【高校数学Ⅰ】3-1-2 三角比の関係｜問題集

\(=\cos 20^{\circ}\)

\(=\sin 15^{\circ}\)

\(\displaystyle =\frac{1}{\tan 32^{\circ}}\)

\(=\cos 10^{\circ}\)

\(=\sin 40^{\circ}\)

\(\displaystyle =\frac{1}{\tan 26^{\circ}}\)

\(\sin^2\theta=1-\cos^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(\frac{2}{7}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{45}{49}\)
\(\sin\theta>0\)より、
\(\displaystyle \sin\theta=\frac{3\sqrt{5}}{7}\)

\(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{3\sqrt{5}}{7}\div\frac{2}{7}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{3\sqrt{5}}{2}\)

\(\displaystyle \cos^2\theta=\frac{1}{1+\tan^2\theta}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{1+(\frac{1}{2})^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{4}{5}\)
\(\cos\theta>0\)より、
\(\displaystyle \cos\theta=\frac{2\sqrt{5}}{5}\)

\(\displaystyle \sin\theta=\cos\theta\tan\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{2\sqrt{5}}{5}\times\frac{1}{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{\sqrt{5}}{5}\)

\(\sin^2\theta=1-\cos^2\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =1-\left(\frac{1}{3}\right)^2\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{8}{9}\)
\(\sin\theta>0\)より、
\(\displaystyle \sin\theta=\frac{2\sqrt{2}}{3}\)

\(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{2\sqrt{2}}{3}\div\frac{1}{3}\)
\(\ \ \ \ \ \ \ \ =2\sqrt{2}\)

\(\displaystyle \cos^2\theta=\frac{1}{1+\tan^2\theta}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{1+(\frac{1}{7})^2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ =\frac{49}{50}\)
\(\cos\theta>0\)より、
\(\displaystyle \cos\theta=\frac{7\sqrt{2}}{10}\)

\(\displaystyle \sin\theta=\cos\theta\tan\theta\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{7\sqrt{2}}{10}\times\frac{1}{7}\)
\(\displaystyle \ \ \ \ \ \ \ \ =\frac{\sqrt{2}}{10}\)