逆三角関数の負角、余角、逆数
逆三角関数の負角
(1)
\[ \sin^{\bullet}\left(-x\right)=-\sin^{\bullet}x \](2)
\[ \cos^{\bullet}\left(-x\right)=\pi-\cos^{\bullet}x \](3)
\[ \tan{}^{\bullet}\left(-x\right)=-\tan^{\bullet}x \](4)
\[ \sin^{-1,\bullet}\left(-x\right)=-\sin^{-1,\bullet}x \](5)
\[ \cos^{-1,\bullet}\left(-x\right)=\pi-\cos^{-1,\bullet}x \](6)
\[ \tan{}^{-1,\bullet}\left(-x\right)=-\tan^{-1,\bullet}x \](1)
奇関数の逆関数は奇関数なので、\[ \sin^{\bullet}\left(-x\right)=-\sin^{\bullet}x \]
(1)-2
\begin{align*} \sin^{\bullet}\left(-x\right) & =\sin^{\bullet}\left(-\sin\left(\sin^{\bullet}x\right)\right)\\ & =\sin^{\bullet}\left(\sin(-\sin^{\bullet}x)\right)\\ & =-\sin^{\bullet}x \end{align*}(2)
\begin{align*} \cos^{\bullet}\left(-x\right) & =\cos^{\bullet}\left(-\cos\left(\cos^{\bullet}x\right)\right)\\ & =\cos^{\bullet}\left(\cos(\pi-\cos^{\bullet}x)\right)\\ & =\pi-\cos^{\bullet}x \end{align*}(3)
奇関数の逆関数は奇関数なので、\[ \tan{}^{\bullet}\left(-x\right)=-\tan^{\bullet}x \]
(3)-2
\begin{align*} \tan^{\bullet}\left(-x\right) & =\tan^{\bullet}\left(-\tan\left(\tan^{\bullet}x\right)\right)\\ & =\tan^{\bullet}\left(\tan(-\tan^{\bullet}x)\right)\\ & =\tan^{\bullet}x \end{align*}(4)
奇関数の逆関数は奇関数なので、\[ \sin^{-1,\bullet}\left(-x\right)=-\sin^{-1,\bullet}x \]
(4)-2
\begin{align*} \sin^{-1,\bullet}\left(-x\right) & =\sin^{-1,\bullet}\left(-\sin^{-1}\left(\sin^{-1,\bullet}x\right)\right)\\ & =\sin^{-1,\bullet}\left(\sin^{-1}(-\sin^{-1\bullet}x)\right)\\ & =-\sin^{-1,\bullet}x \end{align*}(5)
\begin{align*} \cos^{-1,\bullet}\left(-x\right) & =\cos^{-1,\bullet}\left(-\cos^{-1}\left(\cos^{-1,\bullet}x\right)\right)\\ & =\cos^{-1,\bullet}\left(\cos^{-1}(\pi-\cos^{-1,\bullet}x)\right)\\ & =\pi-\cos^{-1,\bullet}x \end{align*}(6)
奇関数の逆関数は奇関数なので、\[ \tan{}^{-1,\bullet}\left(-x\right)=-\tan^{-1,\bullet}x \]
(6)-2
\begin{align*} \tan^{-1,\bullet}\left(-x\right) & =\tan^{-1,\bullet}\left(-\tan^{-1}\left(\tan^{-1,\bullet}x\right)\right)\\ & =\tan^{-1,\bullet}\left(\tan(-\tan^{-1,\bullet}x)\right)\\ & =\tan^{-1,\bullet}x \end{align*}逆三角関数の余角
(1)
\[ \cos^{\bullet}x+\sin^{\bullet}x=\frac{\pi}{2} \](2)
\[ \tan^{\bullet}x+\tan^{-1,\bullet}x=\frac{\pi}{2} \](3)
\[ \sin^{-1,\bullet}x+\cos^{-1,\bullet}x=\frac{\pi}{2} \](1)
\begin{align*} \frac{\pi}{2} & =x+\frac{\pi}{2}-x\\ & =\cos^{\bullet}\cos x+\sin^{\bullet}\sin\left(\frac{\pi}{2}-x\right)\\ & =\cos^{\bullet}y+\sin^{\bullet}y\qquad,\qquad y=\cos x \end{align*}(2)
\begin{align*} \frac{\pi}{2} & =x+\frac{\pi}{2}-x\\ & =\tan^{\bullet}\tan x+\tan^{-1,\bullet}\tan^{-1}\left(\frac{\pi}{2}-x\right)\\ & =\tan^{\bullet}y+\tan^{-1,\bullet}y\qquad,\qquad y=\tan x \end{align*}(3)
\begin{align*} \frac{\pi}{2} & =x+\frac{\pi}{2}-x\\ & =\sin^{-1,\bullet}\sin^{-1}x+\cos^{-1,\bullet}\cos^{-1}\left(\frac{\pi}{2}-x\right)\\ & =\sin^{-1,\bullet}y+\cos^{-1,\bullet}y\qquad,\qquad y=\sin^{-1}x \end{align*}逆三角関数の逆数
(1)
\[ \sin^{\bullet}\left(\frac{1}{x}\right)=\sin^{-1,\bullet}\left(x\right) \](2)
\[ \cos^{\bullet}\left(\frac{1}{x}\right)=\cos^{-1,\bullet}\left(x\right) \](3)
\[ \tan^{\bullet}\left(\frac{1}{x}\right)=\tan^{-1,\bullet}\left(x\right) \](4)
\[ \sin^{-1,\bullet}\left(\frac{1}{x}\right)=\sin^{\bullet}\left(x\right) \](5)
\[ \cos^{-1,\bullet}\left(\frac{1}{x}\right)=\cos^{\bullet}\left(x\right) \](6)
\[ \tan^{-1,\bullet}\left(\frac{1}{x}\right)=\tan^{\bullet}\left(\frac{1}{x}\right) \](1)〜(6)
\[ f\left(\frac{1}{x}\right)=f^{\bullet,-1,\bullet}(x) \] より成り立つ。ページ情報
タイトル | 逆三角関数の負角、余角、逆数 |
URL | https://www.nomuramath.com/xhi31jyt/ |
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三角関数と双曲線関数の実部と虚部
\[
\tan z=\frac{\sin\left(2\Re z\right)+i\sinh\left(2\Im z\right)}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}
\]
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・双曲線関数の微分
\[
\left(\sin x\right)'=\cos x
\]
三角関数の合成
\[
a\sin\theta+b\cos\theta =\sqrt{a^{2}+b^{2}}\sin(\theta+\alpha)
\]