三角関数と双曲線関数の実部と虚部
三角関数の実部と虚部
(1)
\[ \sin z=\sin\left(\Re z\right)\cosh\left(\Im z\right)+i\cos\left(\Re z\right)\sinh\left(\Im z\right) \](2)
\[ \cos z=\cos\left(\Re z\right)\cosh\left(\Im z\right)-i\sin\left(\Re z\right)\sinh\left(\Im z\right) \](3)
\[ \tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \](4)
\[ \sin^{-1}z=2\frac{\sin\left(\Re z\right)\cosh\left(\Im z\right)-i\cos\left(\Re z\right)\sinh\left(\Im z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \](5)
\[ \cos^{-1}z=2\frac{\cos\left(\Re z\right)\cosh\left(\Im z\right)+i\sin\left(\Re z\right)\sinh\left(\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \](6)
\[ \tan^{-1}z=\frac{\sin\left(2\Re z\right)-i\sinh\left(2\Im z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \](1)
\begin{align*} \sin z & =\sin\left(\Re z+i\Im z\right)\\ & =\sin\left(\Re z\right)\cos\left(i\Im z\right)+\cos\left(\Re z\right)\sin\left(i\Im z\right)\\ & =\sin\left(\Re z\right)\cosh\left(\Im z\right)+i\cos\left(\Re z\right)\sinh\left(\Im z\right) \end{align*}(2)
\begin{align*} \cos z & =\cos\left(\Re z+i\Im z\right)\\ & =\cos\left(\Re z\right)\cos\left(i\Im z\right)-\sin\left(\Re z\right)\sin\left(i\Im z\right)\\ & =\cos\left(\Re z\right)\cosh\left(\Im z\right)-i\sin\left(\Re z\right)\sinh\left(\Im z\right) \end{align*}(3)
\begin{align*} \tan z & =\tan\left(\Re z+i\Im z\right)\\ & =\frac{\sin\left(2\Re z\right)+\sin\left(2i\Im z\right)}{\cos\left(2\Re z\right)+\cos\left(2i\Im z\right)}\\ & =\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)} \end{align*}(4)
\begin{align*} \sin^{-1}z & =\sin^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{2\sin\left(\Re z-i\Im z\right)}{-\cos\left(2\Re z\right)+\cos\left(2i\Im z\right)}\\ & =2\frac{\sin\left(\Re z\right)\cos\left(i\Im z\right)-\cos\left(\Re z\right)\sin\left(i\Im z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =2\frac{\sin\left(\Re z\right)\cosh\left(\Im z\right)-i\cos\left(\Re z\right)\sinh\left(\Im z\right)}{-\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \end{align*}(5)
\begin{align*} \cos^{-1}z & =\cos^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{2\cos\left(\Re z-i\Im z\right)}{\cos\left(2\Re z\right)+\cos\left(2i\Im z\right)}\\ & =2\frac{\cos\left(\Re z\right)\cos\left(i\Im z\right)+\sin\left(\Re z\right)\sin\left(i\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\\ & =2\frac{\cos\left(\Re z\right)\cosh\left(\Im z\right)+i\sin\left(\Re z\right)\sinh\left(\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)} \end{align*}(6)
\begin{align*} \tan^{-1}z & =\tan^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{\sin\left(2\Re z\right)-\sin\left(2i\Im z\right)}{-\cos\left(2\Re z\right)+\cos\left(2i\Im z\right)}\\ & =\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)} \end{align*}双曲線関数の実部と虚部
(1)
\[ \sinh z=\sinh\left(\Re z\right)\cos\left(\Im z\right)+i\cosh\left(\Re z\right)\sin\left(\Im z\right) \](2)
\[ \cosh z=\cosh\left(\Re z\right)\cos\left(\Im z\right)+i\sinh\left(\Re z\right)\sin\left(\Im z\right) \](3)
\[ \tanh z=\frac{\sinh\left(2\Re z\right)+i\sin\left(2\Im z\right)}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)} \](4)
\[ \sinh^{-1}z=2\frac{\sinh\left(\Re z\right)\cos\left(\Im z\right)-i\cosh\left(\Re z\right)\sin\left(\Im z\right)}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)} \](5)
\[ \cosh^{-1}z=2\frac{\cosh\left(\Re z\right)\cos\left(\Im z\right)-i\sinh\left(\Re z\right)\sin\left(\Im z\right)}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)} \](6)
\[ \tanh^{-1}z=\frac{\sinh\left(2\Re z\right)-i\sin\left(2\Im z\right)}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)} \](1)
\begin{align*} \sinh z & =\sinh\left(\Re z+i\Im z\right)\\ & =\sinh\left(\Re z\right)\cosh\left(i\Im z\right)+\cosh\left(\Re z\right)\sinh\left(i\Im z\right)\\ & =\sinh\left(\Re z\right)\cos\left(\Im z\right)+i\cosh\left(\Re z\right)\sin\left(\Im z\right) \end{align*}(2)
\begin{align*} \cosh z & =\cosh\left(\Re z+i\Im z\right)\\ & =\cosh\left(\Re z\right)\cosh\left(i\Im z\right)+\sinh\left(\Re z\right)\sinh\left(i\Im z\right)\\ & =\cosh\left(\Re z\right)\cos\left(\Im z\right)+i\sinh\left(\Re z\right)\sin\left(\Im z\right) \end{align*}(3)
\begin{align*} \tanh z & =\tanh\left(\Re z+i\Im z\right)\\ & =\frac{\sinh\left(2\Re z\right)+\sinh\left(2i\Im z\right)}{\cosh\left(2\Re z\right)+\cosh\left(2i\Im z\right)}\\ & =\frac{\sinh\left(2\Re z\right)+i\sin\left(2\Im z\right)}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)} \end{align*}(4)
\begin{align*} \sinh^{-1}z & =\sinh^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{2\sinh\left(\Re z-i\Im z\right)}{\cosh\left(2\Re z\right)-\cosh\left(2i\Im z\right)}\\ & =2\frac{\sinh\left(\Re z\right)\cosh\left(i\Im z\right)-\cosh\left(\Re z\right)\sinh\left(i\Im z\right)}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)}\\ & =2\frac{\sinh\left(\Re z\right)\cos\left(\Im z\right)-i\cosh\left(\Re z\right)\sin\left(\Im z\right)}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)} \end{align*}(5)
\begin{align*} \cosh^{-1}z & =\cosh^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{2\cosh\left(\Re z-i\Im z\right)}{\cosh\left(2\Re z\right)+\cosh\left(2i\Im z\right)}\\ & =2\frac{\cosh\left(\Re z\right)\cosh\left(i\Im z\right)-\sinh\left(\Re z\right)\sinh\left(i\Im z\right)}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)}\\ & =2\frac{\cosh\left(\Re z\right)\cos\left(\Im z\right)-i\sinh\left(\Re z\right)\sin\left(\Im z\right)}{\cosh\left(2\Re z\right)+\cos\left(2\Im z\right)} \end{align*}(6)
\begin{align*} \tanh^{-1}z & =\tanh^{-1}\left(\Re z+i\Im z\right)\\ & =\frac{\sinh\left(2\Re z\right)-\sinh\left(2i\Im z\right)}{\cosh\left(2\Re z\right)-\cosh\left(2i\Im z\right)}\\ & =\frac{\sinh\left(2\Re z\right)-i\sin\left(2\Im z\right)}{\cosh\left(2\Re z\right)-\cos\left(2\Im z\right)} \end{align*}
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三角関数と双曲線関数の対数
\[
\log\sin x=-\log2+\frac{\pi}{2}i-ix-Li_{1}\left(e^{2ix}\right)
\]
3角関数と双曲線関数の加法定理
\[
\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y
\]
x tan(x)とx tanh(x)の積分
\[
\int z\tan^{\pm1}\left(z\right)dz=i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2iz}\right)\right\} +C
\]
逆三角関数と逆双曲線関数の関係
\[
\Sin^{\bullet}\left(iz\right)=i\Sinh^{\bullet}z
\]