三角函数常用公式总结

万能公式

$\sin^2\alpha + \cos^2\alpha = 1$

勾股定理

 

和角公式

$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

$\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

$\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

差角公式

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

$\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$

$\tan(\alpha-\beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$

和角公式差角公式的推导

在单位圆中，用向量$\overrightarrow{OA}$与向量$\overrightarrow{OB}$分别代表角$\alpha,\beta$的终边，$x$轴正半轴为始边，则

$\overrightarrow{OA} = (\cos\alpha, \sin\alpha), \overrightarrow{OB} = (\cos\beta, \sin\beta)$

则 $\overrightarrow{OA}·\overrightarrow{OB} = \cos\alpha\cos\beta + \sin\alpha\sin\beta$

设其夹角为$θ$，则$\overrightarrow{OA}·\overrightarrow{OB} = |\overrightarrow{OA}|·|\overrightarrow{OB}| \cos(θ) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$

因此$\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$

又因为$\cos(\alpha+\beta) = \cos(\alpha-(-\beta))$，因此有$\cos(\alpha+\beta) = \cos\alpha\cos(-\beta) + \sin\alpha\sin(-\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

又因为诱导公式$\sin\alpha = \cos(\dfrac{π}{2}-\alpha)$

因此$\sin(\alpha+\beta) = \cos(\dfrac{π}{2}-\alpha-\beta) = \cos(\dfrac{π}{2}-\alpha)\cos\beta+\sin(\dfrac{π}{2}-\alpha)\sin\beta = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

同理可推得$\sin(\alpha-\beta)$

$\tan(\alpha+\beta) = \dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \dfrac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha\sin\beta}$

上下同时除以$\cos\alpha\cos\beta$，即可得$\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

同理可推得$\tan(\alpha-\beta)$

 

和差化积公式 (一次同名)

$\sin\alpha + \sin\beta = 2\sin(\dfrac{\alpha+\beta}{2})\cos(\dfrac{\alpha-\beta}{2})$

$\sin\alpha - \sin\beta = 2\cos(\dfrac{\alpha+\beta}{2})\sin(\dfrac{\alpha-\beta}{2})$

$\cos\alpha + \cos\beta = 2\cos(\dfrac{\alpha+\beta}{2})\cos(\dfrac{\alpha-\beta}{2})$

$\cos\alpha - \cos\beta = 2\sin(\dfrac{\alpha+\beta}{2})\sin(\dfrac{\alpha-\beta}{2})$

$\tan\alpha + \tan\beta = \dfrac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}$

$\tan\alpha - \tan\beta = \dfrac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta}$

和差化积公式的推导

$\sin\alpha = \sin(\dfrac{\alpha+\beta}{2} +\dfrac{\alpha-\beta}{2} ) = \sin(\dfrac{\alpha+\beta}{2})\cos(\dfrac{\alpha-\beta}{2})+\sin(\dfrac{\alpha-\beta}{2})\cos(\dfrac{\alpha+\beta}{2})$

$\sin\beta = \sin(\dfrac{\alpha+\beta}{2} -\dfrac{\alpha-\beta}{2} ) = \sin(\dfrac{\alpha+\beta}{2})\cos(\dfrac{\alpha-\beta}{2})-\sin(\dfrac{\alpha-\beta}{2})\cos(\dfrac{\alpha+\beta}{2})$

两式相加即可得$\sin\alpha + \sin\beta = 2\sin(\dfrac{\alpha+\beta}{2})\cos(\dfrac{\alpha-\beta}{2})$

同理可推导$\cos\alpha + \cos\beta$与$\cos\alpha - \cos\beta$

$\tan\alpha + \tan\beta = \dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\sin\beta}{\cos\beta}$，通分即可

 

倍角公式 （这里是指二倍角公式）

$\sin(2\alpha) = 2\sin\alpha\cos\alpha$

$\cos(2\alpha) = \cos^2\alpha-\sin^2\alpha$

$\tan(2\alpha) = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$

以上公式利用和角公式证明即可

由于$\sin^2\alpha + \cos^2\alpha = 1$，可得

$\cos(2\alpha) = 2\cos^2\alpha-1$

$\cos(2\alpha) = 1-2\sin^2\alpha$

 

半角公式

$\sin(\dfrac{\alpha}{2}) = ±\sqrt{\dfrac{1-\cos\alpha}{2}}$

$\cos(\dfrac{\alpha}{2}) = ±\sqrt{\dfrac{1+\cos\alpha}{2}}$

$\tan(\dfrac{\alpha}{2}) = ±\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}$

半角公式的推导

由倍角公式$\cos(2\alpha) = 1-2\sin^2\alpha$，将$\alpha$替换为$\dfrac{\alpha}{2}$得$\cos\alpha = 1-2\sin^2(\dfrac{\alpha}{2})$，即$2\sin^2(\dfrac{\alpha}{2}) = 1-\cos\alpha$

同理利用倍角公式$\cos(2\alpha) = 2\cos^2\alpha-1$可推得$\cos(\dfrac{\alpha}{2})$

$\tan(\dfrac{\alpha}{2}) = \dfrac{\sin(\dfrac{\alpha}{2})}{\cos(\dfrac{\alpha}{2})} = \dfrac{±\sqrt{\dfrac{1-\cos\alpha}{2}}}{±\sqrt{\dfrac{1+\cos\alpha}{2}}} = ±\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}} $