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 = \(\frac{\sqrt{3}}{2}\); г) cos α и tg α, если sin α = 1/4.

Решение: а) cos α = 1/2 Так как \(\sin^2 α + cos^2 α = 1\), то \[\sin^2 α = 1 - cos^2 α = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}\] \[\sin α = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\] \[\tan α = \frac{\sin α}{cos α} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\] б) cos α = 2/3 \[\sin^2 α = 1 - cos^2 α = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}\] \[\sin α = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\] \[\tan α = \frac{\sin α}{cos α} = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{2}\] в) sin α = \(\frac{\sqrt{3}}{2}\) \[\cos^2 α = 1 - sin^2 α = 1 - (\frac{\sqrt{3}}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4}\] \[\cos α = \sqrt{\frac{1}{4}} = \frac{1}{2}\] \[\tan α = \frac{\sin α}{cos α} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\] г) sin α = 1/4 \[\cos^2 α = 1 - sin^2 α = 1 - (\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}\] \[\cos α = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\] \[\tan α = \frac{\sin α}{cos α} = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}\]