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)
\[ \sin\Sin^{\bullet}z=z \](2)
\[ \sin\Cos^{\bullet}z=\sqrt{1-z^{2}} \](3)
\[ \sin\Tan^{\bullet}x=\frac{z}{\sqrt{1+z^{2}}} \](4)
\[ \cos\Sin^{\bullet}x=\sqrt{1-z^{2}} \](5)
\[ \cos\Cos^{\bullet}z=z \](6)
\[ \cos\Tan^{\bullet}x=\frac{1}{\sqrt{1+z^{2}}} \](7)
\[ \tan\Sin^{\bullet}z=\frac{z}{\sqrt{1-z^{2}}} \](8)
\[ \tan\Cos^{\bullet}z=\frac{\sqrt{1-z^{2}}}{z} \](9)
\[ \tan\Tan^{\bullet}z=z \](1)
\begin{align*} \sin\Sin^{\bullet}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\[ \sin\Sin^{\bullet}x=x \]
(2)
\begin{align*} \sin\Cos^{\bullet}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^{\bullet}x & =\sqrt{1-\cos^{2}\cos^{\bullet}x}\\ & =\sqrt{1-x^{2}} \end{align*}
(3)
\begin{align*} \sin\Tan^{\bullet}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^{\bullet}x & =\frac{\tan\tan^{\bullet}x}{\sqrt{1+\tan^{2}\tan^{\bullet}x}}\\ & =\frac{x}{\sqrt{1+x^{2}}} \end{align*}
(4)
\begin{align*} \cos\Sin^{\bullet}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^{\bullet}z & =\cos\left(\frac{\pi}{2}-\Cos^{\bullet}z\right)\\ & =\sin\Cos^{\bullet}z\\ & =\sqrt{1-z^{2}} \end{align*}(4)-3
実数の場合\begin{align*} \cos\sin^{\bullet}x & =\sqrt{1-\sin^{2}\sin^{\bullet}x}\\ & =\sqrt{1-x^{2}} \end{align*}
(5)
\begin{align*} \cos\Cos^{\bullet}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\[ \cos\cos^{\bullet}x=x \]
(6)
\begin{align*} \cos\Tan^{\bullet}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^{\bullet}x & =\frac{1}{\sqrt{1+\tan^{2}\tan^{\bullet}x}}\\ & =\frac{1}{\sqrt{1+x^{2}}} \end{align*}
(7)
\begin{align*} \tan\Sin^{\bullet}z & =\frac{\sin\Sin^{\bullet}z}{\cos\Sin^{\bullet}z}\\ & =\frac{z}{\sqrt{1-z^{2}}} \end{align*}(7)-2
実数の場合\begin{align*} \tan\sin^{\bullet}x & =\frac{\sin\sin^{\bullet}x}{\sqrt{1-\sin^{2}\sin^{\bullet}x}}\\ & =\frac{x}{\sqrt{1-x^{2}}} \end{align*}
(8)
\begin{align*} \tan\Cos^{\bullet}z & =\frac{\sin\Cos^{\bullet}z}{\cos\Cos^{\bullet}z}\\ & =\frac{\sqrt{1-z^{2}}}{z} \end{align*}(8)-2
実数の場合\begin{align*} \tan\cos^{\bullet}x & =\frac{\sqrt{1-\cos^{2}\cos^{\bullet}x}}{\cos\cos^{\bullet}x}\\ & =\frac{\sqrt{1-x^{2}}}{x} \end{align*}
(9)
\begin{align*} \tan\Tan^{\bullet}z & =\frac{\sin\Tan^{\bullet}z}{\cos\Tan^{\bullet}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\[ \tan\tan^{\bullet}x=x \]
逆双曲線関数の双曲線関数
(1)
\[ \sinh\Sinh^{\bullet}z=z \](2)
\[ \sinh\Cosh^{\bullet}z=\sqrt{z-1}\sqrt{z+1} \](3)
\[ \sinh\Tanh^{\bullet}z=\frac{z}{\sqrt{1-z^{2}}} \](4)
\[ \cosh\Sinh^{\bullet}z=\sqrt{1+z^{2}} \](5)
\[ \cosh\Cosh^{\bullet}z=z \](6)
\[ \cosh\Tanh^{\bullet}z=\frac{1}{\sqrt{1-z^{2}}} \](7)
\[ \tanh\Sinh^{\bullet}z=\frac{z}{\sqrt{1+z^{2}}} \](8)
\[ \tanh\Cosh^{\bullet}z=\frac{\sqrt{z-1}\sqrt{z+1}}{z} \](9)
\[ \tanh\Tanh^{\bullet}z=z \](1)
\begin{align*} \sinh\Sinh^{\bullet}z & =\sinh\left(-ii\Sinh^{\bullet}z\right)\\ & =-i\sin\Sin^{\bullet}\left(iz\right)\\ & =z \end{align*}(1)-2
実数の場合\[ \sinh\sinh^{\bullet}x=x \]
(2)
\begin{align*} \sinh\Cosh^{\bullet}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^{\bullet}x & =\sqrt{\cosh^{2}\cosh^{\bullet}x-2}\\ & =\sqrt{1-x^{2}} \end{align*}
(3)
\begin{align*} \sinh\Tanh^{\bullet}z & =\sinh\left(-ii\Tanh^{\bullet}z\right)\\ & =-i\sin\Tan^{\bullet}\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^{\bullet}x & =\frac{\tanh\tanh^{\bullet}x}{\sqrt{1-\tanh^{2}\tanh^{\bullet}x}}\\ & =\frac{x}{\sqrt{1-x^{2}}} \end{align*}
(4)
\begin{align*} \cosh\Sinh^{\bullet}z & =\cosh\left(-ii\Sinh^{\bullet}z\right)\\ & =\cos\Sin^{\bullet}\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^{\bullet}x & =\sqrt{1+\sinh^{2}\sinh^{\bullet}x}\\ & =\sqrt{1+x^{2}} \end{align*}
(5)
\begin{align*} \cosh\Cosh^{\bullet}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\[ \cosh\cosh^{\bullet}x=x \]
(6)
\begin{align*} \cosh\Tanh^{\bullet}z & =\cosh\left(-ii\Tanh^{\bullet}z\right)\\ & =\cos\Tan^{\bullet}\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^{\bullet}x & =\frac{1}{\sqrt{1-\tanh^{2}\tanh^{\bullet}x}}\\ & =\frac{1}{\sqrt{1-x^{2}}} \end{align*}
(7)
\begin{align*} \tanh\Sinh^{\bullet}z & =\tanh\left(-ii\Sinh^{\bullet}z\right)\\ & =-i\tan\Sin^{\bullet}\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^{\bullet}x & =\frac{\sinh\sinh^{\bullet}x}{\sqrt{1+\sinh^{2}\sinh^{\bullet}x}}\\ & =\frac{x}{\sqrt{1+x^{2}}} \end{align*}
(8)
\begin{align*} \tanh\Cosh^{\bullet}z & =\frac{\sinh\Cosh^{\bullet}z}{\cosh\Cosh^{\bullet}z}\\ & =\frac{\sqrt{z-1}\sqrt{z+1}}{z} \end{align*}(8)-2
実数の場合\begin{align*} \tanh\cosh^{\bullet}x & =\frac{\sqrt{\cosh^{2}\cosh^{\bullet}x-1}}{\cosh\cosh^{\bullet}x}\\ & =\frac{\sqrt{x^{2}-1}}{x} \end{align*}
(9)
\begin{align*} \tanh\Tanh^{\bullet}z & =\frac{\sinh\Tanh^{\bullet}z}{\cosh\Tanh^{\bullet}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\[ \tanh\tanh^{\bullet}x=x \]
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タイトル | 逆三角関数の三角関数と逆双曲線関数の双曲線関数 |
URL | https://www.nomuramath.com/ksteoj3b/ |
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3角関数3つでの積和公式・和積公式
\[
\sin A+\sin B+\sin C=4\sin\frac{B+C}{2}\sin\frac{C+A}{2}\sin\frac{A+B}{2}+\sin\left(A+B+C\right)
\]
三角関数と双曲線関数の対数
\[
\log\sin x=-\log2+\frac{\pi}{2}i-ix-Li_{1}\left(e^{2ix}\right)
\]
逆三角関数と逆双曲線関数の積分表示
\[
\sin^{\bullet}x=\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}dt
\]
三角関数と双曲線関数の加法定理
\[
\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y
\]