# 三角恒等式本文重定向自 二倍角公式

## 符号

sine sin arcsine arcsin cosecant csc
cosine cos arccosine arccos secant sec
tangent tan arctangent arctan cotangent cot
cotangent cot arccotangent arccot tangent tan
secant sec arcsecant arcsec cosine cos
cosecant csc arccosecant arccsc sine sin

${\displaystyle 0}$ ${\displaystyle {\frac {1}{12}}}$ ${\displaystyle {\frac {1}{8}}}$ ${\displaystyle {\frac {1}{6}}}$ ${\displaystyle {\frac {1}{4}}}$ ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle {\frac {3}{4}}}$ ${\displaystyle 1}$

## 基本关系

 ${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1\,}$ ${\displaystyle \tan ^{2}\theta +1\,=\sec ^{2}\theta }$ ${\displaystyle 1\,+\cot ^{2}\theta =\csc ^{2}\theta }$

${\displaystyle \sin \theta }$ ${\displaystyle \sin \theta \ }$ ${\displaystyle {\sqrt {1-\cos ^{2}\theta }}}$ ${\displaystyle {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}$ ${\displaystyle {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}$ ${\displaystyle {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}$ ${\displaystyle {\frac {1}{\csc \theta }}}$
${\displaystyle \cos \theta }$ ${\displaystyle {\sqrt {1-\sin ^{2}\theta }}}$ ${\displaystyle \cos \theta \ }$ ${\displaystyle {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}$ ${\displaystyle {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}$ ${\displaystyle {\frac {1}{\sec \theta }}}$ ${\displaystyle {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}$
${\displaystyle \tan \theta }$ ${\displaystyle {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}$ ${\displaystyle {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}$ ${\displaystyle \tan \theta \ }$ ${\displaystyle {\frac {1}{\cot \theta }}}$ ${\displaystyle {\sqrt {\sec ^{2}\theta -1}}}$ ${\displaystyle {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}$
${\displaystyle \cot \theta }$ ${\displaystyle {{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }}$ ${\displaystyle {\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}}$ ${\displaystyle {1 \over \tan \theta }}$ ${\displaystyle \cot \theta \ }$ ${\displaystyle {1 \over {\sqrt {\sec ^{2}\theta -1}}}}$ ${\displaystyle {\sqrt {\csc ^{2}\theta -1}}}$
${\displaystyle \sec \theta }$ ${\displaystyle {1 \over {\sqrt {1-\sin ^{2}\theta }}}}$ ${\displaystyle {1 \over \cos \theta }}$ ${\displaystyle {\sqrt {1+\tan ^{2}\theta }}}$ ${\displaystyle {{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }}$ ${\displaystyle \sec \theta \ }$ ${\displaystyle {\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}}$
${\displaystyle \csc \theta }$ ${\displaystyle {1 \over \sin \theta }}$ ${\displaystyle {1 \over {\sqrt {1-\cos ^{2}\theta }}}}$ ${\displaystyle {{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }}$ ${\displaystyle {\sqrt {1+\cot ^{2}\theta }}}$ ${\displaystyle {\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}}$ ${\displaystyle \csc \theta \ }$

### 其他函数的基本关系

${\displaystyle \operatorname {vers} \theta }$
${\displaystyle \operatorname {ver} \theta }$
${\displaystyle 1-\cos \theta }$

${\displaystyle \operatorname {cvs} \theta }$
${\displaystyle 1-\sin \theta }$

cohaversine
${\displaystyle \operatorname {hacoversin} \theta }$ ${\displaystyle {\frac {1-\sin \theta }{2}}}$

cohavercosine
${\displaystyle \operatorname {hacovercosin} \theta }$ ${\displaystyle {\frac {1+\sin \theta }{2}}}$

, chord ${\displaystyle \operatorname {crd} \theta }$ ${\displaystyle 2\sin \left({\frac {\theta }{2}}\right)}$

cosine and imaginary unit sine
${\displaystyle \operatorname {cis} \theta }$ ${\displaystyle \cos \theta +i\;\sin \theta }$

## 对称、移位和周期

### 对称

{\displaystyle {\begin{aligned}\sin(0-\theta )&=-\sin \theta \\\cos(0-\theta )&=+\cos \theta \\\tan(0-\theta )&=-\tan \theta \\\cot(0-\theta )&=-\cot \theta \\\sec(0-\theta )&=+\sec \theta \\\csc(0-\theta )&=-\csc \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\tan(\pi -\theta )&=-\tan \theta \\\cot(\pi -\theta )&=-\cot \theta \\\sec(\pi -\theta )&=-\sec \theta \\\csc(\pi -\theta )&=+\csc \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin({\tfrac {3\pi }{2}}-\theta )&=-\cos \theta \\\cos({\tfrac {3\pi }{2}}-\theta )&=-\sin \theta \\\tan({\tfrac {3\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {3\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {3\pi }{2}}-\theta )&=-\csc \theta \\\csc({\tfrac {3\pi }{2}}-\theta )&=-\sec \theta \end{aligned}}}

### 移位和周期

${\displaystyle \tan }$${\displaystyle \cot }$的周期 ${\displaystyle \sin }$, ${\displaystyle \cos }$, ${\displaystyle \csc }$${\displaystyle \sec }$的周期
{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\\tan(\theta +{\tfrac {\pi }{2}})&=-\cot \theta \\\cot(\theta +{\tfrac {\pi }{2}})&=-\tan \theta \\\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\\cos(\theta +\pi )&=-\cos \theta \\\tan(\theta +\pi )&=+\tan \theta \\\cot(\theta +\pi )&=+\cot \theta \\\sec(\theta +\pi )&=-\sec \theta \\\csc(\theta +\pi )&=-\csc \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {3\pi }{2}})&=-\cos \theta \\\cos(\theta +{\tfrac {3\pi }{2}})&=+\sin \theta \\\tan(\theta +{\tfrac {3\pi }{2}})&=-\cot \theta \\\cot(\theta +{\tfrac {3\pi }{2}})&=-\tan \theta \\\sec(\theta +{\tfrac {3\pi }{2}})&=+\csc \theta \\\csc(\theta +{\tfrac {3\pi }{2}})&=-\sec \theta \end{aligned}}} {\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\\cos(\theta +2\pi )&=+\cos \theta \\\tan(\theta +2\pi )&=+\tan \theta \\\cot(\theta +2\pi )&=+\cot \theta \\\sec(\theta +2\pi )&=+\sec \theta \\\csc(\theta +2\pi )&=+\csc \theta \end{aligned}}}

## 角的和差恒等式

正弦 ${\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,}$ ${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,}$ ${\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}$ ${\displaystyle \cot(\alpha \pm \beta )={\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}$ ${\displaystyle \sec(\alpha \pm \beta )={\frac {\sec \alpha \sec \beta }{1\mp \tan \alpha \tan \beta }}}$ ${\displaystyle \csc(\alpha \pm \beta )={\frac {\csc \alpha \csc \beta }{\cot \beta \pm \cot \alpha }}}$ 注意正负号的对应。 {\displaystyle {\begin{aligned}x\pm y=a\pm b&\Rightarrow \ x+y=a+b\\&{\mbox{and}}\ x-y=a-b\end{aligned}}}{\displaystyle {\begin{aligned}x\pm y=a\mp b&\Rightarrow \ x+y=a-b\\&{\mbox{and}}\ x-y=a+b\end{aligned}}}

### 正弦与余弦的无限多项和

${\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {odd} \ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}$
${\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {even} \ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}$

### 正切的有限多项和

${\displaystyle x_{i}=\tan \theta _{i}}$，对于${\displaystyle i=1,\ldots ,n}$。设${\displaystyle e_{k}}$是变量${\displaystyle x_{i}}$${\displaystyle i=1,\ldots ,n}$${\displaystyle k=0,\ldots ,n}$${\displaystyle k}$基本对称多项式。则

${\displaystyle \tan(\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},}$

{\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2}+\theta _{3})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}

## 多倍角公式

${\displaystyle T_{n}}$是${\displaystyle n}$次切比雪夫多项式 ${\displaystyle \cos n\theta =T_{n}\cos \theta \,}$ ${\displaystyle \sin ^{2}n\theta =S_{n}\sin ^{2}\theta \,}$ ${\displaystyle \cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta )^{n}\,}$
${\displaystyle 1+2\cos x+2\cos 2x+2\cos 3x+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right]}{\sin {\frac {x}{2}}}}}$

（这个${\displaystyle x}$的函数是狄利克雷核。）

### 二倍角、三倍角和半角公式

{\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}} {\displaystyle {\begin{aligned}\cot 2\theta &={\frac {\cot ^{2}\theta -1}{2\cot \theta }}\\&={\frac {\cot \theta -\tan \theta }{2}}\end{aligned}}} {\displaystyle {\begin{aligned}\csc 2\theta &={\frac {\csc ^{2}\theta }{2\cot \theta }}\\&={\frac {\sec \theta \csc \theta }{2}}\end{aligned}}}

${\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,}$ ${\displaystyle \cot 3\theta ={\frac {\cot ^{3}\theta -3\cot \theta }{3\cot ^{2}\theta -1}}}$ ${\displaystyle \csc 3\theta ={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}}$

${\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}}$ {\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {1+\cos \theta \over 1-\cos \theta }}\\&={\frac {\sin \theta }{1-\cos \theta }}\\&={\frac {1+\cos \theta }{\sin \theta }}\\&={\frac {\cos \theta -\sin \theta +1}{\cos \theta +\sin \theta -1}}\end{aligned}}} ${\displaystyle \csc {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta -1}}}}$

### ${\displaystyle n}$倍角公式

${\displaystyle n}$倍角公式
${\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left[{\frac {1}{2}}(n-k)\pi \right]=\sin \theta \sum _{k=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{k}{\binom {n-1-k}{k}}~(2\cos \theta )^{n-1-2k}}$

（第二类切比雪夫多项式

${\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left[{\frac {1}{2}}(n-k)\pi \right]={\frac {1}{2}}\sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{k}{\frac {n}{n-k}}{\binom {n-k}{k}}~(2\cos \theta )^{n-2k}}$

（第一类切比雪夫多项式

${\displaystyle \tan n\theta ={\frac {\displaystyle \sum _{k=1}^{\left[{\frac {n}{2}}\right]}(-1)^{k+1}{\binom {n}{2k-1}}\tan ^{2k-1}\theta }{\displaystyle \sum _{k=1}^{\left[{\frac {n+1}{2}}\right]}(-1)^{k+1}{\binom {n}{2(k-1)}}\tan ^{2(k-1)}\theta }}}$
${\displaystyle n}$倍递回公式
${\displaystyle \tan \,n\theta ={\frac {\tan(n{-}1)\theta +\tan \theta }{1-\tan(n{-}1)\theta \,\tan \theta }}}$${\displaystyle \cot \,n\theta ={\frac {\cot(n{-}1)\theta \,\cot \theta -1}{\cot(n{-}1)\theta +\cot \theta }}}$。(递回关系)

### 其他函数的倍半角公式

• ${\displaystyle \operatorname {versin} 2\theta =2\sin ^{2}\theta ={\frac {(sin2\theta )(sin\theta )}{cos\theta }}=1-\cos 2\theta }$

• ${\displaystyle \operatorname {cvs} 2\theta =(\sin \theta -\cos \theta )^{2}=1-\sin 2\theta }$

## 幂简约公式

${\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}}$ ${\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}$ ${\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}}$
${\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}}$ ${\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}}$ ${\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}}$
${\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}}$ ${\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}}$ ${\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}}$
${\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}}$ ${\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}}$ ${\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}}$

## 数值连乘

${\displaystyle \prod _{k=0}^{n-1}\cos 2^{k}\theta ={\frac {\sin 2^{n}\theta }{2^{n}\sin \theta }}}$
${\displaystyle \prod _{k=0}^{n-1}\sin \left(x+{\frac {k\pi }{n}}\right)={\frac {\sin nx}{2^{n-1}}}}$
${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}}$,${\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}}$,${\displaystyle \prod _{k=1}^{n}\sin \left({\frac {k\pi }{2n+1}}\right)={\frac {\sqrt {2n+1}}{2^{n}}}}$
${\displaystyle \prod _{k=1}^{n-1}\cos \left({\frac {k\pi }{n}}\right)={\frac {\sin {\frac {n\pi }{2}}}{2^{n-1}}}}$,${\displaystyle \prod _{k=1}^{n-1}\cos \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}}$,${\displaystyle \prod _{k=1}^{n}\cos \left({\frac {k\pi }{2n+1}}\right)={\frac {1}{2^{n}}}}$
${\displaystyle \prod _{k=1}^{n-1}\tan \left({\frac {k\pi }{n}}\right)={\frac {n}{\sin {\frac {n\pi }{2}}}}}$,${\displaystyle \prod _{k=1}^{n-1}\tan \left({\frac {k\pi }{2n}}\right)=1}$,${\displaystyle \prod _{k=1}^{n}\tan {\frac {k\pi }{2n+1}}={\sqrt {2n+1}}}$

## 常见的恒等式

### 积化和差与和差化积恒等式

${\displaystyle \sin \alpha \cos \beta ={\sin(\alpha +\beta )+\sin(\alpha -\beta ) \over 2}}$ ${\displaystyle \sin \alpha +\sin \beta =2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$
${\displaystyle \cos \alpha \sin \beta ={\sin(\alpha +\beta )-\sin(\alpha -\beta ) \over 2}}$ ${\displaystyle \sin \alpha -\sin \beta =2\cos {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}}$
${\displaystyle \cos \alpha \cos \beta ={\cos(\alpha +\beta )+\cos(\alpha -\beta ) \over 2}}$ ${\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}}$
${\displaystyle \sin \alpha \sin \beta =-{\cos(\alpha +\beta )-\cos(\alpha -\beta ) \over 2}}$ ${\displaystyle \cos \alpha -\cos \beta =-2\sin {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}}$

### 平方差公式

${\displaystyle \sin(x+y)\sin(x-y)=\sin ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\cos ^{2}{x}\,}$

${\displaystyle \cos(x+y)\cos(x-y)=\cos ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\sin ^{2}{x}\,}$

(可借由积化和差公式+2倍角公式推导而来)

### 其他恒等式

${\displaystyle \cot x\cot y+\cot y\cot z+\cot z\cot x=1}$

${\displaystyle \cot x+\cot y+\cot z=\cot x\cot y\cot z}$

${\displaystyle \sin x+\sin y+\sin z=4\cos {\frac {x}{2}}\cos {\frac {y}{2}}\cos {\frac {z}{2}}}$
${\displaystyle \cos x+\cos y+\cos z=1+4\sin {\frac {x}{2}}\sin {\frac {y}{2}}\sin {\frac {z}{2}}}$

#### 托勒密定理

{\displaystyle {\begin{aligned}\sin(w+x)\sin(x+y)&=\sin(x+y)\sin(y+z)\\&=\sin(y+z)\sin(z+w)\\&{}=\sin(z+w)\sin(w+x)\\&{}=\sin w\sin y+\sin x\sin z\end{aligned}}}

（前三个等式是一般情况；第四个是本质。）

## 三角函数与双曲函数的恒等式

${\displaystyle e^{ix}=\cos x+i\;\sin x,\;e^{-ix}=\cos x-i\;\sin x}$

${\displaystyle e^{x}=\cosh x+\sinh x\!,\;e^{-x}=\cosh x-\sinh x\!}$

${\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x}$

${\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x}$

${\displaystyle \sin \theta =-i\sinh {i\theta }\,}$ ${\displaystyle \sinh {\theta }=i\sin {(-i\theta )}\,}$
${\displaystyle \cos {\theta }=\cosh {i\theta }\,}$ ${\displaystyle \cosh {\theta }=\cos {(-i\theta )}\,}$
${\displaystyle \tan \theta =-i\tanh {i\theta }\,}$ ${\displaystyle \tanh {\theta }=i\tan {(-i\theta )}\,}$
${\displaystyle \cot {\theta }=i\coth {i\theta }\,}$ ${\displaystyle \coth \theta =-i\cot {(-i\theta )}\,}$
${\displaystyle \sec {\theta }=\operatorname {sech} {\,i\theta }\,}$ ${\displaystyle \operatorname {sech} {\theta }=\sec {(-i\theta )}\,}$
${\displaystyle \csc {\theta }=i\;\operatorname {csch} {\,i\theta }\,}$ ${\displaystyle \operatorname {csch} \theta =-i\csc {(-i\theta )}\,}$
• 其他恒等式：
${\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x}$
${\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x}$
${\displaystyle \cosh(x+iy)=\cosh(x)\cos(y)+i\sinh(x)\sin(y)\,}$
${\displaystyle \sinh(x+iy)=\sinh(x)\cos(y)+i\cosh(x)\sin(y)\,}$
${\displaystyle \tanh ix=i\tan x\,}$
${\displaystyle \cosh x=\cos ix\,}$
${\displaystyle \sinh x=-i\sin ix\,}$
${\displaystyle \tanh x=-i\tan ix\,}$

## 线性组合

${\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,}$

${\displaystyle \varphi =\arctan \left({\frac {b}{a}}\right)}$

${\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}$

${\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},}$

${\displaystyle \beta =\operatorname {atan2} \left(b\sin \alpha ,a+b\cos \alpha \right)}$

## 反三角函数

${\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}}\;}$
${\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}.\;}$
${\displaystyle \arctan x+\arctan {\frac {1}{x}}=\left\{{\begin{matrix}{\frac {\pi }{2}},&{\mbox{if }}x>0\\-{\frac {\pi }{2}},&{\mbox{if }}x<0\end{matrix}}\right.}$
${\displaystyle \arctan x+\arctan y=\arctan {\frac {x+y}{1-xy}}+\left\{{\begin{matrix}\pi ,&{\mbox{if }}x,y>0\\-\pi ,&{\mbox{if }}x,y<0\\0,&{\mbox{otherwise }}\end{matrix}}\right.}$

## 微积分

${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}$

${\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos x}{x}}=0}$

${\displaystyle {d \over dx}\sin(x)=\cos(x)}$

{\displaystyle {\begin{aligned}{d \over dx}\sin x&=\cos x,&{d \over dx}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\cos x&=-\sin x,&{d \over dx}\arccos x&=-{1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\tan x&=\sec ^{2}x,&{d \over dx}\arctan x&={1 \over 1+x^{2}}\\\\{d \over dx}\cot x&=-\csc ^{2}x,&{d \over dx}\operatorname {arccot} x&=-{1 \over 1+x^{2}}\\\\{d \over dx}\sec x&=\tan x\sec x,&{d \over dx}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\\\{d \over dx}\csc x&=-\csc x\cot x,&{d \over dx}\operatorname {arccsc} x&=-{1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}}

## 参考文献

1. ^ Abramowitz and Stegun, p. 78, 4.3.147
2. 苏学孟. 求三角函数乘积的常用方法. 中学数学教学. 1995, (6) [2014-12-27]. （原始内容存档于2014-12-27）.
3. ^ Abramowitz and Stegun, p. 75, 4.3.89–90
4. ^ Abramowitz and Stegun, p. 85, 4.5.68–69
• A one-page proof of many trigonometric identities using Euler's formula，by Connelly Barnes.
• Useful Formulae for Hong Kong Advanced Level Examination Pure Mathematics.