反三角函數

在數學中，反三角函數（英語：inverse trigonometric function）是三角函數的反函數。

反三角函數示意圖

幾個反三角函數的圖形，其中，反餘切以複變分析定義，因此在原點處出現不連續斷點

數學符號

符號${\displaystyle \sin ^{-1},\cos ^{-1}}$等常用於${\displaystyle \arcsin ,\arccos }$等。但是這種符號有時在${\displaystyle \sin ^{-1}x}$和${\displaystyle {\frac {1}{\sin x}}}$之間易造成混淆。

在編程中，函數${\displaystyle \arcsin }$, ${\displaystyle \arccos }$, ${\displaystyle \arctan }$通常叫做${\displaystyle \mathrm {asin} }$, ${\displaystyle \mathrm {acos} }$, ${\displaystyle \mathrm {atan} }$。很多編程語言提供兩自變量atan2函數，它計算給定${\displaystyle y}$和${\displaystyle x}$的${\displaystyle {\frac {y}{x}}}$的反正切，但是值域為${\displaystyle [-\pi ,\pi ]}$。

• 
  在笛卡爾平面上${\displaystyle f(x)=\arcsin x}$(紅)和${\displaystyle f(x)=\arccos x}$(綠)函數的常用主值的圖像。
• 
  在笛卡爾平面上${\displaystyle f(x)=\arctan x}$(紅)和${\displaystyle f(x)=\operatorname {arccot} x}$(綠)函數的常用主值的圖像。
• 
  在笛卡爾平面上${\displaystyle f(x)=\operatorname {arccsc} x}$(紅)和${\displaystyle f(x)=\operatorname {arcsec} x}$(綠)函數的常用主值的圖像。

主值

下表列出基本的反三角函數。

名稱	常用符號	定義	定義域	值域
反正弦	${\displaystyle y=\arcsin x}$	${\displaystyle x=\sin y}$	${\displaystyle [-1,1]}$	${\displaystyle [-{\frac {\pi }{2}},{\frac {\pi }{2}}]}$
反餘弦	${\displaystyle y=\arccos x}$	${\displaystyle x=\cos y}$	${\displaystyle [-1,1]}$	${\displaystyle [0,\pi ]}$
反正切	${\displaystyle y=\arctan x}$	${\displaystyle x=\tan y}$	${\displaystyle \mathbb {R} }$	${\displaystyle (-{\frac {\pi }{2}},{\frac {\pi }{2}})}$
反餘切	${\displaystyle y=\operatorname {arccot} x}$	${\displaystyle x=\cot y}$	${\displaystyle \mathbb {R} }$	${\displaystyle (0,\pi )}$
反正割	${\displaystyle y=\operatorname {arcsec} x}$	${\displaystyle x=\sec y}$	${\displaystyle (-\infty ,-1]\cup [1,+\infty )}$	${\displaystyle [0,{\frac {\pi }{2}})\cup ({\frac {\pi }{2}},\pi ]}$
反餘割	${\displaystyle y=\operatorname {arccsc} x}$	${\displaystyle x=\csc y}$	${\displaystyle (-\infty ,-1]\cup [1,+\infty )}$	${\displaystyle [-{\frac {\pi }{2}},0)\cup (0,{\frac {\pi }{2}}]}$

（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${\displaystyle \geq 0}$，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${\displaystyle (-\pi ,-{\frac {\pi }{2}}]\cup (0,{\frac {\pi }{2}}]}$）

如果${\displaystyle x}$允許是複數，則${\displaystyle y}$的值域只適用它的實部。

反三角函數之間的關係

餘角：

${\displaystyle \arccos x={\frac {\pi }{2}}-\arcsin x}$

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

${\displaystyle \operatorname {arccsc} x={\frac {\pi }{2}}-\operatorname {arcsec} x}$

負數參數：

${\displaystyle \arcsin(-x)=-\arcsin x\!}$

${\displaystyle \arccos(-x)=\pi -\arccos x\!}$

${\displaystyle \arctan(-x)=-\arctan x\!}$

${\displaystyle \operatorname {arccot}(-x)=\pi -\operatorname {arccot} x\!}$

${\displaystyle \operatorname {arcsec}(-x)=\pi -\operatorname {arcsec} x\!}$

${\displaystyle \operatorname {arccsc}(-x)=-\operatorname {arccsc} x\!}$

倒數參數：

${\displaystyle \arccos {\frac {1}{x}}\,=\operatorname {arcsec} x}$

${\displaystyle \arcsin {\frac {1}{x}}\,=\operatorname {arccsc} x}$

${\displaystyle \arctan {\frac {1}{x}}={\frac {\pi }{2}}-\arctan x=\operatorname {arccot} x,\ }$ ${\displaystyle \ x>0}$

${\displaystyle \arctan {\frac {1}{x}}=-{\frac {\pi }{2}}-\arctan x=-\pi +\operatorname {arccot} x,\ }$ ${\displaystyle \ x<0}$

${\displaystyle \operatorname {arccot} {\frac {1}{x}}={\frac {\pi }{2}}-\operatorname {arccot} x=\arctan x,\ }$ ${\displaystyle \ x>0}$

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

${\displaystyle \operatorname {arcsec} {\frac {1}{x}}=\arccos x}$

${\displaystyle \operatorname {arccsc} {\frac {1}{x}}=\arcsin x}$

如果有一段正弦表：

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

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

注意只要在使用了複數的平方根的時候，我們選擇正實部的平方根(或者正虛部，如果是負實數的平方根的話)。

從半角公式${\displaystyle \tan {\frac {\theta }{2}}={\frac {\sin \theta }{1+\cos \theta }}}$，可得到：

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

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

${\displaystyle \arctan x=2\arctan {\frac {x}{1+{\sqrt {1+x^{2}}}}}}$

三角函數與反三角函數的關係

通過定義可知：

${\displaystyle \theta }$	${\displaystyle \sin \theta }$	${\displaystyle \cos \theta }$	${\displaystyle \tan \theta }$	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

每個三角函數都周期於它的參數的實部上，在每個${\displaystyle 2\pi }$區間內通過它的所有值兩次。正弦和餘割的周期開始於${\displaystyle 2\pi k-{\frac {\pi }{2}}}$結束於${\displaystyle 2\pi k+{\frac {\pi }{2}}}$（這裡的${\displaystyle k}$是一個整數），在${\displaystyle 2\pi k+{\frac {\pi }{2}}}$到${\displaystyle 2\pi k+{\frac {3\pi }{2}}}$上倒過來。餘弦和正割的周期開始於${\displaystyle 2\pi k}$結束於${\displaystyle 2\pi k+\pi }$，在${\displaystyle 2\pi k+\pi }$到${\displaystyle 2\pi k+2\pi }$上倒過來。正切的周期開始於${\displaystyle 2\pi k-{\frac {\pi }{2}}}$結束於${\displaystyle 2\pi k+{\frac {\pi }{2}}}$，接着(向前)在${\displaystyle 2\pi k+{\frac {\pi }{2}}}$到${\displaystyle 2\pi k+{\frac {3\pi }{2}}}$上重複。餘切的周期開始於${\displaystyle 2\pi k}$結束於${\displaystyle 2\pi k+\pi }$，接着（向前）在${\displaystyle 2\pi k+\pi }$到${\displaystyle 2\pi k+2\pi }$上重複。

這個周期性反應在一般反函數上：

${\displaystyle \sin y=x\ \Leftrightarrow \ (\ y=\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$

${\displaystyle \cos y=x\ \Leftrightarrow \ (\ y=\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$

${\displaystyle \tan y=x\ \Leftrightarrow \ \ y=\arctan x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} }$

${\displaystyle \cot y=x\ \Leftrightarrow \ \ y=\operatorname {arccot} x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} }$

${\displaystyle \sec y=x\ \Leftrightarrow \ (\ y=\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$

${\displaystyle \csc y=x\ \Leftrightarrow \ (\ y=\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )}$

反三角函數的導數

對於實數${\displaystyle x}$的反三角函數的導函數如下：

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

舉例說明，設${\displaystyle \theta =\arcsin x\!}$，得到：

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

因為要使根號內部恆為正，所以在條件加上${\displaystyle |x|<1}$，其他導數公式同理可證^[1]。

表達為定積分

積分其導數並固定在一點上的值給出反三角函數作為定積分的表達式：

${\displaystyle {\begin{aligned}\arcsin x&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arccos x&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arctan x&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arccot} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\end{aligned}}}$

當${\displaystyle x}$等於1時，在有極限的域上的積分是瑕積分，但仍是良好定義的。

無窮級數

如同正弦和餘弦函數，反三角函數可以使用無窮級數計算如下：

${\displaystyle {\begin{aligned}\arcsin z&{}=z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}}$

${\displaystyle {\begin{aligned}\arccos z&{}={\frac {\pi }{2}}-\arcsin z\\&{}={\frac {\pi }{2}}-\left[z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}}$

${\displaystyle {\begin{aligned}\arctan z&{}=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arccot} z&{}={\frac {\pi }{2}}-\arctan z\\&{}={\frac {\pi }{2}}-\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \right)\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arcsec} z&{}=\arccos \left(z^{-1}\right)\\&{}={\frac {\pi }{2}}-\left[z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{(2n+1)}};\qquad \left|z\right|\geq -4\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {arccsc} z&{}=\arcsin \left(z^{-1}\right)\\&{}=z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{2n+1}};\qquad \left|z\right|\geq 1\end{aligned}}}$

歐拉發現了反正切的更有效的級數：

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

（注意對${\displaystyle x=0}$在和中的項是空積1。）

反三角函數的不定積分

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

使用分部積分法和上面的簡單導數很容易得出它們。

舉例

使用${\displaystyle \int u\,\mathrm {d} v=uv-\int v\,\mathrm {d} u}$，設

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

則

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

換元

${\displaystyle k=1-x^{2}.\,}$

則

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

且

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

換元回x得到

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

加法公式和減法公式

.mw-parser-output .serif{font-family:Times,serif}arcsin x + arcsin y

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

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

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

arcsin x - arcsin y

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

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

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

arccos x + arccos y

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

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

arccos x - arccos y

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

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

arctan x + arctan y

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

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

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

arctan x - arctan y

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

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

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

arccot x + arccot y

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

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

arcsin x + arccos y

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

arctan x + arccot y

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

註釋

1. 設${\displaystyle \theta =\arccos x}$，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${\displaystyle |x|<1}$。
   設${\displaystyle \theta =\arctan x}$，得到：

   ${\displaystyle {\frac {d\arctan x}{dx}}={\frac {d\theta }{d\tan \theta }}={\frac {1}{\sec ^{2}\theta }}={\frac {1}{1+\tan ^{2}\theta }}={\frac {1}{1+x^{2}}}}$
   

   設${\displaystyle \theta =\operatorname {arccot} x}$，得到：

   ${\displaystyle {\frac {d\operatorname {arccot} x}{dx}}={\frac {d\theta }{d\cot \theta }}={\frac {-1}{\csc ^{2}\theta }}={\frac {1}{1+\cot ^{2}\theta }}={\frac {-1}{1+x^{2}}}}$
   

   設${\displaystyle \theta =\operatorname {arcsec} x}$，得到：

   ${\displaystyle {\frac {d\operatorname {arcsec} x}{dx}}={\frac {d\theta }{d\sec \theta }}={\frac {1}{\sec \theta \tan \theta }}={\frac {1}{\left|x\right|{\sqrt {x^{2}-1}}}}}$

   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${\displaystyle 0\leq \theta \leq \pi ,\theta \neq {\frac {\pi }{2}}}$，所以${\displaystyle \tan \theta =\pm {\sqrt {x^{2}-1}}}$，所以${\displaystyle x}$要加絕對值。
   設${\displaystyle \theta =\operatorname {arccsc} x}$，得到：

   ${\displaystyle {\frac {d\operatorname {arccsc} x}{dx}}={\frac {d\theta }{d\csc \theta }}={\frac {-1}{\csc \theta \cot \theta }}={\frac {-1}{\left|x\right|{\sqrt {x^{2}-1}}}}}$

   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${\displaystyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}},\theta \neq 0}$。

參見

• 數學主題

• 正切半角公式
• 平方根

外部連結

• 埃里克·韋斯坦因. Inverse Trigonometric Functions. MathWorld.
• http://mathworld.wolfram.com/InverseTangent.html（頁面存檔備份，存於網際網路檔案館）