《2025年同步练习册人民教育出版社高一数学必修第一册人教版》

【变式训练1】计算：
（1）$\sin14^{\circ}\cos16^{\circ} + \sin76^{\circ}\cos74^{\circ}$；
（2）$\frac{\sin7^{\circ} + \cos15^{\circ}\sin8^{\circ}}{\cos7^{\circ} - \sin15^{\circ}\sin8^{\circ}}$。

答案： 解
(1)原式$=\sin14°\cos16°+\sin(90°-$
$14°)\cos(90°-16°)=\sin14°\cos16°+\cos14°\sin16°=$
$\sin(14°+16°)=\sin30°=\frac{1}{2}$.
(2)原式$=\frac{\sin(15°-8°)+\cos15°\sin8°}{\cos(15°-8°)-\sin15°\sin8°}=$
$\frac{\sin15°\cos8°-\cos15°\sin8°+\cos15°\sin8°}{\cos15°\cos8°+\sin15°\sin8°-\sin15°\sin8°}=\frac{\sin15°\cos8°}{\cos15°\cos8°}=$
$\frac{\sin15°}{\cos15°}=\tan15°=\tan(45°-30°)=\frac{\tan45°-\tan30°}{1+\tan45°\tan30°}=$
$1-\frac{\sqrt{3}}{3}1+\frac{\sqrt{3}}{3}=2-\sqrt{3}$.

【例2】已知$\sin(\frac{3\pi}{4} + \alpha) = \frac{5}{13},\cos(\frac{\pi}{4} - \beta) = \frac{3}{5}$，且$0 ＜ \alpha ＜ \frac{\pi}{4} ＜ \beta ＜ \frac{3\pi}{4}$，求$\cos(\alpha + \beta)$。
延伸探究
1. 本例条件不变，如何求$\cos(\alpha - \beta)$的值？
2. 把本例变为“已知$\sin(\alpha + \frac{\pi}{6}) = \frac{4}{5},\cos(\beta - \frac{\pi}{6}) = \frac{12}{13},\alpha \in (0,\frac{\pi}{6}),\beta \in (0,\frac{\pi}{6})$，求$\cos(\alpha + \beta)$”，如何求解？

答案： 解$\because0＜\alpha＜\frac{\pi}{4}＜\beta＜\frac{3\pi}{4}$,
$\therefore\frac{3\pi}{4}＜\frac{3\pi}{4}+\alpha＜\pi,-\frac{\pi}{2}＜\frac{\pi}{4}-\beta＜0$.
又$\sin(\frac{3\pi}{4}+\alpha)=\frac{5}{13},\cos(\frac{\pi}{4}-\beta)=\frac{3}{5}$,
$\therefore\cos(\frac{3\pi}{4}+\alpha)=-\frac{12}{13},\sin(\frac{\pi}{4}-\beta)=-\frac{4}{5}$.
$\therefore\cos(\alpha+\beta)=\sin[\frac{\pi}{2}+(\alpha+\beta)]$
$=\sin[(\frac{3\pi}{4}+\alpha)-(\frac{\pi}{4}-\beta)]$
$=\sin(\frac{3\pi}{4}+\alpha)\cos(\frac{\pi}{4}-\beta)-\cos(\frac{3\pi}{4}+\alpha)·$
$\sin(\frac{\pi}{4}-\beta)=\frac{5}{13}×\frac{3}{5}-(-\frac{12}{13})×(-\frac{4}{5})=-\frac{33}{65}$.
【延伸探究】
1.解$\because0＜\alpha＜\frac{\pi}{4}＜\beta＜\frac{3\pi}{4}$,
$\therefore\frac{3\pi}{4}＜\frac{3\pi}{4}+\alpha＜\pi,-\frac{\pi}{2}＜\frac{\pi}{4}-\beta＜0$.
又$\sin(\frac{3\pi}{4}+\alpha)=\frac{5}{13},\cos(\frac{\pi}{4}-\beta)=\frac{3}{5}$,
$\therefore\cos(\frac{3\pi}{4}+\alpha)=-\frac{12}{13},\sin(\frac{\pi}{4}-\beta)=-\frac{4}{5}$.
$\therefore\cos(\alpha-\beta)=-\cos(\pi+\alpha-\beta)=-\cos[(\frac{3\pi}{4}+$
$\alpha)+(\frac{\pi}{4}-\beta)]=-\cos(\frac{3\pi}{4}+\alpha)\cos(\frac{\pi}{4}-\beta)+$
$\sin(\frac{3\pi}{4}+\alpha)\sin(\frac{\pi}{4}-\beta)=-(-\frac{12}{13})×\frac{3}{5}+\frac{5}{13}×$
$(-\frac{4}{5})=\frac{16}{65}$.
2.解$\because\alpha\in(0,\frac{\pi}{6}),\beta\in(0,\frac{\pi}{6})$,
$\therefore\alpha+\frac{\pi}{6}\in(\frac{\pi}{6},\frac{\pi}{3}),\beta-\frac{\pi}{6}\in(-\frac{\pi}{6},0)$.
又$\sin(\alpha+\frac{\pi}{6})=\frac{4}{5},\cos(\beta-\frac{\pi}{6})=\frac{12}{13}$,
$\therefore\cos(\alpha+\frac{\pi}{6})=\frac{3}{5},\sin(\beta-\frac{\pi}{6})=-\frac{5}{13}$.
$\therefore\cos(\alpha+\beta)=\cos[(\alpha+\frac{\pi}{6})+(\beta-\frac{\pi}{6})]=$
$\cos(\beta-\frac{\pi}{6})\cos(\alpha+\frac{\pi}{6})-\sin(\beta-\frac{\pi}{6})\sin(\alpha+\frac{\pi}{6})=\frac{56}{65}$.

【变式训练2】已知$\alpha,\beta$为第二象限角，$\cos(\alpha - \frac{\pi}{4}) = -\frac{3}{5},\sin(\beta + \frac{\pi}{4}) = \frac{5}{13}$，则$\sin(\alpha + \beta)$的值为(B)。

A.$\frac{33}{65}$
B.$-\frac{63}{65}$
C.$\frac{63}{65}$
D.$-\frac{33}{65}$

答案： B解析$\because\alpha,\beta$为第二象限角,且
$\cos(\alpha-\frac{\pi}{4})=-\frac{3}{5},\sin(\beta+\frac{\pi}{4})=\frac{5}{13}$,
$\therefore\sin(\alpha-\frac{\pi}{4})=\sqrt{1-\cos^2(\alpha-\frac{\pi}{4})}=\frac{4}{5}$,
$\cos(\beta+\frac{\pi}{4})=-\sqrt{1-\sin^2(\beta+\frac{\pi}{4})}=-\frac{12}{13}$.
$\therefore\sin(\alpha+\beta)=\sin[(\alpha-\frac{\pi}{4})+(\beta+\frac{\pi}{4})]=$
$\sin(\alpha-\frac{\pi}{4})\cos(\beta+\frac{\pi}{4})+\cos(\alpha-\frac{\pi}{4})\sin(\beta+\frac{\pi}{4})=$
$\frac{4}{5}×(-\frac{12}{13})+(-\frac{3}{5})×\frac{5}{13}=-\frac{63}{65}$.