《2025年创新设计高考总复习高三数学人教版A版》

角度2 “和”“积”转换
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.$\sin\theta-\cos\theta=\frac{7}{5}$

答案： 例2ABD[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，$
$\therefore\sin\theta - \cos\theta＞0，$
$\therefore\sin\theta - \cos\theta=\sqrt{1 - 2\sin\theta\cos\theta}$
$=\sqrt{1-(-\frac{24}{25})}=\sqrt{\frac{49}{25}}=\frac{7}{5}，$
$\therefore\sin\theta=\frac{4}{5},\cos\theta=-\frac{3}{5}.$
$\therefore\tan\theta=-\frac{4}{3},\thereforeA,B,D$正确.]

(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$1(1)B(2)\frac{\sqrt{5}+1}{2}[(1)$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)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$\tan\theta=\frac{\sqrt{5}+1}{2}.]$

(2)(2025·徐州调研)若$\theta\in\left(0,\frac{\pi}{2}\right)$，$\frac{\sin\theta}{\cos^{2}\theta}=\frac{\sin\theta-\cos\theta}{1-\sin2\theta}$，则$\tan\theta=$\frac{\sqrt{5}+1}{2}。

答案： $(2)\frac{\sqrt{5}+1}{2}[(2)$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$\cos\left(\frac{\pi}{6}-\theta\right)=a(|a|\leqslant1)$，则$\cos\left(\frac{5\pi}{6}+\theta\right)+\sin\left(\frac{2\pi}{3}-\theta\right)=$0。

答案： 例$3(1)0(2)\sin40^{\circ}[(1)$由题知
$\cos(\frac{5\pi}{6}+\theta)=\cos[\pi-(\frac{\pi}{6}-\theta)]$
$=-\cos(\frac{\pi}{6}-\theta)=-a，$
$\sin(\frac{2\pi}{3}-\theta)=\sin[\frac{\pi}{2}+(\frac{\pi}{6}-\theta)]$
$=\cos(\frac{\pi}{6}-\theta)=a，$
$\therefore\cos(\frac{5\pi}{6}+\theta)+\sin(\frac{2\pi}{3}-\theta)=0.$
$(2)\frac{\sin400^{\circ}\sin230^{\circ}}{\cos850^{\circ}\tan50^{\circ}}$
$=\frac{\sin(360^{\circ}+40^{\circ})\sin(180^{\circ}+50^{\circ})}{\cos(2×360^{\circ}+130^{\circ})\tan50^{\circ}}$
$=\frac{-\sin40^{\circ}\sin50^{\circ}-\sin40^{\circ}\sin(180^{\circ}-50^{\circ})}{\cos130^{\circ}\tan50^{\circ}}$
$=\frac{-\sin40^{\circ}\sin50^{\circ}}{\cos(180^{\circ}-50^{\circ})\frac{\sin50^{\circ}}{\cos50^{\circ}}}$
$=\frac{-\sin40^{\circ}\sin50^{\circ}}{-\cos50^{\circ}}×\frac{\cos50^{\circ}}{\sin50^{\circ}}=\sin40^{\circ}.]$

(2)化简：$\frac{\sin400^{\circ}\sin230^{\circ}}{\cos850^{\circ}\tan50^{\circ}}=$\sin40^{\circ}。

答案： $(2)\sin40^{\circ}[(2)\frac{\sin400^{\circ}\sin230^{\circ}}{\cos850^{\circ}\tan50^{\circ}}=\frac{\sin(360^{\circ}+40^{\circ})\sin(180^{\circ}+50^{\circ})}{\cos(2×360^{\circ}+130^{\circ})\tan50^{\circ}}=\frac{-\sin40^{\circ}\sin50^{\circ}-\sin40^{\circ}\sin(180^{\circ}-50^{\circ})}{\cos130^{\circ}\tan50^{\circ}}=\frac{-\sin40^{\circ}\sin50^{\circ}}{\cos(180^{\circ}-50^{\circ})\frac{\sin50^{\circ}}{\cos50^{\circ}}}=\frac{-\sin40^{\circ}\sin50^{\circ}}{-\cos50^{\circ}}×\frac{\cos50^{\circ}}{\sin50^{\circ}}=\sin40^{\circ}.]$

(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.$\cos\left(\frac{5\pi}{2}+3\alpha\right)=\sin3\alpha$
D.$\cos\left(\frac{3\pi}{2}-2\alpha\right)=-\sin2\alpha$

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

(2)求值：$\tan780^{\circ}\cos(-1140^{\circ})-\sin1560^{\circ}\cdot\cos(-1050^{\circ})=$________。

答案： 训练$2(2)\frac{2\sqrt{3}-3}{4}[(2)$原式$=\tan(2×360^{\circ}+60^{\circ})\cos(-3×360^{\circ}-60^{\circ})-\sin(4×360^{\circ}+40^{\circ})\cos(-3×360^{\circ}+30^{\circ})$
$=\tan60^{\circ}\cos(-60^{\circ})-\sin40^{\circ}\cos30^{\circ}$
$=\sqrt{3}×\frac{1}{2}-\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}=\frac{2\sqrt{3}-3}{4}.]$

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。

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$\cos(\frac{\pi}{3}-\alpha)=\frac{1}{3}$
$\therefore\cos^{2}(\frac{\pi}{6}+\alpha)=\cos^{2}[\frac{\pi}{2}-(\frac{\pi}{3}-\alpha)]$
$=\sin^{2}(\frac{\pi}{3}-\alpha)=1-\cos^{2}(\frac{\pi}{3}-\alpha)$
$=1-\frac{1}{9}=\frac{8}{9}，$
$\cos(\frac{2\pi}{3}+\alpha)=\cos[\pi-(\frac{\pi}{3}-\alpha)]$
$=-\cos(\frac{\pi}{3}-\alpha)=-\frac{1}{3}，$
$\therefore\cos^{2}(\frac{\pi}{6}+\alpha)+\cos(\frac{2\pi}{3}+\alpha)$
$=\frac{8}{9}-\frac{1}{3}=\frac{5}{9}.$

(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.$\pm2\sqrt{6}$
C.$\frac{2\sqrt{6}}{5}$
D.$2\sqrt{6}$

答案： 训练3
(1)B
(2)D[
(1)因为$\sin(\frac{2\pi}{7}+\alpha)=\frac{1}{5}，$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$\sin(\frac{3\pi}{14}-\alpha)=\pm\sqrt{1-\cos^{2}(\frac{3\pi}{14}-\alpha)}$
$=\pm\frac{2\sqrt{6}}{5}$
所以$\tan(\frac{3\pi}{14}-\alpha)=\frac{\sin(\frac{3\pi}{14}-\alpha)}{\cos(\frac{3\pi}{14}-\alpha)}=\pm2\sqrt{6}.$
答案解析与规律方法：549
(2)由诱导公式可得$\sin\alpha=\sin(\frac{3\pi}{2}-\alpha)+$
$\cos(\pi-\alpha)=-2\cos\alpha，$所以$\tan\alpha=-2.$
因此，$2\sin^{2}\alpha-\sin\alpha\cos\alpha=\frac{2\sin^{2}\alpha-\sin\alpha\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}$
$=\frac{2\tan^{2}\alpha-\tan\alpha}{\tan^{2}\alpha + 1}=\frac{10}{5}=2.]$

(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：
(2)D[
(2)由诱导公式可得$\sin\alpha=\sin(\frac{3\pi}{2}-\alpha)+\cos(\pi-\alpha)=-2\cos\alpha，$所以$\tan\alpha=-2.$因此，$2\sin^{2}\alpha-\sin\alpha\cos\alpha=\frac{2\sin^{2}\alpha-\sin\alpha\cos\alpha}{\sin^{2}\alpha+\cos^{2}\alpha}=\frac{2\tan^{2}\alpha-\tan\alpha}{\tan^{2}\alpha + 1}=\frac{10}{5}=2.]$