《2025年金版教程高考科学复习解决方案数学》

【巩固迁移】
4. 若$\cos\alpha=\frac{1}{7},\sin(\alpha + \beta)=\frac{5\sqrt{3}}{14},0 ＜ \alpha ＜ \frac{\pi}{2},0 ＜ \beta ＜ \frac{\pi}{2}$，则角$\beta$的值为（ ）
A. $\frac{\pi}{3}$
B. $\frac{5\pi}{12}$
C. $\frac{\pi}{6}$
D. $\frac{\pi}{4}$

答案： A [
∵0 ＜ $\alpha$ ＜ $\frac{\pi}{2}$, 0 ＜ $\beta$ ＜ $\frac{\pi}{2}$,
∴0 ＜ $\alpha + \beta$ ＜ $\pi$, 由$\cos\alpha$ = $\frac{1}{7}$, $\sin(\alpha + \beta)$ = $\frac{5\sqrt{3}}{14}$, 得$\sin\alpha$ = $\frac{4\sqrt{3}}{7}$, $\cos(\alpha + \beta)$ = ±$\frac{11}{14}$. 若$\cos(\alpha + \beta)$ = $\frac{11}{14}$, 则$\sin\beta$ = $\sin[(\alpha + \beta) - \alpha]$ = $\sin(\alpha + \beta)\cos\alpha$ - $\cos(\alpha + \beta)\sin\alpha$ = $\frac{5\sqrt{3}}{14}$×$\frac{1}{7}$ - $\frac{11}{14}$×$\frac{4\sqrt{3}}{7}$ ＜ 0, 与$\sin\beta$ ＞ 0矛盾, 故舍去; 若$\cos(\alpha + \beta)$ = - $\frac{11}{14}$, 则$\cos\beta$ = $\cos[(\alpha + \beta) - \alpha]$ = $\cos(\alpha + \beta)\cos\alpha$ + $\sin(\alpha + \beta)\sin\alpha$ = - $\frac{11}{14}$×$\frac{1}{7}$ + $\frac{5\sqrt{3}}{14}$×$\frac{4\sqrt{3}}{7}$ = $\frac{1}{2}$, 又$\beta$∈(0,$\frac{\pi}{2}$),
∴$\beta$ = $\frac{\pi}{3}$. 故选A.]

例5　设$m$为实数，已知$\sin\alpha-\sqrt{3}\cos\alpha = m - 1$，则$m$的取值范围是________.
[课堂笔记]
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答案： 答案 [-1,3]
解析 $\sin\alpha$ - $\sqrt{3}\cos\alpha$ = 2($\frac{1}{2}\sin\alpha$ - $\frac{\sqrt{3}}{2}\cos\alpha$) = 2$\sin(\alpha - \frac{\pi}{3})$ = $m - 1$, 因为 - 1 ≤ $\sin(\alpha - \frac{\pi}{3})$ ≤ 1, 所以 - 2 ≤ 2$\sin(\alpha - \frac{\pi}{3})$ ≤ 2, 所以 - 2 ≤ $m - 1$ ≤ 2, 解得 - 1 ≤ $m$ ≤ 3, 则$m$的取值范围是[-1,3].

【巩固迁移】
5. 已知函数$f(x)=\frac{\sqrt{2}}{4}\sin(\frac{\pi}{4}-x)+\frac{\sqrt{6}}{4}\cos(\frac{\pi}{4}-x)$
（1）求函数$f(x)$在区间$[\frac{\pi}{4},\frac{3\pi}{2}]$上的最值；
（2）若$\cos\theta=\frac{4}{5},\theta\in(\frac{3\pi}{2},2\pi)$，求$f(2\theta + \frac{\pi}{3})$的值.
提示：同步《课时作业（A本）》P327

答案： 解
(1)由题意得$f(x)$ = $\frac{\sqrt{2}}{4}\sin(\frac{\pi}{4} - x)$ + $\frac{\sqrt{6}}{4}\cos(\frac{\pi}{4} - x)$ = $\frac{\sqrt{2}}{2}$×[$\frac{1}{2}\sin(\frac{\pi}{4} - x)$ + $\frac{\sqrt{3}}{2}\cos(\frac{\pi}{4} - x)$] = - $\frac{\sqrt{2}}{2}\sin(x - \frac{7\pi}{12})$.
因为$x$∈[$\frac{\pi}{4}$,$\frac{3\pi}{2}$], 所以$x - \frac{7\pi}{12}$∈[-$\frac{\pi}{3}$,$\frac{11\pi}{12}$],
所以$\sin(x - \frac{7\pi}{12})$∈[-$\frac{\sqrt{3}}{2}$,1],
所以 - $\frac{\sqrt{2}}{2}\sin(x - \frac{7\pi}{12})$∈[-$\frac{\sqrt{2}}{2}$,$\frac{\sqrt{6}}{4}$],
即函数$f(x)$在区间[$\frac{\pi}{4}$,$\frac{3\pi}{2}$]上的最大值为$\frac{\sqrt{6}}{4}$, 最小值为 - $\frac{\sqrt{2}}{2}$.
(2)因为$\cos\theta$ = $\frac{4}{5}$, $\theta$∈($\frac{3\pi}{2}$,2$\pi$),
所以$\sin\theta$ = - $\frac{3}{5}$, 所以$\sin2\theta$ = 2$\sin\theta\cos\theta$ = - $\frac{24}{25}$,
$\cos2\theta$ = $\cos^{2}\theta$ - $\sin^{2}\theta$ = $\frac{16}{25}$ - $\frac{9}{25}$ = $\frac{7}{25}$,
所以$f(2\theta + \frac{\pi}{3})$ = - $\frac{\sqrt{2}}{2}\sin(2\theta + \frac{\pi}{3} - \frac{7\pi}{12})$ = - $\frac{\sqrt{2}}{2}\sin(2\theta - \frac{\pi}{4})$ = - $\frac{1}{2}$($\sin2\theta$ - $\cos2\theta$) = $\frac{1}{2}$($\cos2\theta$ - $\sin2\theta$) = $\frac{1}{2}$×($\frac{7}{25}$ + $\frac{24}{25}$) = $\frac{31}{50}$.