2.7.2 三角函数的重要公式

2.7.2.1 三角函数间的关系

$$\begin{array}{}\text{(2.82)}& {\sin }^2\alpha +{\cos }^2\alpha =1\end{array}$$$$\begin{array}{}\text{(2.83)}& {\sec }^2\alpha -{\tan }^2\alpha =1,\end{array}$$$$\begin{array}{}\text{(2.84)}& {\csc }^2\alpha -{\cot }^2\alpha =1,\end{array}$$$$\begin{array}{}\text{(2.85)}& \sin \alpha \cdot \csc \alpha =1,\end{array}$$$$\begin{array}{}\text{(2.86)}& \cos \alpha \cdot \sec \alpha =1,\end{array}$$$$\begin{array}{}\text{(2.87)}& \tan \alpha \cdot \cot \alpha =1,\end{array}$$$$\begin{array}{}\text{(2.88)}& \frac{\sin \alpha }{\cos \alpha }=\tan \alpha \end{array}$$$$\begin{array}{}\text{(2.89)}& \frac{\cos \alpha }{\sin \alpha }=\cot \alpha \end{array}$$

当 $0<\alpha <\frac{\pi }{2}$ 时,为了便于查看,表 2.5 总结了它们间的重要关系,该表其他区间平方根的符号依自变量所处的象限决定.

$\alpha$	$\sin \alpha$	$\cos \alpha$	$\tan \alpha$	$\cot \alpha$
$\sin \alpha$		$\sqrt{1-{\cos }^2\alpha }$	$\frac{\tan \alpha }{\sqrt{1+{\tan }^2\alpha }}$	$\frac{1}{\sqrt{1+{\cot }^2\alpha }}$
$\cos \alpha$	$\sqrt{1-{\sin }^2\alpha }$		$\frac{1}{\sqrt{1+{\tan }^2\alpha }}$	$\frac{\cot \alpha }{\sqrt{1+{\cot }^2\alpha }}$
$\tan \alpha$	$\frac{\sin \alpha }{\sqrt{1-{\sin }^2\alpha }}$	$\frac{\sqrt{1-{\cos }^2\alpha }}{\cos \alpha }$		$\frac{1}{\cot \alpha }$
$\cot \alpha$	$\frac{\sqrt{1-{\sin }^2\alpha }}{\sin \alpha }$	$\frac{\cos \alpha }{\sqrt{1-{\cos }^2\alpha }}$	$\frac{1}{\tan \alpha }$

2.7.2.2 两角和与差的三角函数 (加法定理)

$$\begin{array}{}\text{(2.90)}& \sin \left(\alpha \pm \beta \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta ,\end{array}$$$$\begin{array}{}\text{(2.91)}& \cos \left(\alpha \pm \beta \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta ,\end{array}$$$$\begin{array}{}\text{(2.92)}& \tan \left(\alpha \pm \beta \right)=\frac{\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta },\end{array}$$$$\begin{array}{}\text{(2.93)}& \cot \left(\alpha \pm \beta \right)=\frac{\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha },\end{array}$$$$\sin \left(\alpha +\beta +\gamma \right)=\sin \alpha \cos \beta \cos \gamma +\cos \alpha \sin \beta \cos \gamma$$$$\begin{array}{}\text{(2.94)}& +\cos \alpha \cos \beta \sin \gamma -\sin \alpha \sin \beta \sin \gamma ,\end{array}$$$$\cos \left(\alpha +\beta +\gamma \right)=\cos \alpha \cos \beta \cos \gamma -\sin \alpha \sin \beta \cos \gamma$$$$\begin{array}{}\text{(2.95)}& -\sin \alpha \cos \beta \sin \gamma -\cos \alpha \sin \beta \sin \gamma \text{.}\end{array}$$

2.7.2.3 倍角的三角函数

$$\begin{array}{}\text{(2.96)}& \sin 2\alpha =2\sin \alpha \cos \alpha ,\end{array}$$$$\begin{array}{}\text{(2.97)}& \sin 3\alpha =3\sin \alpha -4{\sin }^3\alpha ,\end{array}$$$$\begin{array}{}\text{(2.98)}& \cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha ,\end{array}$$$$\begin{array}{}\text{(2.99)}& \cos 3\alpha =4{\cos }^3\alpha -3\cos \alpha ,\end{array}$$$$\begin{array}{}\text{(2.100)}& \sin 4\alpha =8{\cos }^3\alpha \sin \alpha -4\cos \alpha \sin \alpha ,\end{array}$$$$\begin{array}{}\text{(2.101)}& \cos 4\alpha =8{\cos }^4\alpha -8{\cos }^2\alpha +1,\end{array}$$$$\begin{array}{}\text{(2.102)}& \tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }\end{array}$$$$\begin{array}{}\text{(2.103)}& \tan 3\alpha =\frac{3\tan \alpha -{\tan }^3\alpha }{1-3{\tan }^2\alpha },\end{array}$$$$\begin{array}{}\text{(2.104)}& \tan 4\alpha =\frac{4\tan \alpha -4{\tan }^3\alpha }{1-6{\tan }^2\alpha +{\tan }^4\alpha },\end{array}$$$$\begin{array}{}\text{(2.105)}& \cot 2\alpha =\frac{{\cot }^2\alpha -1}{2\cot \alpha },\end{array}$$$$\begin{array}{}\text{(2.106)}& \cot 3\alpha =\frac{{\cot }^3\alpha -3\cot \alpha }{3{\cot }^2\alpha -1},\end{array}$$$$\begin{array}{}\text{(2.107)}& \cot 4\alpha =\frac{{\cot }^4\alpha -6{\cot }^2\alpha +1}{4{\cot }^3\alpha -4\cot \alpha }.\end{array}$$

当 $n$ 较大时,为了得到 $\sin n\alpha$ 和 $\cos n\alpha$ ,要利用棣莫弗公式(参见第 48 页 1.5.3.5). 利用二项式定理 (参见第 14 页 1.1.6.4), 有

$$\cos n\alpha +\mathrm{i}\sin n\alpha$$$$=\sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right){\mathrm{i}}^k{\cos }^{n-k}\alpha {\sin }^k\alpha ={(\cos \alpha +\mathrm{i}\sin \alpha )}^n$$$$={\cos }^n\alpha +\mathrm{i}n{\cos }^{n-1}\alpha \sin \alpha -\left(\begin{array}{l}n\\ 2\end{array}\right){\cos }^{n-2}\alpha {\sin }^2\alpha$$$$\begin{array}{}\text{(2.108)}& -\mathrm{i}\left(\begin{array}{l}n\\ 3\end{array}\right){\cos }^{n-3}\alpha {\sin }^3\alpha +\left(\begin{array}{l}n\\ 4\end{array}\right){\cos }^{n-4}\alpha {\sin }^4\alpha +\cdots ,\end{array}$$

由此得到

$$\cos n\alpha ={\cos }^n\alpha -\left(\begin{array}{l}n\\ 2\end{array}\right){\cos }^{n-2}\alpha {\sin }^2\alpha +\left(\begin{array}{l}n\\ 4\end{array}\right){\cos }^{n-4}\alpha {\sin }^4\alpha$$$$\begin{array}{}\text{(2.109)}& -\left(\begin{array}{l}n\\ 6\end{array}\right){\cos }^{n-6}\alpha {\sin }^6\alpha +\cdots ,\end{array}$$$$\sin n\alpha =n{\cos }^{n-1}\alpha \sin \alpha -\left(\begin{array}{l}n\\ 3\end{array}\right){\cos }^{n-3}\alpha {\sin }^3\alpha$$$$\begin{array}{}\text{(2.110)}& +\left(\begin{array}{l}n\\ 5\end{array}\right){\cos }^{n-5}\alpha {\sin }^5\alpha -\cdots .\end{array}$$

2.7.2.4 半角的三角函数

下列公式中平方根必须依据半角所处的象限确定正负号.

$$\begin{array}{}\text{(2.111)}& \sin \frac{\alpha }{2}=\sqrt{\frac{1}{2}\left(1-\cos \alpha \right)}\end{array}$$$$\begin{array}{}\text{(2.112)}& \cos \frac{\alpha }{2}=\sqrt{\frac{1}{2}\left(1+\cos \alpha \right)}\end{array}$$$$\begin{array}{}\text{(2.113)}& \tan \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}=\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1+\cos \alpha },\end{array}$$$$\begin{array}{}\text{(2.114)}& \cot \frac{\alpha }{2}=\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}=\frac{1+\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1-\cos \alpha }.\end{array}$$

2.7.2.5 两三角函数的和与差

$$\begin{array}{}\text{(2.115)}& \sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2},\end{array}$$$$\begin{array}{}\text{(2.116)}& \sin \alpha -\sin \beta =2\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2},\end{array}$$$$\begin{array}{}\text{(2.117)}& \cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2},\end{array}$$$$\begin{array}{}\text{(2.118)}& \cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2},\end{array}$$$$\begin{array}{}\text{(2.119)}& \tan \alpha \pm \tan \beta =\frac{\sin \left(\alpha \pm \beta \right)}{\cos \alpha \cos \beta },\end{array}$$$$\begin{array}{}\text{(2.120)}& \cot \alpha \pm \cot \beta =\pm \frac{\sin \left(\alpha \pm \beta \right)}{\sin \alpha \sin \beta },\end{array}$$$$\begin{array}{}\text{(2.121)}& \tan \alpha +\cot \beta =\frac{\cos \left(\alpha -\beta \right)}{\cos \alpha \sin \beta },\end{array}$$$$\begin{array}{}\text{(2.122)}& \cot \alpha -\tan \beta =\frac{\cos \left(\alpha +\beta \right)}{\sin \alpha \cos \beta }.\end{array}$$

2.7.2.6 三角函数的积

$$\begin{array}{}\text{(2.123)}& \sin \alpha \sin \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)\right],\end{array}$$$$\begin{array}{}\text{(2.124)}& \cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right],\end{array}$$$$\begin{array}{}\text{(2.125)}& \sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha -\beta \right)+\sin \left(\alpha +\beta \right)\right],\end{array}$$$$\sin \alpha \sin \beta \sin \gamma =\frac{1}{4}[\sin \left(\alpha +\beta -\gamma \right)+\sin \left(\beta +\gamma -\alpha \right)$$$$\begin{array}{}\text{(2.126)}& +\sin \left(\gamma +\alpha -\beta \right)-\sin \left(\alpha +\beta +\gamma \right)]\end{array}$$$$\sin \alpha \cos \beta \cos \gamma =\frac{1}{4}[\sin \left(\alpha +\beta -\gamma \right)-\sin \left(\beta +\gamma -\alpha \right)$$$$\begin{array}{}\text{(2.127)}& +\sin \left(\gamma +\alpha -\beta \right)+\sin \left(\alpha +\beta +\gamma \right)]\end{array}$$$$\sin \alpha \sin \beta \cos \gamma =\frac{1}{4}[-\cos \left(\alpha +\beta -\gamma \right)+\cos \left(\beta +\gamma -\alpha \right)$$$$\begin{array}{}\text{(2.128)}& +\cos \left(\gamma +\alpha -\beta \right)-\cos \left(\alpha +\beta +\gamma \right)]\end{array}$$$$\cos \alpha \cos \beta \cos \gamma =\frac{1}{4}[\cos \left(\alpha +\beta -\gamma \right)+\cos \left(\beta +\gamma -\alpha \right)$$$$\begin{array}{}\text{(2.129)}& +\cos \left(\gamma +\alpha -\beta \right)+\cos \left(\alpha +\beta +\gamma \right)].\end{array}$$

2.7.2.7 三角函数的幂

$$\begin{array}{}\text{(2.130)}& {\sin }^2\alpha =\frac{1}{2}\left(1-\cos 2\alpha \right)\end{array}$$$$\begin{array}{}\text{(2.131)}& {\cos }^2\alpha =\frac{1}{2}\left(1+\cos 2\alpha \right)\end{array}$$$$\begin{array}{}\text{(2.132)}& {\sin }^3\alpha =\frac{1}{4}\left(3\sin \alpha -\sin 3\alpha \right),\end{array}$$$$\begin{array}{}\text{(2.133)}& {\cos }^3\alpha =\frac{1}{4}\left(\cos 3\alpha +3\cos \alpha \right),\end{array}$$$$\begin{array}{}\text{(2.134)}& {\sin }^4\alpha =\frac{1}{8}\left(\cos 4\alpha -4\cos 2\alpha +3\right),\end{array}$$$$\begin{array}{}\text{(2.135)}& {\cos }^4\alpha =\frac{1}{8}\left(\cos 4\alpha +4\cos 2\alpha +3\right).\end{array}$$

当 $n$ 很大时,可以利用 $\cos n\alpha$ 和 $\sin n\alpha$ 来表示 ${\sin }^n\alpha$ 和 ${\cos }^n\alpha$ (参见第 104 页2.7.2.3).