三角比の相互関係 基本ドリル

$\cos\theta = \dfrac{4}{5}$、$\tan\theta = \dfrac{3}{4}$

$\sin^2\theta + \cos^2\theta = 1$より、$\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$。$\theta$は鋭角で$\cos\theta > 0$なので、$\cos\theta = \dfrac{4}{5}$。また、$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac{3}{4}$。

$\sin\theta = \dfrac{12}{13}$、$\tan\theta = \dfrac{12}{5}$

$\sin^2\theta + \cos^2\theta = 1$より、$\sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$。$\theta$は鋭角で$\sin\theta > 0$なので、$\sin\theta = \dfrac{12}{13}$。また、$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{12/13}{5/13} = \dfrac{12}{5}$。

$\sin\theta = \dfrac{2\sqrt{5}}{5}$、$\cos\theta = \dfrac{\sqrt{5}}{5}$

$1 + \tan^2\theta = \dfrac{1}{\cos^2\theta}$より、$\dfrac{1}{\cos^2\theta} = 1 + 4 = 5$なので、$\cos^2\theta = \dfrac{1}{5}$。$\theta$は鋭角で$\cos\theta > 0$なので、$\cos\theta = \dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$。$\sin\theta = \tan\theta \times \cos\theta = 2 \times \dfrac{\sqrt{5}}{5} = \dfrac{2\sqrt{5}}{5}$。

$\cos\theta = \dfrac{2\sqrt{2}}{3}$、$\tan\theta = \dfrac{\sqrt{2}}{4}$

$\sin^2\theta + \cos^2\theta = 1$より、$\cos^2\theta = 1 - \dfrac{1}{9} = \dfrac{8}{9}$。$\theta$は鋭角で$\cos\theta > 0$なので、$\cos\theta = \dfrac{2\sqrt{2}}{3}$。また、$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{1/3}{2\sqrt{2}/3} = \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{2}}{4}$。

$\sin\theta = \dfrac{\sqrt{5}}{3}$、$\tan\theta = \dfrac{\sqrt{5}}{2}$

$\sin^2\theta + \cos^2\theta = 1$より、$\sin^2\theta = 1 - \dfrac{4}{9} = \dfrac{5}{9}$。$\theta$は鋭角で$\sin\theta > 0$なので、$\sin\theta = \dfrac{\sqrt{5}}{3}$。また、$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\sqrt{5}/3}{2/3} = \dfrac{\sqrt{5}}{2}$。