逆三角関数の三角関数と逆双曲線関数の双曲線関数

逆三角関数の三角関数

(1)

\[ \sin\Sin^{\circ}z=z \]

(2)

\[ \sin\Cos^{\circ}z=\sqrt{1-z^{2}} \]

(3)

\[ \sin\Tan^{\circ}x=\frac{z}{\sqrt{1+z^{2}}} \]

(4)

\[ \cos\Sin^{\circ}x=\sqrt{1-z^{2}} \]

(5)

\[ \cos\Cos^{\circ}z=z \]

(6)

\[ \cos\Tan^{\circ}x=\frac{1}{\sqrt{1+z^{2}}} \]

(7)

\[ \tan\Sin^{\circ}z=\frac{z}{\sqrt{1-z^{2}}} \]

(8)

\[ \tan\Cos^{\circ}z=\frac{\sqrt{1-z^{2}}}{z} \]

(9)

\[ \tan\Tan^{\circ}z=z \]

(1)

\begin{align*} \sin\Sin^{\circ}z & =\sin\left(-i\Log\left(\sqrt{1-z^{2}}+iz\right)\right)\\ & =\frac{e^{\Log\left(\sqrt{1-z^{2}}+iz\right)}-e^{-\Log\left(\sqrt{1-z^{2}}+iz\right)}}{2i}\\ & =\frac{\sqrt{1-z^{2}}+iz-\left(\sqrt{1-z^{2}}+iz\right)^{-1}}{2i}\\ & =\frac{\sqrt{1-z^{2}}+iz-\left(\sqrt{1-z^{2}}-iz\right)}{2i}\\ & =z \end{align*}

(1)-2

実数の場合

\[ \sin\Sin^{\circ}x=x \]

(2)

\begin{align*} \sin\Cos^{\circ}z & =\sin\left(-i\Log\left(z+i\sqrt{1-z^{2}}\right)\right)\\ & =\frac{e^{\Log\left(z+i\sqrt{1-z^{2}}\right)}-e^{-\Log\left(z+i\sqrt{1-z^{2}}\right)}}{2i}\\ & =\frac{z+i\sqrt{1-z^{2}}-\left(z+i\sqrt{1-z^{2}}\right)^{-1}}{2i}\\ & =\frac{z+i\sqrt{1-z^{2}}-\left(z-i\sqrt{1-z^{2}}\right)}{2i}\\ & =\sqrt{1-z^{2}} \end{align*}

(2)-2

実数の場合

\begin{align*} \sin\cos^{\circ}x & =\sqrt{1-\cos^{2}\cos^{\circ}x}\\ & =\sqrt{1-x^{2}} \end{align*}

(3)

\begin{align*} \sin\Tan^{\circ}x & =\sin\left(\frac{i}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)\right)\\ & =\frac{e^{-\frac{1}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)}-e^{\frac{1}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)}}{2i}\\ & =\frac{1}{2i}\left(\frac{\sqrt{1+iz}}{\sqrt{1-iz}}-\frac{\sqrt{1-iz}}{\sqrt{1+iz}}\right)\\ & =\frac{1}{2i}\left(\frac{1+iz}{\sqrt{1+z^{2}}}-\frac{1-iz}{\sqrt{1+z^{2}}}\right)\\ & =\frac{z}{\sqrt{1+z^{2}}} \end{align*}

(3)-2

実数の場合

\begin{align*} \sin\tan^{\circ}x & =\frac{\tan\tan^{\circ}x}{\sqrt{1+\tan^{2}\tan^{\circ}x}}\\ & =\frac{x}{\sqrt{1+x^{2}}} \end{align*}

(4)

\begin{align*} \cos\Sin^{\circ}x & =\cos\left(-i\Log\left(iz+\sqrt{1-z^{2}}\right)\right)\\ & =\frac{e^{\Log\left(iz+\sqrt{1-z^{2}}\right)}+e^{-\Log\left(iz+\sqrt{1-z^{2}}\right)}}{2}\\ & =\frac{\left(iz+\sqrt{1-z^{2}}\right)+\left(iz+\sqrt{1-z^{2}}\right)^{-1}}{2}\\ & =\frac{\left(iz+\sqrt{1-z^{2}}\right)+\left(-iz+\sqrt{1-z^{2}}\right)}{2}\\ & =\sqrt{1-z^{2}} \end{align*}

(4)-2

\begin{align*} \cos\Sin^{\circ}z & =\cos\left(\frac{\pi}{2}-\Cos^{\circ}z\right)\\ & =\sin\Cos^{\circ}z\\ & =\sqrt{1-z^{2}} \end{align*}

(4)-3

実数の場合

\begin{align*} \cos\sin^{\circ}x & =\sqrt{1-\sin^{2}\sin^{\circ}x}\\ & =\sqrt{1-x^{2}} \end{align*}

(5)

\begin{align*} \cos\Cos^{\circ}z & =\cos\left(-i\Log\left(z+i\sqrt{1-z^{2}}\right)\right)\\ & =\frac{e^{\Log\left(z+i\sqrt{1-z^{2}}\right)}+e^{-\Log\left(z+i\sqrt{1-z^{2}}\right)}}{2}\\ & =\frac{\left(z+i\sqrt{1-z^{2}}\right)+\left(z+i\sqrt{1-z^{2}}\right)^{-1}}{2}\\ & =\frac{\left(z+i\sqrt{1-z^{2}}\right)+\left(z-i\sqrt{1-z^{2}}\right)}{2}\\ & =z \end{align*}

(5)-2

実数の場合

\[ \cos\cos^{\circ}x=x \]

(6)

\begin{align*} \cos\Tan^{\circ}x & =\cos\left(\frac{i}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)\right)\\ & =\frac{e^{-\frac{1}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)}+e^{\frac{1}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)}}{2}\\ & =\frac{1}{2}\left(\frac{\sqrt{1+iz}}{\sqrt{1-iz}}+\frac{\sqrt{1-iz}}{\sqrt{1+iz}}\right)\\ & =\frac{1}{2}\left(\frac{1+iz}{\sqrt{1+z^{2}}}+\frac{1-iz}{\sqrt{1+z^{2}}}\right)\\ & =\frac{1}{\sqrt{1+z^{2}}} \end{align*}

(6)-2

実数の場合

\begin{align*} \cos\tan^{\circ}x & =\frac{1}{\sqrt{1+\tan^{2}\tan^{\circ}x}}\\ & =\frac{1}{\sqrt{1+x^{2}}} \end{align*}

(7)

\begin{align*} \tan\Sin^{\circ}z & =\frac{\sin\Sin^{\circ}z}{\cos\Sin^{\circ}z}\\ & =\frac{z}{\sqrt{1-z^{2}}} \end{align*}

(7)-2

実数の場合

\begin{align*} \tan\sin^{\circ}x & =\frac{\sin\sin^{\circ}x}{\sqrt{1-\sin^{2}\sin^{\circ}x}}\\ & =\frac{x}{\sqrt{1-x^{2}}} \end{align*}

(8)

\begin{align*} \tan\Cos^{\circ}z & =\frac{\sin\Cos^{\circ}z}{\cos\Cos^{\circ}z}\\ & =\frac{\sqrt{1-z^{2}}}{z} \end{align*}

(8)-2

実数の場合

\begin{align*} \tan\cos^{\circ}x & =\frac{\sqrt{1-\cos^{2}\cos^{\circ}x}}{\cos\cos^{\circ}x}\\ & =\frac{\sqrt{1-x^{2}}}{x} \end{align*}

(9)

\begin{align*} \tan\Tan^{\circ}z & =\frac{\sin\Tan^{\circ}z}{\cos\Tan^{\circ}z}\\ & =\frac{z}{\sqrt{1+z^{2}}}\left(\frac{1}{\sqrt{1+z^{2}}}\right)^{-1}\\ & =z \end{align*}

(9)-2

実数の場合

\[ \tan\tan^{\circ}x=x \]

逆双曲線関数の双曲線関数

(1)

\[ \sinh\Sinh^{\circ}z=z \]

(2)

\[ \sinh\Cosh^{\circ}z=\sqrt{z-1}\sqrt{z+1} \]

(3)

\[ \sinh\Tanh^{\circ}z=\frac{z}{\sqrt{1-z^{2}}} \]

(4)

\[ \cosh\Sinh^{\circ}z=\sqrt{1+z^{2}} \]

(5)

\[ \cosh\Cosh^{\circ}z=z \]

(6)

\[ \cosh\Tanh^{\circ}z=\frac{1}{\sqrt{1-z^{2}}} \]

(7)

\[ \tanh\Sinh^{\circ}z=\frac{z}{\sqrt{1+z^{2}}} \]

(8)

\[ \tanh\Cosh^{\circ}z=\frac{\sqrt{z-1}\sqrt{z+1}}{z} \]

(9)

\[ \tanh\Tanh^{\circ}z=z \]

(1)

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

(1)-2

実数の場合

\[ \sinh\sinh^{\circ}x=x \]

(2)

\begin{align*} \sinh\Cosh^{\circ}z & =\sinh\left(\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)\right)\\ & =\frac{e^{\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)}-e^{-\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)}}{2}\\ & =\frac{\left(z+\sqrt{z-1}\sqrt{z+1}\right)-\left(z+\sqrt{z-1}\sqrt{z+1}\right)^{-1}}{2}\\ & =\sqrt{z-1}\sqrt{z+1} \end{align*}

(2)-2

実数の場合

\begin{align*} \sinh\cosh^{\circ}x & =\sqrt{\cosh^{2}\cosh^{\circ}x-2}\\ & =\sqrt{1-x^{2}} \end{align*}

(3)

\begin{align*} \sinh\Tanh^{\circ}z & =\sinh\left(-ii\Tanh^{\circ}z\right)\\ & =-i\sin\Tan^{\circ}\left(iz\right)\\ & =-i\frac{iz}{\sqrt{1+\left(iz\right)^{2}}}\\ & =\frac{z}{\sqrt{1-z^{2}}} \end{align*}

(3)-2

実数の場合

\begin{align*} \sinh\tanh^{\circ}x & =\frac{\tanh\tanh^{\circ}x}{\sqrt{1-\tanh^{2}\tanh^{\circ}x}}\\ & =\frac{x}{\sqrt{1-x^{2}}} \end{align*}

(4)

\begin{align*} \cosh\Sinh^{\circ}z & =\cosh\left(-ii\Sinh^{\circ}z\right)\\ & =\cos\Sin^{\circ}\left(iz\right)\\ & =\sqrt{1-\left(iz\right)^{2}}\\ & =\sqrt{1+z^{2}} \end{align*}

(4)-2

実数の場合

\begin{align*} \cosh\sinh^{\circ}x & =\sqrt{1+\sinh^{2}\sinh^{\circ}x}\\ & =\sqrt{1+x^{2}} \end{align*}

(5)

\begin{align*} \cosh\Cosh^{\circ}z & =\cosh\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)\\ & =\frac{e^{\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)}+e^{-\Log\left(z+\sqrt{z-1}\sqrt{z+1}\right)}}{2}\\ & =\frac{\left(z+\sqrt{z-1}\sqrt{z+1}\right)+\left(z+\sqrt{z-1}\sqrt{z+1}\right)^{-1}}{2}\\ & =\frac{\left(z+\sqrt{z-1}\sqrt{z+1}\right)+\left(z-\sqrt{z-1}\sqrt{z+1}\right)}{2}\\ & =z \end{align*}

(5)-2

実数の場合

\[ \cosh\cosh^{\circ}x=x \]

(6)

\begin{align*} \cosh\Tanh^{\circ}z & =\cosh\left(-ii\Tanh^{\circ}z\right)\\ & =\cos\Tan^{\circ}\left(iz\right)\\ & =\frac{1}{\sqrt{1+\left(iz\right)^{2}}}\\ & =\frac{1}{\sqrt{1-z^{2}}} \end{align*}

(6)-2

実数の場合

\begin{align*} \cosh\tanh^{\circ}x & =\frac{1}{\sqrt{1-\tanh^{2}\tanh^{\circ}x}}\\ & =\frac{1}{\sqrt{1-x^{2}}} \end{align*}

(7)

\begin{align*} \tanh\Sinh^{\circ}z & =\tanh\left(-ii\Sinh^{\circ}z\right)\\ & =-i\tan\Sin^{\circ}\left(iz\right)\\ & =-i\frac{\left(iz\right)}{\sqrt{1-\left(iz\right)^{2}}}\\ & =\frac{z}{\sqrt{1+z^{2}}} \end{align*}

(7)-2

実数の場合

\begin{align*} \tanh\sinh^{\circ}x & =\frac{\sinh\sinh^{\circ}x}{\sqrt{1+\sinh^{2}\sinh^{\circ}x}}\\ & =\frac{x}{\sqrt{1+x^{2}}} \end{align*}

(8)

\begin{align*} \tanh\Cosh^{\circ}z & =\frac{\sinh\Cosh^{\circ}z}{\cosh\Cosh^{\circ}z}\\ & =\frac{\sqrt{z-1}\sqrt{z+1}}{z} \end{align*}

(8)-2

実数の場合

\begin{align*} \tanh\cosh^{\circ}x & =\frac{\sqrt{\cosh^{2}\cosh^{\circ}x-1}}{\cosh\cosh^{\circ}x}\\ & =\frac{\sqrt{x^{2}-1}}{x} \end{align*}

(9)

\begin{align*} \tanh\Tanh^{\circ}z & =\frac{\sinh\Tanh^{\circ}z}{\cosh\Tanh^{\circ}z}\\ & =\frac{z}{\sqrt{1-z^{2}}}\left(\frac{1}{\sqrt{1-z^{2}}}\right)^{-1}\\ & =z \end{align*}

(9)-2

実数の場合

\[ \tanh\tanh^{\circ}x=x \]

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逆三角関数の三角関数と逆双曲線関数の双曲線関数

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