2.20 逆三角関数

三角関数の逆関数を 逆三角関数（inverse trigonometric function）と呼び， $\sin x$ , $\cos x$ , $\tan x$ の逆関数をそれぞれ

$\displaystyle y$	$\displaystyle =\sin^{-1}x=\arcsin x\,\qquad (-1\le x\le1)$
$\displaystyle y$	$\displaystyle =\cos^{-1}x=\arccos x\,\qquad (-1\le x\le1)$
$\displaystyle y$	$\displaystyle =\tan^{-1}x=\arctan x\,\qquad (-\infty<x<\infty)$

$\displaystyle y$	$\displaystyle =\mathrm{Sin}^{-1}x=\mathrm{Arcsin}\,x\, \qquad(-1\le x\le1) \qquad\left(-\frac{\pi}{2}\le y\le\frac{\pi}{2}\right)$
$\displaystyle y$	$\displaystyle =\mathrm{Cos}^{-1}x=\mathrm{Arccos}\,x\, \qquad(-1\le x\le1) \qquad(0\le y\le\pi)$
$\displaystyle y$	$\displaystyle =\mathrm{Tan}^{-1}x=\mathrm{Arctan}\,x\, \qquad(-\infty<x<\infty) \qquad\left(-\frac{\pi}{2}<y<\frac{\pi}{2}\right)$

例 2.59 (逆三角関数の値)

$\displaystyle \mathrm{Sin}^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\,, \qquad... ...{6}\,, \qquad \mathrm{Tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\,.$

問 2.60 (逆三角関数の値) 次の値を求めよ．

$\displaystyle \mathrm{Sin}^{-1}\left(0\right)\,,$	$\displaystyle \mathrm{Sin}^{-1}\left(\frac{1}{2}\right)\,,$	$\displaystyle \mathrm{Sin}^{-1}\left(\frac{1}{\sqrt{2}}\right)\,,$	$\displaystyle \mathrm{Sin}^{-1}\left(\frac{\sqrt{3}}{2}\right)\,,$	$\displaystyle \mathrm{Sin}^{-1}\left(1\right)\,,$
$\displaystyle \mathrm{Cos}^{-1}\left(0\right)\,,$	$\displaystyle \mathrm{Cos}^{-1}\left(\frac{1}{2}\right)\,,$	$\displaystyle \mathrm{Cos}^{-1}\left(\frac{1}{\sqrt{2}}\right)\,,$	$\displaystyle \mathrm{Cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)\,,$	$\displaystyle \mathrm{Cos}^{-1}\left(1\right)\,,$
$\displaystyle \mathrm{Tan}^{-1}\left(0\right)\,,$	$\displaystyle \mathrm{Tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)\,,$	$\displaystyle \mathrm{Tan}^{-1}\left(1\right)\,,$	$\displaystyle \mathrm{Tan}^{-1}\left(\sqrt{3}\right)\,.$