699. Найдите: a) sin α и tg α, если cos α = 1/2 б) sin α и tg α, если cos α = 2/3 в) cos α и tg α, если sin α = √3/2 г) cos α и tg α, если sin α = 1/4

a) Если \(cos \alpha = \frac{1}{2}\), то \(\alpha = 60^\circ\). \(sin \alpha = sin 60^\circ = \frac{\sqrt{3}}{2}\) \(tan \alpha = tan 60^\circ = \sqrt{3}\) б) Если \(cos \alpha = \frac{2}{3}\), то \(sin^2 \alpha + cos^2 \alpha = 1\). \(sin \alpha = \sqrt{1 - cos^2 \alpha} = \sqrt{1 - (\frac{2}{3})^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\) \(tan \alpha = \frac{sin \alpha}{cos \alpha} = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{2}\) в) Если \(sin \alpha = \frac{\sqrt{3}}{2}\), то \(\alpha = 60^\circ\). \(cos \alpha = cos 60^\circ = \frac{1}{2}\) \(tan \alpha = tan 60^\circ = \sqrt{3}\) г) Если \(sin \alpha = \frac{1}{4}\), то \(cos^2 \alpha + sin^2 \alpha = 1\). \(cos \alpha = \sqrt{1 - sin^2 \alpha} = \sqrt{1 - (\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\) \(tan \alpha = \frac{sin \alpha}{cos \alpha} = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}\)