\(\cosh{\left(ix\right)}\)
$$\begin{eqnarray}
\cosh{\left(x\right)}&=&\frac{e^x+e^{-x}}{2}
\\\cosh{\left(ix\right)}&=&\frac{e^{ix}+e^{-ix}}{2}
\\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2}
\\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)}
\\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}+\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2}
\\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)}
\\&=&\frac{\cos{\left(x\right)}\color{red}{+i\sin{\left(x\right)}}\color{black}{+\cos{\left(x\right)}}\color{red}{-i\sin{\left(x\right)}}}{2}
\\&=&\frac{\cos{\left(x\right)}+\cos{\left(x\right)}}{2}
\\&=&\frac{2\cos{\left(x\right)}}{2}
\\&=&\cos{\left(x\right)}
\end{eqnarray}$$
\(\sinh{\left(ix\right)}\)
$$\begin{eqnarray}
\sinh{\left(x\right)}&=&\frac{e^x-e^{-x}}{2}
\\\sinh{\left(ix\right)}&=&\frac{e^{ix}-e^{-ix}}{2}
\\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(-x\right)}+i\sin{\left(-x\right)}\right\}}{2}
\\&&\;\ldots\;e^{ix}=\cos{\left(x\right)}+i\sin{\left(x\right)}
\\&=&\frac{\left\{\cos{\left(x\right)}+i\sin{\left(x\right)}\right\}-\left\{\cos{\left(x\right)}-i\sin{\left(x\right)}\right\}}{2}
\\&&\;\ldots\;\cos{\left(-x\right)}=\cos{\left(x\right)},\;\sin{\left(-x\right)}=-\sin{\left(-x\right)}
\\&=&\frac{\color{red}{\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}\color{red}{-\cos{\left(x\right)}}\color{black}{+i\sin{\left(x\right)}}}{2}
\\&=&\frac{i\sin{\left(x\right)}+i\sin{\left(x\right)}}{2}
\\&=&\frac{2i\sin{\left(x\right)}}{2}
\\&=&i\sin{\left(x\right)}
\end{eqnarray}$$
\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))
\(cos{(i x)}, sin{(i x)}\) (純虚数に対する\(\cos, \sin\))
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