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高中三角函数公式

时间:2026-03-20 19:26:35

高中三角函数公式包括基本公式、倍角公式、和差角公式、辅助角公式等。以下是详细内容:

基本公式

锐角三角函数公式

$\sin \alpha = \frac{\text{对边}}{\text{斜边}}$

$\cos \alpha = \frac{\text{邻边}}{\text{斜边}}$

$\tan \alpha = \frac{\text{对边}}{\text{邻边}}$

$\cot \alpha = \frac{\text{邻边}}{\text{对边}}$

诱导公式

$\sin(2k\pi + \alpha) = \sin \alpha$

$\cos(2k\pi + \alpha) = \cos \alpha$

$\tan(2k\pi + \alpha) = \tan \alpha$

$\cot(2k\pi + \alpha) = \cot \alpha$

$\sin(\pi + \alpha) = -\sin \alpha$

$\cos(\pi + \alpha) = -\cos \alpha$

$\tan(\pi + \alpha) = \tan \alpha$

$\cot(\pi + \alpha) = \cot \alpha$

$\sin(-\alpha) = -\sin \alpha$

$\cos(-\alpha) = \cos \alpha$

$\tan(-\alpha) = -\tan \alpha$

$\cot(-\alpha) = -\cot \alpha$

$\sin(\pi - \alpha) = \sin \alpha$

$\cos(\pi - \alpha) = -\cos \alpha$

$\tan(\pi - \alpha) = -\tan \alpha$

$\cot(\pi - \alpha) = -\cot \alpha$

$\sin(2\pi - \alpha) = -\sin \alpha$

$\cos(2\pi - \alpha) = \cos \alpha$

$\tan(2\pi - \alpha) = -\tan \alpha$

$\cot(2\pi - \alpha) = -\cot \alpha$

$\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$

$\cos(\frac{\pi}{2} + \alpha) = -\sin \alpha$

$\tan(\frac{\pi}{2} + \alpha) = -\cot \alpha$

$\cot(\frac{\pi}{2} + \alpha) = -\tan \alpha$

$\sin(\frac{\pi}{2} - \alpha) = \cos \alpha$

$\cos(\frac{\pi}{2} - \alpha) = \sin \alpha$

$\tan(\frac{\pi}{2} - \alpha) = \cot \alpha$

$\cot(\frac{\pi}{2} - \alpha) = \tan \alpha$

倍角公式

1. $\sin 2\alpha = 2\sin \alpha \cos \alpha$

2. $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1$

3. $\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha}$

和差角公式

1. $\sin(A + B) = \sin A \cos B + \cos A \sin B$

2. $\sin(A - B) = \sin A \cos B - \cos A \sin B$

3. $\cos(A + B) = \cos A \cos B - \sin A \sin B$

4. $\cos(A - B) = \cos A \cos B + \sin A \sin B$

5. $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

6. $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

辅助角公式

1. $\sin(A + B) = \sqrt{A^2 + B^2} \sin(A + t)$

2. $\cos(A + B) = \sqrt{A^2 + B^2} \cos(A + t)$

3. $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$