逆三角関数と逆双曲線関数の関係
逆三角関数と逆双曲線関数の関係
(1)
\[ \Sin^{\bullet}\left(iz\right)=i\Sinh^{\bullet}z \](2)
\[ \Cos^{\bullet}z\ne-i\Cosh^{\bullet}z \](3)
\[ \Tan^{\bullet}\left(iz\right)=i\Tanh^{\bullet}z \](4)
\[ \Sin^{-1,\bullet}\left(iz\right)=-i\Sinh^{-1,\bullet}z \](5)
\[ \Cos^{-1,\bullet}z\ne-i\Cosh^{-1,\bullet}z \](6)
\[ \Tan^{-1,\bullet}\left(iz\right)=-i\Tanh^{-1,\bullet}z \](1)
\begin{align*} \Sin^{\bullet}\left(iz\right) & =\Sin^{\bullet}\left(i\sinh\Sinh^{\bullet}z\right)\\ & =\Sin^{\bullet}\sin\left(i\Sinh^{\bullet}z\right)\\ & =i\Sinh^{\bullet}z \end{align*} \(i\Sinh^{\bullet}z\)の値域は\(\Sin^{\bullet}\sin\)が恒等写像になるので成り立つ。(1)-2
\begin{align*} \Sin^{\bullet}\left(iz\right) & =-i\Log\left(i\left(iz\right)+\sqrt{1-\left(iz\right)^{2}}\right)\\ & =-i\Log\left(-z+\sqrt{1+z^{2}}\right)\\ & =-i\Sinh^{\bullet}\left(-z\right)\\ & =i\Sinh^{\bullet}z \end{align*}(2)
\begin{align*} \Cos^{\bullet}z & =\Cos^{\bullet}\cosh\Cosh^{\bullet}z\\ & =\Cos^{\bullet}\cos\left(i\Cosh^{\bullet}z\right)\\ & =\Cos^{\bullet}\cos\left(-i\Cosh^{\bullet}z\right)\\ & \ne-i\Cosh^{\bullet}z \end{align*} \(-i\Cosh^{\bullet}z\)の値域は\(\Cos^{\bullet}\cos\)が恒等写像になっていない。(2)-2
\begin{align*} \Cos^{\bullet}z & =-i\Log\left(z+i\sqrt{1-z^{2}}\right)\\ & =-i\Log\left(z+i\sqrt{1-z}\sqrt{1+z}\right)\\ & \ne-i\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)\\ & =-i\Cosh^{\bullet}z \end{align*}(2)-3 反例
\begin{align*} \Cos^{\bullet}\left(-1\right) & =\pi\\ & =-i\Cosh^{\bullet}\left(-1\right) \end{align*} \begin{align*} \Cos^{\bullet}\left(0\right) & =\frac{\pi}{2}\\ & =-i\Cosh^{\bullet}\left(0\right) \end{align*} となるが、\(\frac{\pi}{2}<\Arg\sqrt{1-z}\;\lor\;\Arg\sqrt{1-z}\leq-\frac{\pi}{2}\)ととると、\begin{align*} \Cos^{\bullet}\left(2\right) & =i\Cosh^{\bullet}\left(2\right)\\ & \ne-i\Cosh^{\bullet}\left(2\right) \end{align*} となるので成り立たない。
(3)
\begin{align*} \Tan^{\bullet}\left(iz\right) & =\Tan^{\bullet}\left(i\tanh\Tanh^{\bullet}z\right)\\ & =\Tan^{\bullet}\tan\left(i\Tanh^{\bullet}z\right)\\ & =i\Tanh^{\bullet}z \end{align*} \(i\Tanh^{\bullet}z\)の値域は\(\Tan^{\bullet}\tan\)が恒等写像になるので成り立つ。(3)-2
\begin{align*} \Tan^{\bullet}\left(iz\right) & =\frac{i}{2}\left(\Log\left(1-i\left(iz\right)\right)-\Log\left(1+i\left(iz\right)\right)\right)\\ & =\frac{i}{2}\left(\Log\left(1+z\right)-\Log\left(1-z\right)\right)\\ & =i\Tanh^{\bullet}z \end{align*}(4)
\begin{align*} \Sin^{-1,\bullet}\left(iz\right) & =\Sin^{-1,\bullet}\left(i\sinh^{-1}\Sinh^{-1,\bullet}z\right)\\ & =\Sin^{-1,\bullet}\left(-\sin^{-1}\left(i\Sinh^{-1,\bullet}z\right)\right)\\ & =\Sin^{-1,\bullet}\sin^{-1}\left(-i\Sinh^{-1,\bullet}z\right)\\ & =-i\Sinh^{-1,\bullet}z \end{align*} \(-i\sinh^{-1,\bullet}z\)の値域は\(\sin^{-1,\bullet}\sin^{-1}\)が恒等写像になるので成り立つ。(4)-2
\begin{align*} \Sin^{-1,\bullet}\left(iz\right) & =\Sin^{\bullet}\left(-\frac{i}{z}\right)\\ & =-i\Sinh^{\bullet}\left(\frac{1}{z}\right)\\ & =-i\Sinh^{-1,\bullet}z \end{align*}(5)
\begin{align*} \Cos^{-1,\bullet}z & =\Cos^{-1,\bullet}\cosh^{-1}\Cosh^{-1,\bullet}z\\ & =\Cos^{-1,\bullet}\cos^{-1}\left(i\Cosh^{-1,\bullet}z\right)\\ & =\Cos^{-1,\bullet}\cos^{-1}\left(-i\Cosh^{-1,\bullet}z\right)\\ & \ne-i\Cosh^{-1,\bullet}z \end{align*} \(-i\Cosh^{-1,\bullet}z\)の値域は\(\Cos^{-1,\bullet}\cos^{-1}\)が恒等写像になっていない。(6)
\begin{align*} \Tan^{-1,\bullet}\left(iz\right) & =\Tan^{-1,\bullet}\left(i\tanh^{-1}\Tanh^{-1,\bullet}z\right)\\ & =\Tan^{-1,\bullet}\left(-\tan^{-1}\left(i\Tanh^{-1,\bullet}z\right)\right)\\ & =\Tan^{-1,\bullet}\tan^{-1}\left(-i\Tanh^{-1,\bullet}z\right)\\ & =-i\Tanh^{-1,\bullet}z \end{align*} \(-i\Tanh^{-1,\bullet}z\)の値域は\(\Tan^{-1,\bullet}\tan^{-1}\)が恒等写像になるので成り立つ。(6)-2
\begin{align*} \Tan^{-1,\bullet}\left(iz\right) & =\Tan^{\bullet}\left(-\frac{i}{z}\right)\\ & =-i\Tanh^{\bullet}\left(\frac{1}{z}\right)\\ & =-i\Tanh^{-1,\bullet}z \end{align*}逆三角関数と逆双曲線関数の関係
(1)
\[ \Sinh^{\bullet}\left(iz\right)=i\Sin^{\bullet}\left(z\right) \](2)
\[ \Cosh^{\bullet}z\ne i\Cos^{\bullet}z \](3)
\[ \Tanh^{\bullet}\left(iz\right)=i\Tan^{\bullet}z \](4)
\[ \Sinh^{-1,\bullet}\left(iz\right)=-i\Sin^{-1,\bullet}z \](5)
\[ \Cosh^{-1,\bullet}z\ne i\Cos^{-1,\bullet}z \](6)
\[ \Tanh^{-1,\bullet}\left(iz\right)=-i\Tan^{-1,\bullet}z \](1)
\begin{align*} \Sinh^{\bullet}\left(iz\right) & =\Sinh^{\bullet}\left(i\sin\Sin^{\bullet}z\right)\\ & =\Sinh^{\bullet}\sinh\left(i\Sin^{\bullet}z\right)\\ & =i\Sin^{\bullet}\left(z\right) \end{align*} \(i\Sin^{\bullet}z\)の値域は\(\Sinh^{\bullet}\sinh\)が恒等写像になるので成り立つ。(1)-2
\begin{align*} \Sinh^{\bullet}\left(iz\right) & =\Log\left(iz+\sqrt{1+\left(iz\right)^{2}}\right)\\ & =i\left(-i\right)\Log\left(iz+\sqrt{1-z^{2}}\right)\\ & =i\Sin^{\bullet}z \end{align*}(2)
\begin{align*} \Cosh^{\bullet}z & =\Cosh^{\bullet}\cos\Cos^{\bullet}z\\ & =\Cosh^{\bullet}\cosh\left(i\Cos^{\bullet}z\right)\\ & \ne i\Cos^{\bullet}z \end{align*} \(i\Cos^{\bullet}z\)の値域は\(\Cosh^{\bullet}\cosh\)が恒等写像になっていない。(2)-2
\begin{align*} \Cosh^{\bullet}z & =\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)\\ & \ne\Log\left(z+i\sqrt{1-z}\sqrt{1+z}\right)\\ & =\Log\left(z+i\sqrt{1-z^{2}}\right)\\ & =i\left(-i\right)\Log\left(z+i\sqrt{1-z^{2}}\right)\\ & =i\Cos^{\bullet}z \end{align*}(3)
\begin{align*} \Tanh^{\bullet}\left(iz\right) & =\Tanh^{\bullet}\left(i\tan\Tan^{\bullet}z\right)\\ & =\Tanh^{\bullet}\tanh\left(i\Tan^{\bullet}z\right)\\ & =i\Tan^{\bullet}z \end{align*} \(i\Tan^{\bullet}z\)の値域は\(\Tanh^{\bullet}\tanh\)が恒等写像になるので成り立つ。(3)-2
\begin{align*} \Tanh^{\bullet}\left(iz\right) & =\frac{1}{2}\left(\Log\left(1+iz\right)-\Log\left(1-iz\right)\right)\\ & =i\frac{i}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)\\ & =i\Tanh^{\bullet}z \end{align*}(4)
\begin{align*} \Sinh^{-1,\bullet}\left(iz\right) & =\Sinh^{-1,\bullet}\left(i\sin^{-1}\Sin^{-1,\bullet}z\right)\\ & =\Sinh^{-1,\bullet}\left(-\sinh^{-1}\left(i\Sin^{-1,\bullet}z\right)\right)\\ & =\Sinh^{-1,\bullet}\sinh^{-1}\left(-i\Sin^{-1,\bullet}z\right)\\ & =-i\Sin^{-1,\bullet}z \end{align*} \(-i\Sin^{-1,\bullet}z\)の値域は\(\Sinh^{-1,\bullet}\sinh^{-1}\)が恒等写像になるので成り立つ。(4)-2
\begin{align*} \Sinh^{-1,\bullet}\left(iz\right) & =\Sinh^{\bullet}\left(-\frac{i}{z}\right)\\ & =-i\Sin^{\bullet}\left(\frac{1}{z}\right)\\ & =-i\Sin^{-1,\bullet}\left(z\right) \end{align*}(5)
\begin{align*} \Cosh^{-1,\bullet}z & =\Cosh^{-1,\bullet}\cos^{-1}\Cos^{-1,\bullet}z\\ & =\Cosh^{-1,\bullet}\cosh^{-1}\left(i\Cos^{-1,\bullet}z\right)\\ & \ne i\Cos^{-1,\bullet}z \end{align*} \(i\Cos^{-1,\bullet}z\)の値域は\(\Cosh^{-1,\bullet}\cosh^{-1}\)が恒等写像になっていない。(5)-2
\begin{align*} \Cosh^{-1,\bullet}\left(z\right) & =\Cosh^{\bullet}\left(\frac{1}{z}\right)\\ & \ne i\Cos^{\bullet}\left(\frac{1}{z}\right)\\ & =i\Cos^{-1,\bullet}\left(z\right) \end{align*}(6)
\begin{align*} \Tanh^{-1,\bullet}\left(iz\right) & =\Tanh^{-1,\bullet}\left(i\tan^{-1}\Tan^{-1,\bullet}z\right)\\ & =\Tanh^{-1,\bullet}\left(-\tanh^{-1}\left(i\Tan^{-1,\bullet}z\right)\right)\\ & =\Tanh^{-1,\bullet}\tanh^{-1}\left(-i\Tan^{-1,\bullet}z\right)\\ & =-i\Tan^{-1,\bullet}z \end{align*} \(-i\Tan^{-1,\bullet}z\)の値域は\(\Tanh^{-1,\bullet}\tanh^{-1}\)が恒等写像になるので成り立つ。(6)-2
\begin{align*} \Tanh^{-1,\bullet}\left(iz\right) & =\Tanh^{\bullet}\left(-\frac{i}{z}\right)\\ & =-i\Tan^{\bullet}\left(\frac{1}{z}\right)\\ & =-i\Tan^{-1,\bullet}\left(z\right) \end{align*}ページ情報
| タイトル | 逆三角関数と逆双曲線関数の関係 |
| URL | https://www.nomuramath.com/o64nr5tw/ |
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三角関数と双曲線関数のn乗積分
\[
\int\sin^{2n+m_{\pm}}xdx=\frac{\Gamma\left(n+\frac{1}{2}+\frac{m_{\pm}}{2}\right)}{\Gamma\left(n+1+\frac{m_{\pm}}{2}\right)}\left\{ -\frac{1}{2}\sum_{k=0}^{n-1}\left(\frac{\Gamma\left(k+1+\frac{m_{\pm}}{2}\right)}{\Gamma\left(k+\frac{3}{2}+\frac{m_{\pm}}{2}\right)}\cos x\sin^{2k+1+m_{\pm}}x\right)+\frac{\Gamma\left(1+\frac{m_{\pm}}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{m_{\pm}}{2}\right)}\int\sin^{m_{\pm}}xdx\right\}
\]
逆三角関数と逆双曲線関数の冪乗積分漸化式
\[
\int\sin^{\bullet,n}xdx=x\sin^{\bullet,n}x+n\sqrt{1-x^{2}}\sin^{\bullet,n-1}x-n(n-1)\int\sin^{\bullet,n-2}xdx
\]
三角関数の還元(負角・余角・補角)公式
\[
\sin(-x)=-\sin x
\]
三角関数と双曲線関数の実部と虚部
\[
\tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}
\]