《2025年优化探究同步导学案高中数学必修第一册人教版》

2. 求证：$\frac{2\sin(\theta-\frac{3\pi}{2})\cos(\theta+\frac{\pi}{2}) - 1\tan(9\pi+\theta)+1}{1 - 2\sin^2(\pi+\theta)}=\frac{}{\tan(\pi+\theta)-1}$.

答案： 跟踪训练 2. 证明：左边 =
$2\sin(\theta+\frac{\pi}{2})(-\sin\theta)-1$
$=\frac{1 - 2\sin^{2}\theta}{1-2\sin^{2}\theta}$
$=\frac{-2\sin\theta\cos\theta-(\sin^{2}\theta+\cos^{2}\theta)}{\sin^{2}\theta+\cos^{2}\theta - 2\sin^{2}\theta}$
$=\frac{-(\sin\theta+\cos\theta)^{2}}{\cos^{2}\theta-\sin^{2}\theta}$
$=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}$ (原解析此步骤跳跃，按原图呈现)
右边$=\frac{\tan(\pi+\theta)+1}{\tan(\pi+\theta)-1}=\frac{\tan\theta + 1}{\tan\theta - 1}=\frac{\frac{\sin\theta}{\cos\theta}+1}{\frac{\sin\theta}{\cos\theta}-1}=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}$
所以左边=右边，故原等式成立.

[例3] 已知$f(\alpha)=\frac{\sin(\alpha-\frac{\pi}{2})\cos(\frac{3\pi}{2}-\alpha)\tan(\pi+\alpha)\cos(\frac{\pi}{2}+\alpha)}{\sin(2\pi-\alpha)\tan(-\alpha-\pi)\sin(-\alpha-\pi)}$，化简$f(\alpha)$.
[分析] 观察三角函数式中角的形式，灵活运用诱导公式一～六进行化简.

答案： [例3] [解] $f(\alpha)=$
$\frac{-\sin(\frac{\pi}{2}-\alpha)(-\sin\alpha)\tan\alpha(-\sin\alpha)}{-\sin(-\alpha)(-\tan\alpha)[-\sin(\pi+\alpha)]}$
$=\frac{-\cos\alpha(-\sin\alpha)\tan\alpha(-\sin\alpha)}{-\sin\alpha(-\tan\alpha)\sin\alpha}$
$=-\cos\alpha$，

3. 化简：$\frac{\sin(\theta - 5\pi)\cos(-\frac{\pi}{2}-\theta)\cos(8\pi - \theta)}{\sin(\theta-\frac{3\pi}{2})\sin(-\theta - 4\pi)}$.

答案： 跟踪训练 3.解：原式=
$\frac{-\sin(-\theta + 5\pi)\cos(\frac{\pi}{2}+\theta)\cos\theta}{-\sin(-\theta+\frac{3\pi}{2})[-\sin(\theta + 4\pi)]}$
$=\frac{-\sin(-\theta+\pi)(-\sin\theta)\cos\theta}{\cos\theta(-\sin\theta)}$
$=-\sin\theta$.