【微分積分】2-3-3 逆三角関数｜問題集

\(\theta=\tan^{-1}0\)とおくと、
\(\displaystyle \tan\theta=0\ \ \ \left(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\)
\(\theta=0\)
よって、
\(\tan^{-1}0=0\)

\(\displaystyle \theta=\sin^{-1}\left(\frac{-1}{2}\right)\)とおくと、
\(\displaystyle \sin\theta=-\frac{1}{2}\ \ \ \left(-\frac{\pi}{2}\leqq\theta\leqq\frac{\pi}{2}\right)\)
\(\displaystyle \theta=-\frac{\pi}{6}\)
よって、
\(\displaystyle \sin^{-1}\left(\frac{-1}{2}\right)=-\frac{\pi}{6}\)

\(\displaystyle \theta=\cos^{-1}(-1)\)とおくと、
\(\cos\theta=-1\ \ \ (0\leqq\theta\leqq\pi)\)
\(\theta=\pi\)
よって、
\(\cos^{-1}(-1)=\pi\)

\(\displaystyle \theta=\tan^{-1}(-1)\)とおくと、
\(\displaystyle \tan\theta=-1\ \ \ \left(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\)
\(\displaystyle \theta=-\frac{\pi}{4}\)
よって、
\(\displaystyle \tan^{-1}(-1)=-\frac{\pi}{4}\)

\(\displaystyle \theta=\sin^{-1}\frac{\sqrt{3}-1}{2\sqrt{2}}\)とおくと、
\(\displaystyle \sin\theta=\frac{\sqrt{3}-1}{2\sqrt{2}}\ \ \ \left(-\frac{\pi}{2}\leqq\theta\leqq\frac{\pi}{2}\right)\)
\(\displaystyle \sin\theta=\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\)
\(\displaystyle \theta=\frac{\pi}{12}\)
よって、
\(\displaystyle \sin^{-1}\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\pi}{12}\)

二倍角の公式より
\(\displaystyle 2\tan^{-1}\frac{1}{3}\)
\(\displaystyle =\tan^{-1}\frac{2・\frac{1}{3}}{1-(\frac{1}{3})^2}\)
\(\displaystyle =\tan^{-1}\frac{\frac{2}{3}}{\frac{8}{9}}\)
\(\displaystyle =\tan^{-1}\frac{3}{4}\)
\(\displaystyle \alpha=\tan^{-1}\frac{3}{4},\beta=\tan^{-1}\frac{1}{7}\)とおくと、
\(\displaystyle \tan\alpha=\frac{3}{4}\ \ \ \left(-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\right)\)
\(\displaystyle \tan\beta=\frac{1}{7}\ \ \ \left(-\frac{\pi}{2}<\beta<\frac{\pi}{2}\right)\)
加法定理より
\(\displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}・\frac{1}{7}}=1\)
ここで、\(0<\tan\beta<\tan\alpha<1\)なので、\(\displaystyle 0<\beta<\alpha<\frac{\pi}{4}\)
\(\displaystyle \tan(\alpha+\beta)=1\ \ \ \left(0<\alpha+\beta<\frac{\pi}{2}\right)\)
\(\displaystyle \alpha+\beta=\frac{\pi}{4}\)
よって、
\(\displaystyle 2\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}=\frac{\pi}{4}\)

\(\displaystyle \alpha=\sin^{-1}\frac{3}{5},\beta=\sin^{-1}\frac{4}{5}\)とおくと、
\(\displaystyle \sin\alpha=\frac{3}{5}\ \ \ \left(-\frac{\pi}{2}\leqq\alpha\leqq\frac{\pi}{2}\right)\)
\(\displaystyle \sin\beta=\frac{4}{5}\ \ \ \left(-\frac{\pi}{2}\leqq\beta\leqq\frac{\pi}{2}\right)\)
\(\sin^2\alpha+\sin^2\beta=1\)より
\(\sin^2\alpha=1-\sin^2\beta=\cos^2\beta\)
\(\displaystyle \sin\alpha=\pm\cos\beta=\pm\sin\left(\frac{\pi}{2}-\beta\right)\)
\(\displaystyle \alpha=\frac{\pi}{2}-\beta\)
よって、
\(\displaystyle \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{4}{5}=\alpha+\beta=\frac{\pi}{2}\)

\(\displaystyle \alpha=\cos^{-1}\frac{1}{2}\)とおくと、
\(\displaystyle \cos\alpha=\frac{1}{2}\ \ \ (0\leqq\alpha\leqq\pi)\)
\(\displaystyle \alpha=\frac{\pi}{3}\)
よって、
\(\displaystyle \sin\left(\cos^{-1}\frac{1}{2}\right)=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

\(\alpha=\tan^{-1}x,\beta=\cot^{-1}x\)とおくと、
\(\displaystyle \tan\alpha=x\ \ \ \left(-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\right)\)
\(\displaystyle \cot\beta=\tan\left(\frac{\pi}{2}-\beta\right)=x\ \ \ \left(-\frac{\pi}{2}<\beta<\frac{\pi}{2}\right)\)
\(\displaystyle \alpha=\frac{\pi}{2}-\beta\)
よって、
\(\displaystyle \tan^{-1}x+\cot^{-1}x=\alpha+\beta=\frac{\pi}{2}\)

\(\displaystyle \alpha=\sin^{-1}x=\cos^{-1}\frac{4}{5}\)とおくと、
\(\displaystyle \sin\alpha=x\ \ \ \left(-\frac{\pi}{2}\leqq\alpha\leqq\frac{\pi}{2}\right)\)
\(\displaystyle \cos\alpha=\frac{4}{5}\ \ \ (0\leqq\alpha\leqq\pi)\)
\(\displaystyle 0\leqq\alpha\leqq\frac{\pi}{2}\)より、
\(\displaystyle x=\sin\alpha=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\)