# 三角恆等式

## 符號

sine sin cosecant csc
cosine cos secant sec
tangent tan cotangent cot

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

$\mathbf{0}$ $\tfrac{1}{12}$ $\tfrac{1}{8}$ $\tfrac{1}{6}$ $\tfrac{1}{4}$ $\tfrac{1}{2}$ $\tfrac{3}{4}$ $\mathbf{1}$

## 基本關係

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

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

### 其他函數的基本關係

$\operatorname{vers}(\theta)$
$\operatorname{ver}(\theta)$
$1 - \cos (\theta)$

$\operatorname{cvs}(\theta)$
$1 - \sin(\theta)$

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

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

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

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

## 對稱、移位和周期

### 對稱

\begin{align} \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{align} \begin{align} \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{align} \begin{align} \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{align} \begin{align} \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{align}

### 移位和周期

$\tan$$\cot$ 的周期

$\sin$, $\cos$, $\csc$$\sec$ 的周期
\begin{align} \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{align} \begin{align} \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{align} \begin{align} \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{align} \begin{align} \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{align}

## 角的和差恆等式

正弦 $\sin(\theta\pm\psi)=\sin\theta\cos\psi\pm\cos\theta\sin\psi\,$ $\cos(\theta\pm\psi)=\cos\theta\cos\psi\mp\sin\theta\sin\psi\,$ $\tan(\theta\pm\psi)=\frac{\tan\theta\pm\tan\psi}{1\mp\tan\theta\tan\psi}$ $\cot(\theta\pm\psi)=\frac{\cot\theta\cot\psi\mp1}{\cot\psi\pm\cot\theta}$ $\sec(\theta\pm\psi)=\frac{\sec\theta\sec\psi}{1\mp\tan\theta\tan\psi}$ $\csc(\theta\pm\psi)=\frac{\csc\theta\csc\psi}{\cot\psi\pm\cot\theta}$ 注意正負號的對應。 \begin{align}x \pm y = a \pm b &\Rightarrow \ x + y = a + b \\ &\mbox{and} \ x -y = a -b \end{align} \begin{align} x \pm y = a \mp b &\Rightarrow \ x + y = a - b \\ &\mbox{and}\ x - y = a + b\end{align}

### 正弦與餘弦的無限多項和

$\sin\left(\sum_{i=1}^\infty \theta_i\right) =\sum_{\mathrm{odd}\ k \ge 1} (-1)^{(k-1)/2} \sum_{ |A| = k } \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)$
$\cos\left(\sum_{i=1}^\infty \theta_i\right) =\sum_{\mathrm{even}\ k \ge 0} ~ (-1)^{k/2} ~~ \sum_{ |A| = k } \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right)$

### 正切的有限多項和

xi = tan(θi )，對於 i = 1, ..., n。設 ek 是變數 xi, i = 1, ..., n, k = 0, ..., nk基本對稱多項式。則

$\tan(\theta_1+\cdots+\theta_n) = \frac{e_1 - e_3 + e_5 -\cdots}{e_0 - e_2 + e_4 - \cdots},$

\begin{align} \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{align}

## 多倍角公式

Tn 是 n 次切比雪夫多項式 $\cos n\theta =T_n \cos \theta \,$ $\sin^2 n\theta = S_n \sin^2\theta\,$ $\cos n\theta +i\sin n\theta=(\cos(\theta)+i\sin(\theta))^n \,$
$1+2\cos(x) + 2\cos(2x) + 2\cos(3x) + \cdots + 2\cos(nx) = \frac{\sin\left(\left(n +\frac{1}{2}\right)x\right)}{\sin(x/2)}.$

(這個 x 的函數是狄利克雷核。)

## 倍角公式和半形公式

$\cos \frac{\theta}{2} = \pm\, \sqrt{\frac{1 + \cos\theta}{2}}$ \begin{align} \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{align} $\csc \frac{\theta}{2} = \pm\, \sqrt{\frac{2\sec\theta}{\sec\theta - 1}}$

\begin{align} \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{align} $\cot 2\theta = \frac{\cot \theta - \tan \theta}{2}\,$ $\csc 2\theta = \frac{\csc^2\theta}{2\cot\theta}$
$=\frac{\sec\theta\csc\theta}{2}$

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

$\cos4\theta=8\cos^4\theta-8\cos^2\theta+1\,$

$\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta \,$

$\sin 6\theta=32\cos\theta\sin^5\theta-32\cos\theta\sin^3\theta+6\cos\theta\sin\theta\,$
$\cos6\theta=32\cos^6\theta-48\cos^4\theta+18\cos^2\theta-1\,$
$\tan 6\theta = \frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}\,$

$\sin 7\theta=7\sin\theta-56\sin^3\theta+112\sin^5\theta-64\sin^7\theta \,$
$\cos 7\theta=64\cos^7\theta-112\cos^5\theta+56\cos^3\theta-7\cos\theta \,$
$\tan 7\theta = \frac{\tan^7\theta-21\tan^5\theta+35\tan^3\theta-7\tan\theta}{7\tan^6\theta-35\tan^4\theta+21\tan^2\theta-1}\,$[來源請求]

$\sin 8\theta=192\cos\theta\sin^5\theta-128\cos\theta\sin^7\theta-80\cos\theta\sin^3\theta+8\cos\theta\sin\theta \,$
$\cos 8\theta=128\cos^8\theta-256\cos^6\theta+160\cos^4\theta-32\cos^2\theta+1 \,$
$\tan 8\theta = \frac{8\tan\theta-56\tan^3\theta+56\tan^5\theta-8\tan^7\theta}{1-28\tan^2\theta+70\tan^4\theta-28\tan^6\theta+\tan^8\theta}\,$[來源請求]

$\sin 9\theta=256\sin^9\theta-576\sin^7\theta+432\sin^5\theta-120\sin^3\theta+9\sin\theta \,$
$\cos 9\theta=256\cos^9\theta-576\cos^7\theta+432\cos^5\theta-120\cos^3\theta+9\cos\theta \,$
$\tan 9\theta = \frac{9\tan\theta-84\tan^3\theta+126\tan^5\theta-36\tan^7\theta+\tan^9\theta}{1-36\tan^2\theta+126\tan^4\theta-84\tan^6\theta+9\tan^8\theta}\,$[來源請求]

$\sin 10\theta=512\sin^9\theta\cos\theta-1024\sin^7\theta\cos\theta+672\sin^5\theta\cos\theta-160\sin^3\theta\cos\theta+10\sin\theta\cos\theta \,$
$\cos 10\theta=512\cos^{10}\theta-1280\cos^8\theta+1102\cos^6\theta-400\cos^4\theta+50\cos\theta -1\,$
$\tan 10\theta = \frac{10\tan\theta-120\tan^3\theta+252\tan^5\theta-120\tan^7\theta+10\tan^9\theta}{1-45\tan^2\theta+210\tan^4\theta-210\tan^6\theta+45\tan^8\theta-\tan^{10}\theta}\,$[來源請求]
n倍角公式
$\sin n\theta = \sum_{k=0}^n \binom{n}{k} \cos^k \theta\,\sin^{n-k} \theta\,\sin\left(\frac{1}{2}(n-k)\pi\right) =\sin \theta \sum_{k=0}^{\lfloor \frac{n-1}{2}\rfloor}(-1)^k \binom{n-1-k}{k}~(2\cos \theta)^{n-1-2k}$（第二類切比雪夫多項式
$\cos n\theta = \sum_{k=0}^n \binom{n}{k} \cos^k \theta\,\sin^{n-k} \theta\,\cos\left(\frac{1}{2}(n-k)\pi\right) =\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}(-1)^k \frac{n}{n-k} \binom{n-k}{k}~(2\cos \theta)^{n-2k}$（第一類切比雪夫多項式
$\tan n\theta = \frac{\displaystyle \sum_{k=1}^{\left[\frac{n}{2}\right]} (-1)^{k+1} \binom{n}{2k-1} \tan^{2k-1}\theta}{\displaystyle \sum_{k=1}^{\left[\frac{n+1}{2}\right]} (-1)^{k+1} \binom{n}{2(k-1)} \tan^{2(k-1)}\theta}$
n倍遞迴公式
$\tan\,n\theta = \frac{\tan (n{-}1)\theta + \tan \theta}{1 - \tan (n{-}1)\theta\,\tan \theta}$$\cot\,n\theta = \frac{\cot (n{-}1)\theta\,\cot \theta - 1}{\cot (n{-}1)\theta + \cot \theta}.$(遞迴關係)

### 其他函數的倍半形公式

• $\operatorname{versin} 2\theta = 2\sin^2\theta$

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

## 三分之一角公式

• 利用三倍角公式
• $\sin 3\theta = 3 \sin \theta- 4 \sin^3\theta \,$
• $\cos 3\theta = 4 \cos^3\theta - 3 \cos \theta \,$

• $\sin \theta = 3\sin\frac{\theta}{3}- 4\sin^3\frac{1}{3}\theta\,$
• $\cos \theta = 4\cos^3\frac{\theta}{3} - 3\cos\frac{1}{3}\theta\,$

$\cos \frac{\theta}{3}\,$當成未知數，$\cos \theta \,$當成常數項，解一元三次方程式即可求出

• $x_1=\frac{1}{2 \sqrt[3]{-\sin\theta +\sqrt{\sin^2\theta -1}}} + \frac{\sqrt[3]{-\sin\theta +\sqrt{\sin^2\theta -1}}}{2}\,$
• $x_2= - \frac{1 + i \sqrt{3}}{4 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} - \frac{(1 - i \sqrt{3}) \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{4}\,$
• $x_3= - \frac{1 - i \sqrt{3}}{4 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} - \frac{(1 + i \sqrt{3}) \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{4}\,$
• 當-90°≤$\theta \,$≤90°時$\sin \frac{1}{3}\theta =x_3= - \frac{1 - i \sqrt{3}}{4 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} - \frac{(1 + i \sqrt{3}) \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{4}\,$
• 當90°≤$\theta \,$≤450°時$\sin \frac{1}{3}\theta =x_1= \frac{1}{2 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} + \frac{ \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{2}\,$
• 當450°≤$\theta \,$≤630°時$\sin \frac{1}{3}\theta =x_3= - \frac{1 - i \sqrt{3}}{4 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} - \frac{(1 + i \sqrt{3}) \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{4}\,$
• 當630°≤$\theta \,$≤990°時$\sin \frac{1}{3}\theta =x_2= - \frac{1 + i \sqrt{3}}{4 \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}} - \frac{(1 - i \sqrt{3}) \sqrt[3]{-\sin \theta + \sqrt{\sin^2\theta -1}}}{4}\,$

### 簡化

$\cos\frac{\theta}{n} = \Re\left(\sqrt[n]{\cos\theta+i\sin\theta}\right) = \frac{1}{2}\left(\sqrt[n]{\cos\theta+i\sin\theta}+\sqrt[n]{\cos\theta-i\sin\theta}\right)$
$\sin\frac{\theta}{n} = \Im\left(\sqrt[n]{\cos\theta+i\sin\theta}\right) = \frac{1}{2i}\left(\sqrt[n]{\cos\theta+i\sin\theta}-\sqrt[n]{\cos\theta-i\sin\theta}\right)$

$\cos\frac{\theta}{3}=\frac{1}{2}\left(\sqrt[3]{\cos\theta+i\sin\theta}+\sqrt[3]{\cos\theta-i\sin\theta}\right)$
$\sin\frac{\theta}{3}=\frac{1}{2i}\left(\sqrt[3]{\cos\theta+i\sin\theta}-\sqrt[3]{\cos\theta-i\sin\theta}\right)$

## 冪簡約公式

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

## 常見的恆等式

### 積化和差與和差化積恆等式

$\sin\theta\sin\phi={\cos(\theta+\phi)-\cos(\theta-\phi)\over-2}$
$\cos\theta\cos\phi={\cos(\theta-\phi)+\cos(\theta+\phi)\over2}$
$\sin\theta\cos\phi={\sin(\theta+\phi)+\sin(\theta-\phi)\over2}$
$\cos\theta\sin\phi={\sin(\theta+\phi)-\sin(\theta-\phi)\over2}$

$\sin\theta+\sin\phi=2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$
$\cos\theta+\cos\phi=2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}$
$\cos\theta-\cos\phi=-2\sin{\theta+\phi\over2}\sin{\theta-\phi\over2}$
$\sin\theta-\sin\phi=2\cos{\theta+\phi\over2}\sin{\theta-\phi\over2}$

### 其他恆等式

$\cot x\cot y + \cot y\cot z + \cot z\cot x = 1.\,$

$\cot x + \cot y + \cot z = \cot x\cot y\cot z.\,$

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

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

### 托勒密定理

$\mbox{If }w + x + y + z = \pi = \mbox{half circle,} \,$
\begin{align} \mbox{then } & \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{align}

（前三個等式是平凡的；第四個是這個恆等式的實質。）本質上這是使用三角學語言的托勒密定理

## 三角函數和雙曲函數的恆等式

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

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

$\cosh ix = \tfrac12(e^{i x} + e^{-i x}) = \cos x$

$\sinh ix = \tfrac12(e^{i x} - e^{-i x}) = i \sin x$

$\sin \theta =-i\sinh{i\theta}\,$ $\sinh{\theta}=i\sin{(-i\theta)}\,$
$\cos{\theta}=\cosh{i\theta}\,$ $\cosh{\theta}=\cos{(-i\theta)}\,$
$\tan \theta =-i\tanh{i\theta}\,$ $\tanh{\theta}=i\tan{(-i\theta)}\,$
$\cot{\theta}=i\coth{i\theta}\,$ $\coth \theta =-i\cot{(-i\theta)}\,$
$\sec{\theta}=\operatorname{sech}{\,i\theta}\,$ $\operatorname{sech}{\theta}=\sec{(-i\theta)}\,$
$\csc{\theta}=i\;\operatorname{csch}{\, i\theta}\,$ $\operatorname{csch} \theta =-i\csc{(-i\theta)}\,$
• 其他恆等式:
$\cosh ix = \tfrac12(e^{i x} + e^{-i x}) = \cos x$
$\sinh ix = \tfrac12(e^{i x} - e^{-i x}) = i \sin x$
$\cosh(x+iy) = \cosh(x) \cos(y) + i \sinh(x) \sin(y) \,$
$\sinh(x+iy) = \sinh(x) \cos(y) + i \cosh(x) \sin(y) \,$
$\tanh ix = i \tan x \,$
$\cosh x = \cos ix \,$
$\sinh x = -i \sin ix \,$
$\tanh x = -i \tan ix \,$

## 線性組合

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

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

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

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

$\beta = {\rm arctan} \left(\frac{b\sin \alpha}{a + b\cos \alpha}\right).$

## 反三角函數

$\arcsin x+\arccos x=\frac{\pi}{2}\;$
$\arctan x+\arccot x=\frac{\pi}{2}.\;$
$\arctan x+\arctan \frac{1}{x}=\left\{\begin{matrix} \frac{\pi}{2}, & \mbox{if }x > 0 \\ -\frac{\pi}{2}, & \mbox{if }x < 0 \end{matrix}\right.$
$\arctan x+\arctan y=\arctan \frac{x+y}{1-xy}+\left\{\begin{matrix} \pi, & \mbox{if }x,y>0 \\ -\pi, & \mbox{if }x,y<0 \\ 0, & \mbox{otherwise } \end{matrix}\right.$
 $\sin ( \arccos x)=\sqrt{1-x^2} \,$ $\sin ( \arctan x)=\frac{x}{\sqrt{1+x^2}}$ $\cos ( \arctan x)=\frac{1}{\sqrt{1+x^2}}$ $\cos ( \arcsin x)=\sqrt{1-x^2} \,$ $\tan ( \arcsin x)=\frac{x}{\sqrt{1 - x^2}}$ $\tan ( \arccos x)=\frac{\sqrt{1 - x^2}}{x}$

## 無限乘積公式

 $\sin x = x \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2 n^2}\right)$ $\sinh x = x \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2 n^2}\right)$ $\frac{\sin x}{x} = \prod_{n = 1}^\infty\cos\left(\frac{x}{2^n}\right)$ $\cos x = \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$ $\cosh x = \prod_{n = 1}^\infty\left(1 + \frac{x^2}{\pi^2(n - \frac{1}{2})^2}\right)$ $|\sin x| = \frac1{2}\prod_{n = 0}^\infty \sqrt[2^{n+1}]{\left|\tan\left(2^n x\right)\right|}$

## 微積分

$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$

$\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=0$

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

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

## 指數定義

$\sin \theta = \frac{e^{{\mathrm{i}}\theta} - e^{-{\mathrm{i}}\theta}}{2{\mathrm{i}}} \,$ $\arcsin x = -{\mathrm{i}}\ln \left({\mathrm{i}}x + \sqrt{1 - x^2}\right) \,$
$\cos \theta = \frac{e^{{\mathrm{i}}\theta} + e^{-{\mathrm{i}}\theta}}{2} \,$ $\arccos x = -{\mathrm{i}}\ln \left(x + \sqrt{x^2 - 1}\right) \,$

$\tan \theta = \frac{e^{{\mathrm{i}}\theta} - e^{-{\mathrm{i}}\theta}}{{\mathrm{i}}(e^{{\mathrm{i}}\theta} + e^{-{\mathrm{i}}\theta})} \,$ $\arctan x = \frac{\mathrm{i}}{2}\ln \left(\frac{{\mathrm{i}}+ x}{{\mathrm{i}}- x}\right)\,$
$\csc \theta = \frac{2{\mathrm{i}}}{e^{{\mathrm{i}}\theta} - e^{-{\mathrm{i}}\theta}} \,$ $\arccsc x = -{\mathrm{i}}\ln \left(\tfrac{\mathrm{i}}{x} + \sqrt{1 - \tfrac{1}{x^2}}\right) \,$
$\sec \theta = \frac{2}{e^{{\mathrm{i}}\theta} + e^{-{\mathrm{i}}\theta}} \,$ $\arcsec x = -{\mathrm{i}}\ln \left(\tfrac{1}{x} + \sqrt{1 - \tfrac{\mathrm{i}}{x^2}}\right) \,$
$\cot \theta = \frac{{\mathrm{i}}(e^{{\mathrm{i}}\theta} + e^{-{\mathrm{i}}\theta})}{e^{{\mathrm{i}}\theta} - e^{-{\mathrm{i}}\theta}} \,$ $\arccot x = \frac{\mathrm{i}}{2}\ln \left(\frac{{\mathrm{i}}- x}{{\mathrm{i}}+ x}\right) \,$
$\operatorname{cis} \, \theta = e^{{\mathrm{i}}\theta} \,$ $\operatorname{arccis} \, x =-{\mathrm{i}} \ln x \,$

## 參考文獻

1. ^ Abramowitz and Stegun, p. 78, 4.3.147
2. ^ Abramowitz and Stegun, p. 75, 4.3.89–90
3. ^ Abramowitz and Stegun, p. 85, 4.5.68–69