} x + \cos^{4} x = \frac{1}{2}$，则$\sin x - \cos x =$(B)

A.$\sqrt{2}$
B.$-\sqrt{2}$
C.$\frac{\sqrt{2}}{2}$
D.$-\frac{\sqrt{2}}{2}$

答案： 训练1
(1)B [
(1)因为$\sin^{4}x+\cos^{4}x$
$=(\sin^{2}x+\cos^{2}x)^{2}-2\sin^{2}x\cos^{2}x$
$=1 - 2\sin^{2}x\cos^{2}x=\frac{1}{2}$，
所以$\sin^{2}x\cos^{2}x=\frac{1}{4}$，
又$x\in(-\frac{\pi}{2},0)$，
所以$\sin x＜0,\cos x＞0,\sin x - \cos x＜0$，
所以$\sin x\cos x=-\frac{1}{2}$，
$\sin x - \cos x=-\sqrt{(\sin x - \cos x)^{2}}$
$=-\sqrt{1 - 2\sin x\cos x}$
$=-\sqrt{1 - 2×(-\frac{1}{2})}=-\sqrt{2}$.]

(2)(2025·徐州调研)若$\theta \in \left( 0, \frac{\pi}{2} \right)$，$\frac{\sin \theta}{\cos^{2} \theta} = \frac{\sin \theta - \cos \theta}{1 - \sin 2\theta}$，则$\tan \theta =$$\frac{\sqrt{5}+1}{4}$。

答案：
(2)$\frac{\sqrt{5}+1}{4}$ [
(2)因为$\frac{\sin\theta}{\cos^{2}\theta}=\frac{\sin\theta - \cos\theta}{1 - \sin2\theta}$，
所以$\frac{\sin\theta}{\cos^{2}\theta}=\frac{\sin\theta - \cos\theta}{(\sin\theta - \cos\theta)^{2}}$，
所以$\frac{\sin\theta}{\cos^{2}\theta}=\frac{1}{\sin\theta - \cos\theta}$，
即$\sin^{2}\theta - \sin\theta\cos\theta - \cos^{2}\theta = 0$，
因为$\theta\in(0,\frac{\pi}{2})$，所以$\cos\theta\neq0,\tan\theta＞0$，
所以$\tan^{2}\theta - \tan\theta - 1 = 0$，
得$\tan\theta=\frac{\sqrt{5}+1}{2}$.]

例3(1)已知$\cos \left( \frac{\pi}{6} - \theta \right) = a (|a| \leq 1)$，则$\cos \left( \frac{5\pi}{6} + \theta \right) + \sin \left( \frac{2\pi}{3} - \theta \right) =$0。

答案： 例3
(1)0 [
(1)由题知
$\cos(\frac{5\pi}{6}+\theta)=\cos[\pi - (\frac{\pi}{6}-\theta)]$
$=-\cos(\frac{\pi}{6}-\theta)=-a$，
$\sin(\frac{2\pi}{3}-\theta)=\sin[\frac{\pi}{2}+(\frac{\pi}{6}-\theta)]$
$=\cos(\frac{\pi}{6}-\theta)=a$，
$\therefore\cos(\frac{5\pi}{6}+\theta)+\sin(\frac{2\pi}{3}-\theta)=0$.]

(2)化简：$\frac{\sin 400^{\circ} \sin 230^{\circ}}{\cos 850^{\circ} \tan 50^{\circ}} =$$\sin40^{\circ}$。

答案：
(2)$\sin40^{\circ}$ [
(2)$\frac{\sin400^{\circ}\sin230^{\circ}}{\cos850^{\circ}\tan50^{\circ}}$
$=\frac{\sin(360^{\circ}+40^{\circ})\sin(180^{\circ}+50^{\circ})}{\cos(2×360^{\circ}+130^{\circ})\tan50^{\circ}}$
$=\frac{-\sin40^{\circ}\sin130^{\circ}}{-\sin40^{\circ}\sin(180^{\circ}-50^{\circ})÷\frac{\sin50^{\circ}}{\cos50^{\circ}}}$
$=\frac{-\sin40^{\circ}\sin50^{\circ}}{-\cos50^{\circ}\cdot\frac{\sin50^{\circ}}{\cos50^{\circ}}}=\sin40^{\circ}$.]

训练2(1)已知$\alpha \in \mathbf{R}$，则下列等式恒成立的是(D)
A. $\sin(3\pi - \alpha) = -\sin \alpha$
B. $\sin \frac{\pi - \alpha}{2} = -\cos \frac{\alpha}{2}$
C. $\cos \left( \frac{5\pi}{2} + 3\alpha \right) = \sin 3\alpha$
D. $\cos \left( \frac{3\pi}{2} - 2\alpha \right) = -\sin 2\alpha$D

答案： 训练2
(1)D [
(1)$\sin(3\pi - \alpha)$
$=\sin(\pi - \alpha)=\sin\alpha$，
$\sin\frac{\pi}{2}=\sin(\frac{\pi}{2}-\frac{\alpha}{2})=\cos\frac{\alpha}{2}$，
$\cos(\frac{5\pi}{2}+3\alpha)=\cos(\frac{\pi}{2}+3\alpha)=-\sin3\alpha$，
$\cos(\frac{3\pi}{2}-2\alpha)=-\sin2\alpha$.]

（2）求值：$\tan 780^{\circ} \cos (-1140^{\circ}) - \sin 1560^{\circ} \cdot \cos (-1050^{\circ}) =$__________。

答案： （2）$\frac{2\sqrt{3}-3}{4}$ [
(2)原式$=\tan(2×360^{\circ}+60^{\circ})\cos(-3×360^{\circ}-60^{\circ})-\sin(4×360^{\circ}+120^{\circ})\cos(-3×360^{\circ}+30^{\circ})$
$=\tan60^{\circ}\cos(-60^{\circ})-\sin120^{\circ}\cos30^{\circ}$
$=\sqrt{3}×\frac{1}{2}-\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}=\frac{2\sqrt{3}-3}{4}$.]

例4已知$f(\alpha) = \frac{\sin \left( \frac{\pi}{2} + \alpha \right) \cos (\pi + \alpha) \sin (-\alpha)}{\sin \left( \frac{3\pi}{2} - \alpha \right) \cos (2\pi - \alpha) \tan (\pi - \alpha)}$。
（1）化简$f(\alpha)$；
（2）若$f \left( \frac{\pi}{3} - \alpha \right) = \frac{1}{3}$，求$\cos^{2} \left( \frac{\pi}{6} + \alpha \right) + \cos \left( \frac{2\pi}{3} + \alpha \right)$的值。$f(\alpha)=\cos\alpha$$\frac{5}{9}$

答案： 例4 解
(1)由题意得，
$f(\alpha)=\frac{\cos\alpha\cdot(-\cos\alpha)\cdot(-\sin\alpha)}{-\cos\alpha\cdot\cos\alpha\cdot(-\tan\alpha)}=\cos\alpha$.
(2)$f(\frac{\pi}{3}-\alpha)=\frac{1}{3}$，即$\cos(\frac{\pi}{3}-\alpha)=\frac{1}{3}$，
$\therefore\cos^{2}(\frac{\pi}{6}+\alpha)=\cos^{2}[\frac{\pi}{2}-(\frac{\pi}{3}-\alpha)]$
$=\sin^{2}(\frac{\pi}{3}-\alpha)=1 - \cos^{2}(\frac{\pi}{3}-\alpha)$
$=1-\frac{1}{9}=\frac{8}{9}$，
$\cos(\frac{2\pi}{3}+\alpha)=\cos[\pi - (\frac{\pi}{3}-\alpha)]$
$=-\cos(\frac{\pi}{3}-\alpha)=-\frac{1}{3}$，
$\therefore\cos^{2}(\frac{\pi}{6}+\alpha) + \cos(\frac{2\pi}{3}+\alpha)$
$=\frac{8}{9}-\frac{1}{3}=\frac{5}{9}$.

训练3(1)(2025·丽水模拟)已知$\sin \left( \frac{2\pi}{7} + \alpha \right) = \frac{1}{5}$，那么$\tan \left( \frac{3\pi}{14} - \alpha \right) =$(B)

A.$-\frac{1}{5}$
B.$\pm 2\sqrt{6}$
C.$\frac{2\sqrt{6}}{5}$
D.$2\sqrt{6}$

答案： 训练3
(1)B [
(1)因为$\frac{3\pi}{14}-\alpha=\frac{\pi}{2}-(\frac{2\pi}{7}+\alpha)$，所以$\cos(\frac{3\pi}{14}-\alpha)=\cos[\frac{\pi}{2}-(\frac{2\pi}{7}+\alpha)]=\sin(\frac{2\pi}{7}+\alpha)=\frac{1}{5}$，则$\sin(\frac{3\pi}{14}-\alpha)=\pm\sqrt{1 - \cos^{2}(\frac{3\pi}{14}-\alpha)}=\pm\sqrt{1 - (\frac{1}{5})^{2}}=\pm\frac{2\sqrt{6}}{5}$，所以$\tan(\frac{3\pi}{14}-\alpha)=\frac{\sin(\frac{3\pi}{14}-\alpha)}{\cos(\frac{3\pi}{14}-\alpha)}=\pm2\sqrt{6}$.]

(2)(2024·衡水模拟)已知$\sin \left( \frac{3\pi}{2} - \alpha \right) + \cos (\pi - \alpha) = \sin \alpha$，则$2\sin^{2} \alpha - \sin \alpha \cos \alpha =$(D)

A.$\frac{21}{10}$
B.$\frac{3}{2}$
C.$\frac{\sqrt{3}}{2}$
D.$2$

答案： 训练3
(2)D [
(2)由诱导公式可得$\sin(\frac{3\pi}{2}-\alpha)+\cos(\pi - \alpha)=-\cos\alpha - \cos\alpha=-2\cos\alpha=\sin\alpha$，所以$\tan\alpha=-2$，则$2\sin^{2}\alpha - \sin\alpha\cos\alpha=\frac{2\sin^{2}\alpha - \sin\alpha\cos\alpha}{\sin^{2}\alpha + \cos^{2}\alpha}=\frac{2\tan^{2}\alpha - \tan\alpha}{\tan^{2}\alpha + 1}=\frac{2×(-2)^{2}-(-2)}{(-2)^{2}+1}=\frac{8 + 2}{5}=2$.]