3角関数・双曲線関数の還元公式(負角・余角・補角)

by nomura · 2019年11月13日

3角関数・双曲線関数の還元公式(負角・余角・補角)

3角関数
負角

(1)

\[ \sin\left(-z\right)=-\sin z \]

(2)

\[ \cos\left(-z\right)=\cos z \]

(3)

\[ \tan\left(-z\right)=-\tan z \]
余角

(4)

\[ \sin\left(\frac{\pi}{2}\pm z\right)=\cos\left(z\right) \]

(5)

\[ \cos\left(\frac{\pi}{2}\pm z\right)=\mp\sin\left(z\right) \]

(6)

\[ \tan\left(\frac{\pi}{2}\pm z\right)=\mp\tan^{-1}\left(z\right) \]
補角

(7)

\[ \sin\left(z\pm\pi\right)=-\sin\left(z\right) \]

(8)

\[ \cos\left(z\pm\pi\right)=-\cos\left(z\right) \]

(9)

\[ \tan\left(z\pm\pi\right)=\tan\left(z\right) \]
双曲線関数
負角

(10)

\[ \sinh\left(-z\right)=-\sinh z \]

(11)

\[ \cosh\left(-z\right)=\cosh z \]

(12)

\[ \tanh\left(-z\right)=-\tanh z \]
余角

(13)

\[ \sinh\left(\frac{\pi}{2}i\pm z\right)=i\cosh z \]

(14)

\[ \cosh\left(\frac{\pi}{2}i\pm z\right)=\pm i\sinh z \]

(15)

\[ \tanh\left(\frac{\pi}{2}\pm z\right)=\pm\tanh^{-1}z \]
補角

(16)

\[ \sinh\left(z\pm\pi i\right)=-\sinh z \]

(17)

\[ \cosh\left(z\pm\pi i\right)=-\cosh\left(z\right) \]

(18)

\[ \tanh\left(z\pm\pi i\right)=\tanh\left(z\right) \]

(1)

\begin{align*} \sin\left(-z\right) & =\frac{e^{-iz}-e^{iz}}{2i}\\ & =-\frac{e^{iz}-e^{-iz}}{2i}\\ & =-\sin z \end{align*}

(2)

\begin{align*} \cos\left(-z\right) & =\frac{e^{-iz}+e^{iz}}{2}\\ & =\cos z \end{align*}

(3)

\begin{align*} \tan\left(-z\right) & =\frac{\sin\left(-z\right)}{\cos\left(-z\right)}\\ & =\frac{-\sin z}{\cos z}\\ & =-\tan z \end{align*}

(4)

\begin{align*} \sin\left(\frac{\pi}{2}\pm z\right) & =\sin\left(\frac{\pi}{2}\right)\cos\left(\pm z\right)\pm\cos\left(\frac{\pi}{2}\right)\sin\left(\pm z\right)\\ & =\cos\left(z\right) \end{align*}

(5)

\begin{align*} \cos\left(\frac{\pi}{2}\pm z\right) & =\cos\left(\frac{\pi}{2}\right)\cos\left(\pm z\right)-\sin\left(\frac{\pi}{2}\right)\sin\left(\pm z\right)\\ & =\mp\sin\left(z\right) \end{align*}

(6)

\begin{align*} \tan\left(\frac{\pi}{2}\pm z\right) & =\frac{\sin\left(\frac{\pi}{2}\pm z\right)}{\cos\left(\frac{\pi}{2}\pm z\right)}\\ & =\frac{\cos\left(z\right)}{\mp\sin\left(z\right)}\\ & =\mp\tan^{-1}\left(z\right) \end{align*}

(7)

\begin{align*} \sin\left(z\pm\pi\right) & =\sin\left(z\right)\cos\left(\pi\right)\pm\cos\left(z\right)\sin\left(\pi\right)\\ & =-\sin\left(z\right) \end{align*}

(8)

\begin{align*} \cos\left(z\pm\pi\right) & =\cos\left(z\right)\cos\left(\pi\right)\mp\sin\left(z\right)\sin\left(\pi\right)\\ & =-\cos\left(z\right) \end{align*}

(9)

\begin{align*} \tan\left(z\pm\pi\right) & =\frac{\sin\left(z\pm\pi\right)}{\cos\left(z\pm\pi\right)}\\ & =\frac{-\sin\left(z\right)}{-\cos\left(z\right)}\\ & =\tan\left(z\right) \end{align*}

(10)

\begin{align*} \sinh\left(-z\right) & =-i\sin\left(-iz\right)\\ & =i\sin\left(iz\right)\cmt{\because\sin\left(-z\right)=-\sin z}\\ & =-\sinh z \end{align*}

(11)

\begin{align*} \cosh\left(-z\right) & =\cos\left(-iz\right)\\ & =\cos\left(iz\right)\cmt{\because\cos\left(-z\right)=\cos z}\\ & =\cosh z \end{align*}

(12)

\begin{align*} \tanh\left(-z\right) & =\frac{\sinh\left(-z\right)}{\cosh\left(-z\right)}\\ & =\frac{-\sinh z}{\cosh z}\\ & =-\tanh z \end{align*}

(13)

\begin{align*} \sinh\left(\frac{\pi}{2}i\pm z\right) & =i\sin\left(\frac{\pi}{2}\mp zi\right)\\ & =i\cos\left(zi\right)\cmt{\because\sin\left(\frac{\pi}{2}\pm z\right)=\cos\left(z\right)}\\ & =i\cosh z \end{align*}

(14)

\begin{align*} \cosh\left(\frac{\pi}{2}i\pm z\right) & =\cos\left(\frac{\pi}{2}\mp zi\right)\\ & =\pm\sin\left(zi\right)\cmt{\because\cos\left(\frac{\pi}{2}\pm z\right)=\mp\sin\left(z\right)}\\ & =\pm i\sinh z \end{align*}

(15)

\begin{align*} \tanh\left(\frac{\pi}{2}\pm z\right) & =\frac{\sinh\left(\frac{\pi}{2}i\pm z\right)}{\cosh\left(\frac{\pi}{2}i\pm z\right)}\\ & =\frac{i\cosh z}{\pm i\sinh z}\\ & =\pm\tanh^{-1}z \end{align*}

(16)

\begin{align*} \sinh\left(z\pm\pi i\right) & =i\sin\left(-zi\pm\pi\right)\\ & =-i\sin\left(-zi\right)\cmt{\because\sin\left(z\pm\pi\right)=-\sin\left(z\right)}\\ & =-\sinh z \end{align*}

(17)

\begin{align*} \cosh\left(z\pm\pi i\right) & =\cos\left(-zi\pm\pi\right)\\ & =-\cos\left(-zi\right)\cmt{\because\cos\left(z\pm\pi\right)=-\cos\left(z\right)}\\ & =-\cosh\left(z\right) \end{align*}

(18)

\begin{align*} \tanh\left(z\pm\pi i\right) & =\frac{\sinh\left(z\pm\pi i\right)}{\cosh\left(z\pm\pi i\right)}\\ & =\frac{-\sinh z}{-\cosh\left(z\right)}\\ & =\tanh\left(z\right) \end{align*}