5.5.1 两角和与差的正弦、余弦和正切公式:修订间差异
来自高中数学
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[[5.5 三角恒等变换]] | [[5.5 三角恒等变换]] | ||
==知识要点== | ==知识要点== | ||
* $C_{(\alpha-\beta)}$ | |||
$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$ | |||
* $C_{(\alpha+\beta)}$、$S_{(\alpha+\beta)}$、$S_{(\alpha-\beta)}$ | |||
$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ | |||
$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$ | |||
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$ | |||
* $T_{(\alpha+\beta)}$、$T_{(\alpha-\beta)}$ | |||
$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ | |||
$\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$ | |||
* $S_{2\alpha}$、$C_{2\alpha}$、$T_{2\alpha}$ | |||
$\sin2\alpha=2\sin\alpha\cos\alpha$ | |||
$\cos2\alpha=\cos^2\alpha-\sin^2\alpha=1-2\sin^2\alpha=2\cos^2\alpha-1$ | |||
$\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$ | |||
==例题== | ==例题== | ||
==练习p217== | ==练习p217== | ||
==练习p220== | ==练习p220== | ||
==练习p223== | ==练习p223== |
2024年4月16日 (二) 10:16的最新版本
知识要点
- $C_{(\alpha-\beta)}$
$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$
- $C_{(\alpha+\beta)}$、$S_{(\alpha+\beta)}$、$S_{(\alpha-\beta)}$
$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
- $T_{(\alpha+\beta)}$、$T_{(\alpha-\beta)}$
$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
- $S_{2\alpha}$、$C_{2\alpha}$、$T_{2\alpha}$
$\sin2\alpha=2\sin\alpha\cos\alpha$
$\cos2\alpha=\cos^2\alpha-\sin^2\alpha=1-2\sin^2\alpha=2\cos^2\alpha-1$
$\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$