逆三角関数と逆双曲線関数の負角
逆三角関数の負角
(1)
\[ \Sin^{\bullet}\left(-z\right)=-\Sin^{\bullet}z \](2)
\[ \Cos^{\bullet}\left(-z\right)=-\Cos^{\bullet}\left(z\right)+\pi \](3)
\[ \Tan^{\bullet}\left(-z\right)=-\Tan^{\bullet}z \](4)
\[ \Sin^{-1,\bullet}\left(-z\right)=-\Sin^{-1,\bullet}z \](5)
\[ \Cos^{-1,\bullet}\left(-z\right)=-\Cos^{-1,\bullet}z+\pi \](6)
\[ \Tan^{-1,\bullet}\left(-z\right)=-\Tan^{-1,\bullet}z \](1)
\begin{align*} \Sin^{\bullet}\left(-z\right) & =\Sin^{\bullet}\left(-\sin\Sin^{\bullet}z\right)\\ & =\Sin^{\bullet}\sin\left(-\Sin^{\bullet}z\right)\\ & =-\Sin^{\bullet}z \end{align*}(1)-2
\begin{align*} \Sin^{\bullet}\left(-z\right) & =-i\Log\left(-iz+\sqrt{1-\left(-z\right)^{2}}\right)\\ & =-i\Log\frac{\left(-iz+\sqrt{1-z^{2}}\right)\left(iz+\sqrt{1-z^{2}}\right)}{\left(iz+\sqrt{1-z^{2}}\right)}\\ & =-i\Log\left(iz+\sqrt{1-z^{2}}\right)^{-1}\\ & =-i\left\{ -\Log\left(iz+\sqrt{1-z^{2}}\right)+2\pi i\delta_{\pi,\Arg\left(iz+\sqrt{1-z^{2}}\right)}\right\} \\ & =i\Log\left(iz+\sqrt{1-z^{2}}\right)\\ & =-\Sin^{\bullet}z \end{align*}(2)
\begin{align*} \Cos^{\bullet}\left(-z\right) & =-\Sin^{\bullet}\left(-z\right)+\frac{\pi}{2}\\ & =\Sin^{\bullet}\left(z\right)+\frac{\pi}{2}\\ & =-\Cos^{\bullet}\left(z\right)+\pi \end{align*}(3)
\begin{align*} \Tan^{\bullet}\left(-z\right) & =\Tan^{\bullet}\left(-\tan\Tan^{\bullet}z\right)\\ & =\Tan^{\bullet}\tan\left(-\Tan^{\bullet}z\right)\\ & =-\Tan^{\bullet}z \end{align*}(3)-2
\begin{align*} \Tan^{\bullet}\left(-z\right) & =\frac{i}{2}\left(\Log\left(1-i\left(-z\right)\right)-\Log\left(1+i\left(-z\right)\right)\right)\\ & =\frac{i}{2}\left(\Log\left(1+iz\right)-\Log\left(1-iz\right)\right)\\ & =-\frac{i}{2}\left(\Log\left(1-iz\right)-\Log\left(1+iz\right)\right)\\ & =-\Tan\left(z\right) \end{align*}(4)
\begin{align*} \Sin^{-1,\bullet}\left(-z\right) & =\Sin^{-1,\bullet}\left(-\sin^{-1}\Sin^{-1,\bullet}z\right)\\ & =\Sin^{-1,\bullet}\sin^{-1}\left(-\Sin^{-1,\bullet}z\right)\\ & =-\Sin^{-1,\bullet}z \end{align*}(4)-2
\begin{align*} \Sin^{-1,\bullet}\left(-z\right) & =\Sin^{\bullet}\left(-\frac{1}{z}\right)\\ & =-\Sin^{\bullet}\frac{1}{z}\\ & =-\Sin^{-1,\bullet}z \end{align*}(5)
\begin{align*} \Cos^{-1,\bullet}\left(-z\right) & =\Cos^{\bullet}\left(-\frac{1}{z}\right)\\ & =-\Cos^{\bullet}\left(\frac{1}{z}\right)+\pi\\ & =-\Cos^{-1,\bullet}z+\pi \end{align*}(6)
\begin{align*} \Tan^{-1,\bullet}\left(-z\right) & =\Tan^{-1,\bullet}\left(-\tan^{-1}\Tan^{-1,\bullet}z\right)\\ & =\Tan^{-1,\bullet}\tan^{-1}\left(-\Tan^{-1,\bullet}z\right)\\ & =-\Tan^{-1,\bullet}z \end{align*}(6)-2
\begin{align*} \Tan^{-1,\bullet}\left(-z\right) & =\Tan^{\bullet}\left(-\frac{1}{z}\right)\\ & =-\Tan^{\bullet}\frac{1}{z}\\ & =-\Tan^{-1,\bullet}z \end{align*}逆双曲線関数の負角
(1)
\[ \Sinh^{\bullet}\left(-z\right)=-\Sinh^{\bullet}z \](2)
\[ \Tanh^{\bullet}\left(-z\right)=-\Tanh^{\bullet}z \](3)
\[ \Sinh^{-1,\bullet}\left(-z\right)=-\Sinh^{-1,\bullet}z \](4)
\[ \Tanh^{-1,\bullet}\left(-z\right)=-\Tanh^{-1,\bullet}z \](1)
\begin{align*} \Sinh^{\bullet}\left(-z\right) & =\Sinh^{\bullet}\left(-\sinh\Sinh^{\bullet}z\right)\\ & =\Sinh^{\bullet}\sinh\left(-\Sinh^{\bullet}z\right)\\ & =-\Sinh^{\bullet}z \end{align*}(1)-2
\begin{align*} \sinh^{\bullet}\left(-z\right) & =\Log\left(-z+\sqrt{1+\left(-z\right)^{2}}\right)\\ & =\Log\frac{\left(-z+\sqrt{1+z^{2}}\right)\left(z+\sqrt{1+z^{2}}\right)}{\left(z+\sqrt{1+z^{2}}\right)}\\ & =\Log\left(z+\sqrt{1+z^{2}}\right)^{-1}\\ & =-\Log\left(z+\sqrt{1+z^{2}}\right)+2\pi i\delta_{\pi,\Arg\left(z+\sqrt{1+z^{2}}\right)}\\ & =-\Log\left(z+\sqrt{1+z^{2}}\right)\\ & =-\sinh^{\bullet}z \end{align*}(2)
\begin{align*} \Tanh^{\bullet}\left(-z\right) & =\Tanh^{\bullet}\left(-\tanh\Tanh^{\bullet}z\right)\\ & =\Tanh^{\bullet}\tanh\left(-\Tanh^{\bullet}z\right)\\ & =-\Tanh^{\bullet}z \end{align*}(2)-2
\begin{align*} \Tanh^{\bullet}\left(-z\right) & =\frac{1}{2}\left(\Log\left(1+\left(-z\right)\right)-\Log\left(1-\left(-z\right)\right)\right)\\ & =\frac{1}{2}\left(\Log\left(1-z\right)-\Log\left(1+z\right)\right)\\ & =-\frac{1}{2}\left(\Log\left(1+z\right)-\Log\left(1-z\right)\right)\\ & =-\Tanh^{\bullet}z \end{align*}(3)
\begin{align*} \Sinh^{-1,\bullet}\left(-z\right) & =\Sinh^{-1,\bullet}\left(-\sinh^{-1}\Sinh^{-1,\bullet}z\right)\\ & =\Sinh^{-1,\bullet}\sinh^{-1}\left(-\Sinh^{-1,\bullet}z\right)\\ & =-\Sinh^{-1,\bullet}z \end{align*}(3)-2
\begin{align*} \Sinh^{-1,\bullet}\left(-z\right) & =\Sinh^{\bullet}\left(-\frac{1}{z}\right)\\ & =-\Sinh^{\bullet}\frac{1}{z}\\ & =-\Sinh^{-1,\bullet}z \end{align*}(4)
\begin{align*} \Tanh^{-1,\bullet}\left(-z\right) & =\Tanh^{-1,\bullet}\left(-\tanh^{-1}\Tanh^{-1,\bullet}z\right)\\ & =\Tanh^{-1,\bullet}\tanh^{-1}\left(-\Tanh^{-1,\bullet}z\right)\\ & =-\Tanh^{-1,\bullet}z \end{align*}(4)-2
\begin{align*} \Tanh^{-1,\bullet}\left(-z\right) & =\Tanh^{\bullet}\left(-\frac{1}{z}\right)\\ & =-\Tanh^{\bullet}\frac{1}{z}\\ & =-\Tanh^{-1,\bullet}z \end{align*}ページ情報
| タイトル | 逆三角関数と逆双曲線関数の負角 |
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逆三角関数と逆双曲線関数の冪乗積分漸化式
\[
\int\sin^{\bullet,n}xdx=x\sin^{\bullet,n}x+n\sqrt{1-x^{2}}\sin^{\bullet,n-1}x-n(n-1)\int\sin^{\bullet,n-2}xdx
\]
三角関数と双曲線関数の加法定理
\[
\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y
\]
三角関数と双曲線関数の積分
\[
\int\cos xdx=\sin x
\]
三角関数と双曲線関数の冪乗積分漸化式
\[
\int\sin^{n}xdx=-\frac{1}{n}\cos x\sin^{n-1}x+\frac{n-1}{n}\int\sin^{n-2}xdx\qquad(n\ne0)
\]