逆双曲線関数
arcosh  ( 2 x 2 − 1 ) = 2 arcosh  ( x ) for x ≥ 1 arcosh  ( 8 x 4 − 8 x 2 + 1 ) = 4 arcosh  ( x ) for x ≥ 1 arcosh  ( 2 x 2 + 1 ) = 2 arsinh  ( x ) for x ≥ 0 arcosh  ( 8 x 4 + 8 x 2 + 1 ) = 4 arsinh  ( x ) for x ≥ 0 {\displaystyle {\begin{aligned}\operatorname {arcosh} (2x^{2}-1)&=2\operatorname {arcosh} (x)&{\text{ for }}x\geq 1\\\operatorname {arcosh} (8x^{4}-8x^{2}+1)&=4\operatorname {arcosh} (x)&{\text{ for }}x\geq 1\\\operatorname {arcosh} (2x^{2}+1)&=2\operatorname {arsinh} (x)&{\text{ for }}x\geq 0\\\operatorname {arcosh} (8x^{4}+8x^{2}+1)&=4\operatorname {arsinh} (x)&{\text{ for }}x\geq 0\end{aligned}}}
arsinh  u ± arsinh  v = arsinh  ( u 1 + v 2 ± v 1 + u 2 ) arcosh  u ± arcosh  v = arcosh  ( u v ± ( u 2 − 1 ) ( v 2 − 1 ) ) artanh  u ± artanh  v = artanh  ( u ± v 1 ± u v ) arsinh  u + arcosh  v = arsinh  ( u v + ( 1 + u 2 ) ( v 2 − 1 ) ) = arcosh  ( v 1 + u 2 + u v 2 − 1 ) {\displaystyle {\begin{aligned}\operatorname {arsinh} u\pm \operatorname {arsinh} v&=\operatorname {arsinh} \left(u{\sqrt {1+v^{2}}}\pm v{\sqrt {1+u^{2}}}\right)\\\operatorname {arcosh} u\pm \operatorname {arcosh} v&=\operatorname {arcosh} \left(uv\pm {\sqrt {\left(u^{2}-1\right)\left(v^{2}-1\right)}}\right)\\\operatorname {artanh} u\pm \operatorname {artanh} v&=\operatorname {artanh} \left({\frac {u\pm v}{1\pm uv}}\right)\\\operatorname {arsinh} u+\operatorname {arcosh} v&=\operatorname {arsinh} \left(uv+{\sqrt {\left(1+u^{2}\right)\left(v^{2}-1\right)}}\right)\\&=\operatorname {arcosh} \left(v{\sqrt {1+u^{2}}}+u{\sqrt {v^{2}-1}}\right)\end{aligned}}}
5ex]\operatorname {artanh} \,z&={\tfrac {1}{2}}\ln \left({\frac {1+z}{1-z}}\right)\\\operatorname {arcoth} \,z&={\tfrac {1}{2}}\ln \left({\frac {z+1}{z-1}}\right)\\\operatorname {arcsch} \,z&=\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z^{2}}}+1}}\,\right)\\\operatorname {arsech} \,z&=\ln \left({\frac {1}{z}}+{\sqrt {{\frac {1}{z}}+1}}\,{\sqrt {{\frac {1}{z}}-1}}\,\right)\end{aligned}}}
上記の平方根は正の平方根であり、対数関数は複素対数である。実数の引数、例えばz = xは実数値を返すが、一定の簡素化を行うことが可能であり、例えば x + 1 x − 1 = x 2 − 1 {\displaystyle {\sqrt {x+1}}{\sqrt {x-1}}={\sqrt {x^{2}-1}}} は正の平方根を使うとき、一般に真ではない。
演算子は複素数平面で次のように定義される。
arsinh z = ln  ( z + z 2 + 1 ) arcosh z = ln  ( z + z + 1 z − 1 ) artanh z = 1 2 ln  ( 1 + z 1 − z ) arcoth z = 1 2 ln  ( z + 1 z − 1 ) arcsch z = ln  ( 1 z + 1 z 2 + 1 ) arsech z = ln  ( 1 z + 1 z + 1 1 z − 1 ) {\displaystyle {\begin{aligned}\operatorname {arsinh} \,z&=\ln(z+{\sqrt {z^{2}+1}}\,)\\[2.5ex]\operatorname {arcosh} \,z&=\ln(z+{\sqrt {z+1}}{\sqrt {z-1}}\,)\\[1.
}{2^{2n}(n!)^{2}}}\right){\frac {x^{2n}}{2n}},\qquad 0<x\leq 1\end{aligned}}}
arcoth x = artanh  1 x = x − 1 + x − 3 3 + x − 5 5 + x − 7 7 + ⋯ = ∑ n = 0 ∞ x − ( 2 n + 1 ) ( 2 n + 1 ) , | x | > 1 {\displaystyle {\begin{aligned}\operatorname {arcoth} \,x=\operatorname {artanh} {\frac {1}{x}}&=x^{-1}+{\frac {x^{-3}}{3}}+{\frac {x^{-5}}{5}}+{\frac {x^{-7}}{7}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {x^{-(2n+1)}}{(2n+1)}},\qquad \left|x\right|>1\end{aligned}}}
またオイラーによるarctanの展開の類似も成り立つ。
artanh x = ∑ n = 0 ∞ 2 2 n ( n ! ) 2 ( 2 n + 1 ) !