# 反三角函数

## 主值

（注意：某些數學教科書的作者將arcsec的值域定為$[0,\frac{\pi}{2})\cup[\pi,\frac{3\pi}{2})$因為當tan的定義域落在此區間時，tan的值域≧0，如果arcsec的值域仍定為$[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$，將會造成tan(arcsec(x)) = ± x2 − 1，如果希望tan(arcsec(x)) = x2 − 1，那就必須將arcsec的值域定為$[0,\frac{\pi}{2})\cup[\pi,\frac{3\pi}{2})$，基於類似的理由arccsc的值域定為$(-\pi,-\frac{\pi}{2}]\cup(0,\frac{\pi}{2}]$

## 反三角函数之间的关系

$\arccos x = \frac{\pi}{2} - \arcsin x$
$\arccot x = \frac{\pi}{2} - \arctan x$
$\arccsc x = \frac{\pi}{2} - \arcsec x$

$\arcsin (-x) = - \arcsin x \!$
$\arccos (-x) = \pi - \arccos x \!$
$\arctan (-x) = - \arctan x \!$
$\arccot (-x) = \pi - \arccot x \!$
$\arcsec (-x) = \pi - \arcsec x \!$
$\arccsc (-x) = - \arccsc x \!$

$\arccos \frac{1}{x} \,= \arcsec x$
$\arcsin \frac{1}{x} \,= \arccsc x$
$\arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x =\arccot x, \$ $\ x > 0$
$\arctan \frac{1}{x} = -\frac{\pi}{2} - \arctan x = -\pi + \arccot x, \$ $\ x < 0$
$\arccot \frac{1}{x} = \frac{\pi}{2} - \arccot x =\arctan x, \$ $\ x > 0$
$\arccot \frac{1}{x} = \frac{3\pi}{2} - \arccot x = \pi + \arctan x,\$ $\ x < 0$
$\arcsec \frac{1}{x} = \arccos x$
$\arccsc \frac{1}{x} = \arcsin x$

$\arccos x = \arcsin \sqrt{1-x^2},$ $\ 0 \leq x \leq 1$
$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$

$\arcsin x = 2 \arctan \frac{x}{1+\sqrt{1-x^2}}$
$\arccos x = 2 \arctan \frac{\sqrt{1-x^2}}{1+x},$ $-1 < x \leq +1$
$\arctan x = 2 \arctan \frac{x}{1+\sqrt{1+x^2}}$

## 三角函數與反三角函數的關係

$\theta$ $\sin \theta$ $\cos \theta$ $\tan \theta$ 圖示
$\arcsin x$ $\sin (\arcsin x) = x$ $\cos (\arcsin x) = \sqrt{1-x^2}$ $\tan (\arcsin x) = \frac{x}{\sqrt{1-x^2}}$
$\arccos x$ $\sin (\arccos x) = \sqrt{1-x^2}$ $\cos (\arccos x) = x$ $\tan (\arccos x) = \frac{\sqrt{1-x^2}}{x}$
$\arctan x$ $\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}}$ $\cos (\arctan x) = \frac{1}{\sqrt{1+x^2}}$ $\tan (\arctan x) = x$
$\arccot x$ $\sin (\arccot x) = \frac{1}{\sqrt{1+x^2}}$ $\cos (\arccot x) = \frac{x}{\sqrt{1+x^2}}$ $\tan (\arccot x) = \frac{1}{x}$
$\arcsec x$ $\sin (\arcsec x) = \frac{\sqrt{x^2-1}}{x}$ $\cos (\arcsec x) = \frac{1}{x}$ $\tan (\arcsec x) = \sqrt{x^2-1}$
$\arccsc x$ $\sin (\arccsc x) = \frac{1}{x}$ $\cos (\arccsc x) = \frac{\sqrt{x^2-1}}{x}$ $\tan (\arccsc x) = \frac{1}{\sqrt{x^2-1}}$

## 一般解

$\sin y= x \ \Leftrightarrow\ (\ y = \arcsin x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z} \ \lor\ y= \pi - \arcsin x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z}\ )$
$\cos y= x \ \Leftrightarrow\ (\ y = \arccos x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z} \ \lor\ y = 2\pi - \arccos x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z}\ )$
$\tan y= x \ \Leftrightarrow\ \ y = \arctan x+ k\pi \text{ } \forall \text{ } k \in \mathbb{Z}$
$\cot y= x \ \Leftrightarrow\ \ y = \arccot x+ k\pi \text{ } \forall \text{ } k \in \mathbb{Z}$
$\sec y= x \ \Leftrightarrow\ (\ y = \arcsec x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z} \ \lor\ y = 2\pi - \arcsec x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z}\ )$
$\csc y= x \ \Leftrightarrow\ (\ y = \arccsc x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z} \ \lor\ y = \pi - \arccsc x+ 2k\pi \text{ } \forall \text{ } k \in \mathbb{Z}\ )$

## 反三角函数的导数

\begin{align} \frac{d}{dx} \arcsin x & {}= \frac{1}{\sqrt{1-x^2}}; \qquad |x| < 1\\ \frac{d}{dx} \arccos x & {}= \frac{-1}{\sqrt{1-x^2}}; \qquad |x| < 1\\ \frac{d}{dx} \arctan x & {}= \frac{1}{1+x^2}\\ \frac{d}{dx} \arccot x & {}= \frac{-1}{1+x^2}\\ \frac{d}{dx} \arcsec x & {}= \frac{1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1\\ \end{align}

$\frac{d \arcsin x}{dx} = \frac{d \theta}{d \sin \theta} = \frac{1} {\cos \theta} = \frac{1} {\sqrt{1-\sin^2 \theta}} = \frac{1}{\sqrt{1-x^2}}$

## 表达为定积分

\begin{align} \arcsin x &{}= \int_0^x \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arccos x &{}= \int_x^1 \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arctan x &{}= \int_0^x \frac 1 {z^2 + 1}\,dz,\\ \arccot x &{}= \int_x^\infty \frac {1} {z^2 + 1}\,dz,\\ \arcsec x &{}= \int_1^x \frac 1 {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1\\ \arccsc x &{}= \int_x^\infty \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1 \end{align}

x等于1时，在有极限的域上的积分是瑕积分，但仍是良好定义的。

## 无穷级数

\begin{align} \arcsin z & {}= z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots\\ & {}= \sum_{n=0}^\infty \left[ \frac {(2n)!} {2^{2n}(n!)^2} \right] \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}
\begin{align} \arccos z & {}= \frac {\pi} {2} - \arcsin z \\ & {}= \frac {\pi} {2} - \left[z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots \right] \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left[ \frac {(2n)!} {2^{2n}(n!)^2} \right] \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}
\begin{align} \arctan z & {}= z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \qquad z \neq i,-i \end{align}
\begin{align} \arccot z & {}= \frac {\pi} {2} - \arctan z \\ & {}= \frac {\pi} {2} - \left( z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots \right) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \qquad z \neq i,-i \end{align}
\begin{align} \arcsec z & {}= \arccos\left(z^{-1}\right) \\ & {}= \frac {\pi} {2} -\left[z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^{-5}} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^{-7}} {7} + \cdots \right] \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left[ \frac {(2n)!} {2^{2n}(n!)^2} \right] \frac {z^{-(2n+1)}} {(2n+1)} ; \qquad \left| z \right| \ge 1 \end{align}
\begin{align} \arccsc z & {}= \arcsin\left(z^{-1}\right) \\ & {}= z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4 } \right) \frac {z^{-5}} {5} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {z^{-7}} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \left[ \frac {(2n)!} {2^{2n}(n!)^2} \right] \frac {z^{-(2n+1)}} {2n+1} ; \qquad \left| z \right| \ge 1 \end{align}

$\arctan x = \frac{x}{1+x^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k x^2}{(2k+1)(1+x^2)}$

（注意对x= 0在和中的项是空积1。）

## 反三角函数的不定积分

\begin{align} \int \arcsin x\,dx &{}= x\,\arcsin x + \sqrt{1-x^2} + C, \qquad x\le 1\\ \int \arccos x\,dx &{}= x\,\arccos x - \sqrt{1-x^2} + C, \qquad x\le 1\\ \int \arctan x\,dx &{}= x\,\arctan x - \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arccot x\,dx &{}= x\,\arccot x + \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arcsec x\,dx &{}= x\,\arcsec x - \ln\left(\left|x\right|+\sqrt{x^2-1}\right) + C= x\,\arcsec x - \operatorname{arcosh}\left|x\right| + C, \qquad x>1 \\ \int \arccsc x\,dx &{}= x\,\arccsc x + \ln\left(\left|x\right|+\sqrt{x^2-1}\right) + C= x\,\arccsc x + \operatorname{arcosh}\left|x\right| + C, \qquad x> 1 \end{align}

### 舉例

\begin{align} u &{}=&\arcsin x &\quad\quad\mathrm{d}v = \mathrm{d}x\\ \mathrm{d}u &{}=&\frac{\mathrm{d}x}{\sqrt{1-x^2}}&\quad\quad{}v = x \end{align}

$\int \arcsin(x)\,\mathrm{d}x = x \arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x$

$k = 1 - x^2.\,$

$\mathrm{d}k = -2x\,\mathrm{d}x$

$\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x = -\frac{1}{2}\int \frac{\mathrm{d}k}{\sqrt{k}} = -\sqrt{k}$

$\int \arcsin(x)\, \mathrm{d}x = x \arcsin x + \sqrt{1-x^2}+C$

## 加法公式和減法公式

### arcsin x + arcsin y

$\arcsin x + \arcsin y = \arcsin\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), xy \leq 0 \or x^2 + y^2\leq 1$
$\arcsin x + \arcsin y = \pi - \arcsin\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), x > 0, y > 0, x^2 + y^2 > 1$
$\arcsin x + \arcsin y = - \pi - \arcsin\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), x < 0 , y < 0, x^2 + y^2 > 1$

### arcsin x - arcsin y

$\arcsin x - \arcsin y = \arcsin\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right), xy \geq 0 \or x^2 + y^2\leq 1$
$\arcsin x - \arcsin y = \pi - \arcsin\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right), x > 0, y < 0, x^2 + y^2 > 1$
$\arcsin x - \arcsin y = - \pi - \arcsin\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right), x < 0, y > 0, x^2 + y^2 > 1$

### arccos x + arccos y

$\arccos x + \arccos y = \arccos\left(xy - \sqrt{1-x^2}\cdot\sqrt{1-y^2}\right), x + y \geq 0$
$\arccos x + \arccos y = 2\pi - \arccos\left(xy - \sqrt{1-x^2}\cdot\sqrt{1-y^2}\right), x + y < 0$

### arccos x - arccos y

$\arccos x - \arccos y = -\arccos\left(xy + \sqrt{1-x^2}\cdot\sqrt{1-y^2}\right), x \geq y$
$\arccos x - \arccos y = \arccos\left(xy + \sqrt{1-x^2}\cdot\sqrt{1-y^2}\right), x < y$

### arctan x + arctan y

$\arctan\,x + \arctan\,y =\arctan\,{\frac{x+y}{1-xy}}, xy < 1$
$\arctan\,x + \arctan\,y =\pi + \arctan\,{\frac{x+y}{1-xy}}, x > 0, xy > 1$
$\arctan\,x + \arctan\,y =-\pi + \arctan\,{\frac{x+y}{1-xy}}, x < 0, xy > 1$

### arctan x - arctan y

$\arctan x - \arctan y =\arctan{\frac{x-y}{1+xy}}, xy > -1$
$\arctan x - \arctan y =\pi + \arctan{\frac{x-y}{1+xy}}, x > 0, xy < -1$
$\arctan x - \arctan y =-\pi + \arctan {\frac{x-y}{1+xy}}, x < 0, xy < -1$

### arccot x + arccot y

$\arccot x + \arccot y =\arccot{\frac{xy-1}{x+y}}, x > -y$
$\arccot x + \arccot y =\arccot {\frac{xy-1}{x+y}}+\pi, x < -y$

### arcsin x + arccos x

$\arcsin x + \arccos x =\frac{\pi}{2}, |x|\leq1$

### arctan x + arccot x

$\arctan x + \arccot x =\frac{\pi}{2}$

## 註釋與引用

1. ^ $\theta =\arccos x$，得到：
$\frac{d\arccos x}{dx}=\frac{d\theta }{d\cos \theta }=\frac{-1}{\sin \theta }=\frac{1}{\sqrt{1-\cos ^{2}\theta }}=\frac{-1}{\sqrt{1-x^{2}}}$
因為要使根號內部恆為正，所以在條件加上$|x|<1$$\theta =\arctan x$，得到：
$\frac{d\arctan x}{dx}=\frac{d\theta }{d\tan \theta }=\frac{1}{\sec ^{2}\theta }=\frac{1}{1+\tan ^{2}\theta }=\frac{1}{1+x^{2}}$
$\theta =\arccot x$，得到：
$\frac{d\operatorname{arc}\cot x}{dx}=\frac{d\theta }{d\cot \theta }=\frac{-1}{\csc ^{2}\theta }=\frac{1}{1+\cot ^{2}\theta }=\frac{-1}{1+x^{2}}$
$\theta =\arcsec x$，得到：
$\frac{d\operatorname{arc}\sec x}{dx}=\frac{d\theta }{d\sec \theta }=\frac{1}{\sec \theta \tan \theta }=\frac{1}{\left| x \right|\sqrt{x^{2}-1}}$
因為要使根號內部恆為正，所以在條件加上$|x|>1$，比較容易被忽略是$\sec \theta$產生的絕對值 $\sec ^{-1}\theta$的定義域是$0\le \theta \le \pi$，其所產生的反函數皆為正，所以需要加上絕對值。
$\theta =\arccsc x$，得到：
$\frac{d\operatorname{arc}\csc x}{dx}=\frac{d\theta }{d\csc \theta }=\frac{-1}{\csc \theta \cot \theta }=\frac{-1}{\left| x \right|\sqrt{x^{2}-1}}$
因為要使根號內部恆為正，所以在條件加上$|x|>1$，比較容易被忽略是$\csc \theta$產生的絕對值 $\csc ^{-1}\theta$的定義域是$-\frac{\pi}{2}\le \theta \le \frac{\pi}{2},\theta\ne0$，其所產生的反函數皆為負，所以需要加上絕對值。