Công thức bổ sung Hàm hyperbolic ngược


arsinh ⁡ u ± arsinh ⁡ v = arsinh ⁡ ( u 1 + v 2 ± v 1 + u 2 ) {\displaystyle \operatorname {arsinh} u\pm \operatorname {arsinh} v=\operatorname {arsinh} \left(u{\sqrt {1+v^{2}}}\pm v{\sqrt {1+u^{2}}}\right)} arcosh ⁡ u ± arcosh ⁡ v = arcosh ⁡ ( u v ± ( u 2 − 1 ) ( v 2 − 1 ) ) {\displaystyle \operatorname {arcosh} u\pm \operatorname {arcosh} v=\operatorname {arcosh} \left(uv\pm {\sqrt {(u^{2}-1)(v^{2}-1)}}\right)} artanh ⁡ u ± artanh ⁡ v = artanh ⁡ ( u ± v 1 ± u v ) {\displaystyle \operatorname {artanh} u\pm \operatorname {artanh} v=\operatorname {artanh} \left({\frac {u\pm v}{1\pm uv}}\right)} 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+\operatorname {arcosh} v&=\operatorname {arsinh} \left(uv+{\sqrt {(1+u^{2})(v^{2}-1)}}\right)\\&=\operatorname {arcosh} \left(v{\sqrt {1+u^{2}}}+u{\sqrt {v^{2}-1}}\right)\end{aligned}}}

Tài liệu tham khảo

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