# 反三角函数

（重定向自反余割

## 主值

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

## 反三角函数之间的关系

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

${\displaystyle \arcsin(-x)=-\arcsin x\!}$
${\displaystyle \arccos(-x)=\pi -\arccos x\!}$
${\displaystyle \arctan(-x)=-\arctan x\!}$
${\displaystyle \operatorname {arccot}(-x)=\pi -\operatorname {arccot} x\!}$
${\displaystyle \operatorname {arcsec}(-x)=\pi -\operatorname {arcsec} x\!}$
${\displaystyle \operatorname {arccsc}(-x)=-\operatorname {arccsc} x\!}$

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

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

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

## 三角函數與反三角函數的關係

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

## 一般解

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

## 反三角函数的导数

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

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

## 表达为定积分

{\displaystyle {\begin{aligned}\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,\\\operatorname {arccot} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\end{aligned}}}

${\displaystyle x}$ 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

{\displaystyle {\begin{aligned}\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|\leq 1\end{aligned}}}
{\displaystyle {\begin{aligned}\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|\leq 1\end{aligned}}}
{\displaystyle {\begin{aligned}\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|\leq 1\qquad z\neq i,-i\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {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|\leq 1\qquad z\neq i,-i\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {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|\geq -4\end{aligned}}}
{\displaystyle {\begin{aligned}\operatorname {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|\geq 1\end{aligned}}}

${\displaystyle \arctan x=\infty {x}{1+x^{2}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kx^{2}}{(2k+1)(1+x^{2})}}}$

（注意对x= 0在和中的项是空积1。）

## 反三角函数的不定积分

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

### 舉例

{\displaystyle {\begin{aligned}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{aligned}}}

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

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

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

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

## 加法公式和減法公式

### ${\displaystyle \arcsin x+\arcsin y}$

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

### ${\displaystyle \arcsin x-\arcsin y}$

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

### ${\displaystyle \arccos x+\arccos y}$

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

### ${\displaystyle \arccos x-\arccos y}$

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

### ${\displaystyle \arctan x+\arctan y}$

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

### ${\displaystyle \arctan x-\arctan y}$

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

### ${\displaystyle \operatorname {arccot} x+\operatorname {arccot} y}$

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

### ${\displaystyle \arcsin x+\operatorname {arccot} x}$

${\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}},|x|\leq 1}$

### ${\displaystyle \arctan x+\operatorname {arccot} y}$

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

## 註釋與引用

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