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.

Ответ: a) sin α = √2/2, tg α = 1; б) sin α = √5/3, tg α = √5/2; в) cos α = 1/2, tg α = √3; г) cos α = √15/4, tg α = √15/15

\[sin^2 α + cos^2 α = 1\]\[sin^2 α = 1 - cos^2 α = 1 - (\frac{1}{\sqrt{2}})^2 = 1 - \frac{1}{2} = \frac{1}{2}\]\[sin α = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}\]\[tg α = \frac{sin α}{cos α} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1\]\[sin^2 α + cos^2 α = 1\]\[sin^2 α = 1 - cos^2 α = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}\]\[sin α = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\]\[tg α = \frac{sin α}{cos α} = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{3} \cdot \frac{3}{2} = \frac{\sqrt{5}}{2}\]\[sin^2 α + cos^2 α = 1\]\[cos^2 α = 1 - sin^2 α = 1 - (\frac{\sqrt{3}}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4}\]\[cos α = \sqrt{\frac{1}{4}} = \frac{1}{2}\]\[tg α = \frac{sin α}{cos α} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \cdot 2 = \sqrt{3}\]\[sin^2 α + cos^2 α = 1\]\[cos^2 α = 1 - sin^2 α = 1 - (\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}\]\[cos α = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\]\[tg α = \frac{sin α}{cos α} = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = \frac{1}{4} \cdot \frac{4}{\sqrt{15}} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}\]

Ответ: a) sin α = √2/2, tg α = 1; б) sin α = √5/3, tg α = √5/2; в) cos α = 1/2, tg α = √3; г) cos α = √15/4, tg α = √15/15