逆三角関数
ここでは複素数の平方根を、正の実部(あるいは平方が負の実数であれば正の虚部)を持つように選ぶ。
半角公式(英語版) tan  θ 2 = sin  θ 1 + cos  θ {\displaystyle \tan {\frac {\theta }{2}}={\frac {\sin \theta }{1+\cos \theta }}} から、次を得る:
arcsin  x = 2 arctan  x 1 + 1 − x 2 arccos  x = 2 arctan  1 − x 2 1 + x , if − 1 < x ≤ + 1 arctan  x = 2 arctan  x 1 + 1 + x 2 {\displaystyle {\begin{aligned}\arcsin x&=2\arctan {\frac {x}{1+{\sqrt {1-x^{2}}}}}\\[1ex]\arccos x&=2\arctan {\frac {\sqrt {1-x^{2}}}{1+x}},{\text{ if }}-1<x\leq +1\\[1ex]\arctan x&=2\arctan {\frac {x}{1+{\sqrt {1+x^{2}}}}}\end{aligned}}}
z の複素数値の導関数は次の通りである:
d d z arcsin  z = 1 1 − z 2 ; z ≠ ± 1 d d z arccos  z = − 1 1 − z 2 ; z ≠ ± 1 d d z arctan  z = 1 1 + z 2 ; z ≠ ± i d d z arccot  z = − 1 1 + z 2 ; z ≠ ± i d d z arcsec  z = 1 z 2 1 − z − 2 ; z ≠ 0 , ± 1 d d z arccsc  z = − 1 z 2 1 − z − 2 ; z ≠ 0 , ± 1 {\displaystyle {\begin{aligned}{\frac {d}{dz}}\arcsin z&={\frac {1}{\sqrt {1-z^{2}}}};\quad z\neq \pm 1\\{\frac {d}{dz}}\arccos z&={\frac {-1}{\sqrt {1-z^{2}}}};\quad z\neq \pm 1\\{\frac {d}{dz}}\arctan z&={\frac {1}{1+z^{2}}};\quad z\neq \pm i\\{\frac {d}{dz}}\operatorname {arccot} z&={\frac {-1}{1+z^{2}}};\quad z\neq \pm i\\{\frac {d}{dz}}\operatorname {arcsec} z&={\frac {1}{z^{2}{\sqrt {1-z^{-2}}}}};\quad z\neq 0,\pm 1\\{\frac {d}{dz}}\operatorname {arccsc} z&={\frac {-1}{z^{2}{\sqrt {1-z^{-2}}}}};\quad z\neq 0,\pm 1\end{aligned}}}
arccos  z = π 2 − arcsin  z = π 2 − ∑ n = 0 ∞ ( 2 n n ) 4 n ( 2 n + 1 ) z 2 n + 1 = π 2 − ( z + ( 1 2 ) z 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) z 5 5 + ⋯ ) ; | z | ≤ 1 {\displaystyle {\begin{aligned}\arccos z&={\frac {\pi }{2}}-\arcsin z\\&={\frac {\pi }{2}}-\textstyle \sum \limits _{n=0}^{\infty }{\dfrac {\binom {2n}{n}}{4^{n}(2n+1)}}z^{2n+1}\\&={\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}}+\dotsb \right);\quad |z|\leq 1\end{aligned}}}
1つの方程式に書けば: sin  y = x ⇔ y = ( − 1 ) k arcsin  x + k π {\displaystyle \sin y=x\ \Leftrightarrow \ y=(-1)^{k}\arcsin x+k\pi }
cos  y = x ⇔ y = arccos  x + 2 k π or y = 2 π − arccos  x + 2 k π {\displaystyle \cos y=x\ \Leftrightarrow \ y=\arccos x+2k\pi {\text{ or }}y=2\pi -\arccos x+2k\pi }
1つの方程式に書けば: cos  y = x ⇔ y = ± arccos  x + 2 k π {\displaystyle \cos y=x\ \Leftrightarrow \ y=\pm \arccos x+2k\pi }
tan  y = x ⇔ y = arctan  x + k π {\displaystyle \tan y=x\ \Leftrightarrow \ y=\arctan x+k\pi }