《2025年优化探究同步导学案高中数学必修第一册人教版》

诱导公式五、六的作用：实现正弦函数与余弦函数的相互转化.

答案： 正弦函数 余弦函数

[例4] (1)已知$\cos(\frac{\pi}{6}-\alpha)=\frac{1}{3}$，则$\cos(\frac{5\pi}{6}+\alpha)=$$-\frac{1}{3}$，$\sin(\frac{2\pi}{3}-\alpha)=$$\frac{1}{3}$；
(2)已知$\alpha$为锐角，且$\cos(\alpha+\frac{\pi}{3})=-\frac{3}{4}$，则$\sin(\frac{7\pi}{6}-\alpha)+\sin(\alpha-\frac{2\pi}{3})=$$\frac{3 - \sqrt{7}}{4}$.

答案： [例4] [答案]
(1)$-\frac{1}{3}$
(2)$\frac{3 - \sqrt{7}}{4}$
[解析]
(1)$\cos(\frac{5\pi}{6}+\alpha)=\cos[\pi-(\frac{\pi}{6}-\alpha)]=-\cos(\frac{\pi}{6}-\alpha)=-\frac{1}{3}$.
$\sin(\frac{2\pi}{3}-\alpha)=\sin[\frac{\pi}{2}+(\frac{\pi}{6}-\alpha)]=\cos(\frac{\pi}{6}-\alpha)=\frac{1}{3}$.
(2)因为$\alpha$为锐角，所以$\alpha+\frac{\pi}{3}\in[\frac{\pi}{3},\frac{5\pi}{6}]$,
$\sin(\alpha+\frac{\pi}{3})=\sqrt{1-\cos^{2}(\alpha+\frac{\pi}{3})}=\frac{\sqrt{7}}{4}$.
$\sin(\frac{7\pi}{6}-\alpha)+\sin(\alpha-\frac{2\pi}{3})$
$=\sin[\frac{3\pi}{2}-(\alpha+\frac{\pi}{3})]+\sin[(\alpha+\frac{\pi}{3})-\pi]$
$=-\cos(\alpha+\frac{\pi}{3})-\sin(\alpha+\frac{\pi}{3})$
$=\frac{3 - \sqrt{7}}{4}$.

4. 已知$\sin(\frac{5\pi}{2}+\alpha)=\frac{1}{3}$，则$\cos(\alpha-\frac{\pi}{2})$等于 (D)

A.$-\frac{1}{3}$
B.$-\frac{2\sqrt{2}}{3}$
C.$\frac{2\sqrt{2}}{3}$
D.$\pm\frac{2\sqrt{2}}{3}$

答案： 跟踪训练 4. D $\sin(\frac{5\pi}{2}+\alpha)=\sin(\frac{\pi}{2}+\alpha)=\cos\alpha=\frac{1}{3}$,
$\therefore\sin\alpha=\pm\sqrt{1-\cos^{2}\alpha}=\pm\frac{2\sqrt{2}}{3}$
$\therefore\cos(\alpha-\frac{\pi}{2})=\cos(\frac{\pi}{2}-\alpha)=\sin\alpha=\pm\frac{2\sqrt{2}}{3}$.

1. 已知$\sin\alpha=\frac{5}{13}$，则$\cos(\frac{\pi}{2}+\alpha)$等于 (C)

A.$\frac{5}{13}$
B.$\frac{12}{13}$
C.$-\frac{5}{13}$
D.$-\frac{12}{13}$

答案： 1.C $\cos(\frac{\pi}{6}+\alpha)=-\sin\alpha=-\frac{5}{13}$.

2. (多选)以下与$\sin(\theta-\frac{\pi}{2})$的值恒相等的式子为 (CD)

A.$\sin(\theta+\frac{\pi}{2})$
B.$\cos(\theta+\frac{\pi}{2})$
C.$\cos(\pi-\theta)$
D.$\sin(\theta+\frac{3\pi}{2})$

答案： 2.CD $\sin(\theta-\frac{\pi}{2})=-\sin(\frac{\pi}{2}-\theta)=-\cos\theta$,
因为$\sin(\theta+\frac{\pi}{2})=\cos\theta,\cos(\theta+\frac{\pi}{2})=-\sin\theta$,
$\cos(\pi-\theta)=-\cos\theta,\sin(\theta+\frac{3\pi}{2})=-\cos\theta$,
所以A,B错误,C,D正确.

3. 已知角$\alpha$的终边过点$(3m, -2)$，$\tan(\pi+\alpha)=\frac{1}{3}$，则实数$m =$$-2$；$\cos(\frac{\pi}{2}+\alpha)=$$\frac{\sqrt{10}}{10}$.

答案： 3.答案$ -2\frac{\sqrt{10}}{10}$
解析：由$\tan(\pi+\alpha)=\frac{1}{3}$,得$\tan\alpha=\frac{1}{3}$,
$\therefore\tan\alpha=\frac{1}{3}=\frac{-2}{3m}$
$\therefore m = - 2$,
$\therefore\cos(\frac{\pi}{2}+\alpha)=-\sin\alpha=-\frac{-2}{\sqrt{36 + 4}}=\frac{\sqrt{10}}{10}$.

4. 化简：$\frac{\cos(6\pi+\theta)·\sin(-2\pi-\theta)·\tan(2\pi-\theta)}{\cos(\frac{3\pi}{2}+\theta)·\sin(\frac{3\pi}{2}+\theta)}=$
$=$$-\tan\theta$.

答案： 4.答案$ -\tan\theta$
解析：原式 =
$\frac{\cos\theta·\sin(-\theta)·\tan(-\theta)}{\cos(\pi+\frac{\pi}{2}+\theta)·\sin(\pi+\frac{\pi}{2}+\theta)}$
$=\frac{\cos\theta·(-\sin\theta)·(-\tan\theta)}{-\cos(\frac{\pi}{2}+\theta)·[-\sin(\frac{\pi}{2}+\theta)]}$
$=\frac{\cos\theta·\sin\theta·\tan\theta}{\sin\theta·(-\cos\theta)}$
$=-\tan\theta$.