《2025年金版教程高考科学复习解决方案数学》

【巩固迁移】
1.（2024·江苏无锡天一中学高三模拟）已知$0 ＜ \theta ＜ \pi$，则$\frac{(1 + \sin\theta + \cos\theta)(\sin\frac{\theta}{2}-\cos\frac{\theta}{2})}{\sqrt{2 + 2\cos\theta}}=$________.

答案： 答案 - $\cos\theta$
解析 原式 = $\frac{(2\sin\frac{\theta}{2}\cos\frac{\theta}{2} + 2\cos^{2}\frac{\theta}{2})(\sin\frac{\theta}{2} - \cos\frac{\theta}{2})}{\sqrt{4\cos^{2}\frac{\theta}{2}}}$ = $\cos\frac{\theta}{2}$·$\frac{\sin^{2}\frac{\theta}{2} - \cos^{2}\frac{\theta}{2}}{|\cos\frac{\theta}{2}|}$ = $\frac{-\cos\frac{\theta}{2}\cos\theta}{|\cos\frac{\theta}{2}|}$. 因为0 ＜ $\theta$ ＜ $\pi$, 所以0 ＜ $\frac{\theta}{2}$ ＜ $\frac{\pi}{2}$, 即$\cos\frac{\theta}{2}$ ＞ 0, 所以原式 = - $\cos\theta$.

例2　（1）$\cos(-75^{\circ})$的值是（ ）
A. $\frac{\sqrt{6}-\sqrt{2}}{2}$
B. $\frac{\sqrt{6}+\sqrt{2}}{2}$
C. $\frac{\sqrt{6}-\sqrt{2}}{4}$
D. $\frac{\sqrt{6}+\sqrt{2}}{4}$

答案：
(1)C [$\cos(-75^{\circ})$ = $\cos(45^{\circ} - 120^{\circ})$ = $\cos45^{\circ}\cos120^{\circ}$ + $\sin45^{\circ}\sin120^{\circ}$ = $\frac{\sqrt{2}}{2}$×(-$\frac{1}{2}$) + $\frac{\sqrt{2}}{2}$×$\frac{\sqrt{3}}{2}$ = $\frac{\sqrt{6} - \sqrt{2}}{4}$. 故选C.]

（2）化简：$\frac{\cos40^{\circ}}{\cos25^{\circ}\sqrt{1 - \cos50^{\circ}}}=$（ ）
A. $\sqrt{2}$
B. $2\sqrt{2}$
C. $\sqrt{3}$
D. $\sqrt{3}-1$

答案：
(2)A [$\frac{\cos40^{\circ}}{\cos25^{\circ}\sqrt{1 - \cos50^{\circ}}}$ = $\frac{\cos(90^{\circ} - 50^{\circ})}{\cos25^{\circ}\sqrt{1 - (1 - 2\sin^{2}25^{\circ})}}$ = $\frac{\sin50^{\circ}}{\sqrt{2}\cos25^{\circ}\sin25^{\circ}}$ = $\frac{2\sin50^{\circ}}{\sqrt{2}\sin50^{\circ}}$ = $\sqrt{2}$. 故选A.]

（3）$\cos20^{\circ}\cos40^{\circ}\cos100^{\circ}=$________.
[课堂笔记]
　______________________________

答案：
(3)答案 - $\frac{1}{8}$
解析 $\cos20^{\circ}\cos40^{\circ}\cos100^{\circ}$ = - $\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}$ = - $\frac{\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{\sin20^{\circ}}$ = - $\frac{\frac{1}{2}\sin40^{\circ}\cos40^{\circ}\cos80^{\circ}}{\sin20^{\circ}}$ = - $\frac{\frac{1}{4}\sin80^{\circ}\cos80^{\circ}}{\sin20^{\circ}}$ = - $\frac{\frac{1}{8}\sin160^{\circ}}{\sin20^{\circ}}$ = - $\frac{\frac{1}{8}\sin20^{\circ}}{\sin20^{\circ}}$ = - $\frac{1}{8}$.

【巩固迁移】
2. 若$\tan\alpha = 2\tan10^{\circ}$，则$\frac{\cos(\alpha - 80^{\circ})}{\sin(\alpha - 10^{\circ})}=$________.

答案： 答案 3
解析
∵$\tan\alpha$ = 2$\tan10^{\circ}$,
∴$\frac{\cos(\alpha - 80^{\circ})}{\sin(\alpha - 10^{\circ})}$ = $\frac{\cos(\alpha + 10^{\circ} - 90^{\circ})}{\sin(\alpha - 10^{\circ})}$ = $\frac{\sin(\alpha + 10^{\circ})}{\sin(\alpha - 10^{\circ})}$ = $\frac{\sin\alpha\cos10^{\circ} + \cos\alpha\sin10^{\circ}}{\sin\alpha\cos10^{\circ} - \cos\alpha\sin10^{\circ}}$ = $\frac{\tan\alpha + \tan10^{\circ}}{\tan\alpha - \tan10^{\circ}}$ = $\frac{3\tan10^{\circ}}{\tan10^{\circ}}$ = 3.

例3　已知$\cos(\frac{\pi}{4}+\alpha)=-\frac{3}{5},-\pi ＜ \alpha ＜ 0$，则$\cos\alpha =$（ ）
A. $\frac{\sqrt{2}}{10}$ B. $-\frac{\sqrt{2}}{10}$ C. $\frac{7\sqrt{2}}{10}$ D. $-\frac{7\sqrt{2}}{10}$
[课堂笔记]　____________________________

答案： D [因为 - $\pi$ ＜ $\alpha$ ＜ 0, 所以 - $\frac{3\pi}{4}$ ＜ $\frac{\pi}{4}$ + $\alpha$ ＜ $\frac{\pi}{4}$, 又$\cos(\frac{\pi}{4} + \alpha)$ = - $\frac{3}{5}$ ＜ 0, 所以 - $\frac{3\pi}{4}$ ＜ $\frac{\pi}{4}$ + $\alpha$ ＜ - $\frac{\pi}{2}$, 所以$\sin(\frac{\pi}{4} + \alpha)$ = - $\frac{4}{5}$, 所以$\cos\alpha$ = $\cos[(\frac{\pi}{4} + \alpha) - \frac{\pi}{4}]$ = $\cos(\frac{\pi}{4} + \alpha)\cos\frac{\pi}{4}$ + $\sin(\frac{\pi}{4} + \alpha)\sin\frac{\pi}{4}$ = (- $\frac{3}{5}$)×$\frac{\sqrt{2}}{2}$ + (- $\frac{4}{5}$)×$\frac{\sqrt{2}}{2}$ = - $\frac{7\sqrt{2}}{10}$. 故选D.]

【巩固迁移】
3. 已知$\alpha\in(0,\frac{\pi}{4}),\cos2\alpha=\frac{4}{5}$，则$\sin^{2}(\alpha + \frac{\pi}{4})=$（ ）
A. $\frac{1}{5}$
B. $\frac{2}{5}$
C. $\frac{3}{5}$
D. $\frac{4}{5}$

答案： D [
∵$\alpha$∈(0,$\frac{\pi}{4}$),
∴2$\alpha$∈(0,$\frac{\pi}{2}$), 又$\cos2\alpha$ = $\frac{4}{5}$,
∴$\sin2\alpha$ = $\sqrt{1 - \cos^{2}2\alpha}$ = $\sqrt{1 - (\frac{4}{5})^{2}}$ = $\frac{3}{5}$,
∴$\sin^{2}(\alpha + \frac{\pi}{4})$ = $\frac{1 - \cos(2\alpha + \frac{\pi}{2})}{2}$ = $\frac{1 + \sin2\alpha}{2}$ = $\frac{1 + \frac{3}{5}}{2}$ = $\frac{4}{5}$. 故选D.]

例4　已知$\cos\alpha=\frac{2\sqrt{5}}{5},\sin\beta=\frac{\sqrt{10}}{10}$，且$\alpha\in(0,\frac{\pi}{2}),\beta\in(0,\frac{\pi}{2})$，则$\alpha + \beta$的值是（ ）
A. $\frac{3\pi}{4}$ B. $\frac{\pi}{4}$ C. $\frac{7\pi}{4}$ D. $\frac{5\pi}{4}$
[课堂笔记]　____________________________

答案： B [
∵$\alpha$∈(0,$\frac{\pi}{2}$), $\beta$∈(0,$\frac{\pi}{2}$),
∴$\sin\alpha$ = $\sqrt{1 - \cos^{2}\alpha}$ = $\frac{\sqrt{5}}{5}$, $\cos\beta$ = $\sqrt{1 - \sin^{2}\beta}$ = $\frac{3\sqrt{10}}{10}$,
∴$\cos(\alpha + \beta)$ = $\cos\alpha\cos\beta$ - $\sin\alpha\sin\beta$ = $\frac{2\sqrt{5}}{5}$×$\frac{3\sqrt{10}}{10}$ - $\frac{\sqrt{5}}{5}$×$\frac{\sqrt{10}}{10}$ = $\frac{\sqrt{2}}{2}$, 又$\alpha + \beta$∈(0,$\pi$),
∴$\alpha + \beta$ = $\frac{\pi}{4}$. 故选B.]