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)

\[ \Sin^{\circ}\left(iz\right)=i\Sinh^{\circ}z \]

(2)

\[ \Cos^{\circ}z\ne-i\Cosh^{\circ}z \]

(3)

\[ \Tan^{\circ}\left(iz\right)=i\Tanh^{\circ}z \]

(4)

\[ \Sin^{-1,\circ}\left(iz\right)=-i\Sinh^{-1,\circ}z \]

(5)

\[ \Cos^{-1,\circ}z\ne-i\Cosh^{-1,\circ}z \]

(6)

\[ \Tan^{-1,\circ}\left(iz\right)=-i\Tanh^{-1,\circ}z \]

(1)

\begin{align*} \Sin^{\circ}\left(iz\right) & =\Sin^{\circ}\left(i\sinh\Sinh^{\circ}z\right)\\ & =\Sin^{\circ}\sin\left(i\Sinh^{\circ}z\right)\\ & =i\Sinh^{\circ}z \end{align*}

\(i\Sinh^{\circ}z\)の値域は\(\Sin^{\circ}\sin\)が恒等写像になるので成り立つ。

(1)-2

\begin{align*} \Sin^{\circ}\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^{\circ}\left(-z\right)\\ & =i\Sinh^{\circ}z \end{align*}

(2)

\begin{align*} \Cos^{\circ}z & =\Cos^{\circ}\cosh\Cosh^{\circ}z\\ & =\Cos^{\circ}\cos\left(i\Cosh^{\circ}z\right)\\ & =\Cos^{\circ}\cos\left(-i\Cosh^{\circ}z\right)\\ & \ne-i\Cosh^{\circ}z \end{align*}

\(-i\Cosh^{\circ}z\)の値域は\(\Cos^{\circ}\cos\)が恒等写像になっていない。

(2)-2

\begin{align*} \Cos^{\circ}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^{\circ}z \end{align*}

(2)-3 反例

\begin{align*} \Cos^{\circ}\left(-1\right) & =\pi\\ & =-i\Cosh^{\circ}\left(-1\right) \end{align*}

\begin{align*} \Cos^{\circ}\left(0\right) & =\frac{\pi}{2}\\ & =-i\Cosh^{\circ}\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^{\circ}\left(2\right) & =i\Cosh^{\circ}\left(2\right)\\ & \ne-i\Cosh^{\circ}\left(2\right) \end{align*}

となるので成り立たない。

(3)

\begin{align*} \Tan^{\circ}\left(iz\right) & =\Tan^{\circ}\left(i\tanh\Tanh^{\circ}z\right)\\ & =\Tan^{\circ}\tan\left(i\Tanh^{\circ}z\right)\\ & =i\Tanh^{\circ}z \end{align*}

\(i\Tanh^{\circ}z\)の値域は\(\Tan^{\circ}\tan\)が恒等写像になるので成り立つ。

(3)-2

\begin{align*} \Tan^{\circ}\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^{\circ}z \end{align*}

(4)

\begin{align*} \Sin^{-1,\circ}\left(iz\right) & =\Sin^{-1,\circ}\left(i\sinh^{-1}\Sinh^{-1,\circ}z\right)\\ & =\Sin^{-1,\circ}\left(-\sin^{-1}\left(i\Sinh^{-1,\circ}z\right)\right)\\ & =\Sin^{-1,\circ}\sin^{-1}\left(-i\Sinh^{-1,\circ}z\right)\\ & =-i\Sinh^{-1,\circ}z \end{align*}

\(-i\sinh^{-1,\circ}z\)の値域は\(\sin^{-1,\circ}\sin^{-1}\)が恒等写像になるので成り立つ。

(4)-2

\begin{align*} \Sin^{-1,\circ}\left(iz\right) & =\Sin^{\circ}\left(-\frac{i}{z}\right)\\ & =-i\Sinh^{\circ}\left(\frac{1}{z}\right)\\ & =-i\Sinh^{-1,\circ}z \end{align*}

(5)

\begin{align*} \Cos^{-1,\circ}z & =\Cos^{-1,\circ}\cosh^{-1}\Cosh^{-1,\circ}z\\ & =\Cos^{-1,\circ}\cos^{-1}\left(i\Cosh^{-1,\circ}z\right)\\ & =\Cos^{-1,\circ}\cos^{-1}\left(-i\Cosh^{-1,\circ}z\right)\\ & \ne-i\Cosh^{-1,\circ}z \end{align*}

\(-i\Cosh^{-1,\circ}z\)の値域は\(\Cos^{-1,\circ}\cos^{-1}\)が恒等写像になっていない。

(6)

\begin{align*} \Tan^{-1,\circ}\left(iz\right) & =\Tan^{-1,\circ}\left(i\tanh^{-1}\Tanh^{-1,\circ}z\right)\\ & =\Tan^{-1,\circ}\left(-\tan^{-1}\left(i\Tanh^{-1,\circ}z\right)\right)\\ & =\Tan^{-1,\circ}\tan^{-1}\left(-i\Tanh^{-1,\circ}z\right)\\ & =-i\Tanh^{-1,\circ}z \end{align*}

\(-i\Tanh^{-1,\circ}z\)の値域は\(\Tan^{-1,\circ}\tan^{-1}\)が恒等写像になるので成り立つ。

(6)-2

\begin{align*} \Tan^{-1,\circ}\left(iz\right) & =\Tan^{\circ}\left(-\frac{i}{z}\right)\\ & =-i\Tanh^{\circ}\left(\frac{1}{z}\right)\\ & =-i\Tanh^{-1,\circ}z \end{align*}

逆三角関数と逆双曲線関数の関係

(1)

\[ \Sinh^{\circ}\left(iz\right)=i\Sin^{\circ}\left(z\right) \]

(2)

\[ \Cosh^{\circ}z\ne i\Cos^{\circ}z \]

(3)

\[ \Tanh^{\circ}\left(iz\right)=i\Tan^{\circ}z \]

(4)

\[ \Sinh^{-1,\circ}\left(iz\right)=-i\Sin^{-1,\circ}z \]

(5)

\[ \Cosh^{-1,\circ}z\ne i\Cos^{-1,\circ}z \]

(6)

\[ \Tanh^{-1,\circ}\left(iz\right)=-i\Tan^{-1,\circ}z \]

(1)

\begin{align*} \Sinh^{\circ}\left(iz\right) & =\Sinh^{\circ}\left(i\sin\Sin^{\circ}z\right)\\ & =\Sinh^{\circ}\sinh\left(i\Sin^{\circ}z\right)\\ & =i\Sin^{\circ}\left(z\right) \end{align*}

\(i\Sin^{\circ}z\)の値域は\(\Sinh^{\circ}\sinh\)が恒等写像になるので成り立つ。

(1)-2

\begin{align*} \Sinh^{\circ}\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^{\circ}z \end{align*}

(2)

\begin{align*} \Cosh^{\circ}z & =\Cosh^{\circ}\cos\Cos^{\circ}z\\ & =\Cosh^{\circ}\cosh\left(i\Cos^{\circ}z\right)\\ & \ne i\Cos^{\circ}z \end{align*}

\(i\Cos^{\circ}z\)の値域は\(\Cosh^{\circ}\cosh\)が恒等写像になっていない。

(2)-2

\begin{align*} \Cosh^{\circ}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^{\circ}z \end{align*}

(3)

\begin{align*} \Tanh^{\circ}\left(iz\right) & =\Tanh^{\circ}\left(i\tan\Tan^{\circ}z\right)\\ & =\Tanh^{\circ}\tanh\left(i\Tan^{\circ}z\right)\\ & =i\Tan^{\circ}z \end{align*}

\(i\Tan^{\circ}z\)の値域は\(\Tanh^{\circ}\tanh\)が恒等写像になるので成り立つ。

(3)-2

\begin{align*} \Tanh^{\circ}\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^{\circ}z \end{align*}

(4)

\begin{align*} \Sinh^{-1,\circ}\left(iz\right) & =\Sinh^{-1,\circ}\left(i\sin^{-1}\Sin^{-1,\circ}z\right)\\ & =\Sinh^{-1,\circ}\left(-\sinh^{-1}\left(i\Sin^{-1,\circ}z\right)\right)\\ & =\Sinh^{-1,\circ}\sinh^{-1}\left(-i\Sin^{-1,\circ}z\right)\\ & =-i\Sin^{-1,\circ}z \end{align*}

\(-i\Sin^{-1,\circ}z\)の値域は\(\Sinh^{-1,\circ}\sinh^{-1}\)が恒等写像になるので成り立つ。

(4)-2

\begin{align*} \Sinh^{-1,\circ}\left(iz\right) & =\Sinh^{\circ}\left(-\frac{i}{z}\right)\\ & =-i\Sin^{\circ}\left(\frac{1}{z}\right)\\ & =-i\Sin^{-1,\circ}\left(z\right) \end{align*}

(5)

\begin{align*} \Cosh^{-1,\circ}z & =\Cosh^{-1,\circ}\cos^{-1}\Cos^{-1,\circ}z\\ & =\Cosh^{-1,\circ}\cosh^{-1}\left(i\Cos^{-1,\circ}z\right)\\ & \ne i\Cos^{-1,\circ}z \end{align*}

\(i\Cos^{-1,\circ}z\)の値域は\(\Cosh^{-1,\circ}\cosh^{-1}\)が恒等写像になっていない。

(5)-2

\begin{align*} \Cosh^{-1,\circ}\left(z\right) & =\Cosh^{\circ}\left(\frac{1}{z}\right)\\ & \ne i\Cos^{\circ}\left(\frac{1}{z}\right)\\ & =i\Cos^{-1,\circ}\left(z\right) \end{align*}

(6)

\begin{align*} \Tanh^{-1,\circ}\left(iz\right) & =\Tanh^{-1,\circ}\left(i\tan^{-1}\Tan^{-1,\circ}z\right)\\ & =\Tanh^{-1,\circ}\left(-\tanh^{-1}\left(i\Tan^{-1,\circ}z\right)\right)\\ & =\Tanh^{-1,\circ}\tanh^{-1}\left(-i\Tan^{-1,\circ}z\right)\\ & =-i\Tan^{-1,\circ}z \end{align*}

\(-i\Tan^{-1,\circ}z\)の値域は\(\Tanh^{-1,\circ}\tanh^{-1}\)が恒等写像になるので成り立つ。

(6)-2

\begin{align*} \Tanh^{-1,\circ}\left(iz\right) & =\Tanh^{\circ}\left(-\frac{i}{z}\right)\\ & =-i\Tan^{\circ}\left(\frac{1}{z}\right)\\ & =-i\Tan^{-1,\circ}\left(z\right) \end{align*}

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逆三角関数と逆双曲線関数の関係

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