\[\boxed{\mathbf{593.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Основное\ тригонометрическое\ \]
\[тождество\ :\]
\[\sin^{2}\alpha + \cos^{2}\alpha = 1.\]
\[\textbf{а)}\cos\alpha = \frac{1}{2}:\]
\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{1}{4}} =\]
\[= \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]
\[\textbf{б)}\cos\alpha = \frac{2}{3}:\]
\[\sin\alpha = \sqrt{1 - \cos^{2}\alpha} = \sqrt{1 - \frac{4}{9}} =\]
\[= \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{5}}{3} \bullet \frac{3}{2} = \frac{\sqrt{5}}{2}.\]
\[\textbf{в)}\sin\alpha = \frac{\sqrt{3}}{2}:\]
\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{3}{4}} =\]
\[= \sqrt{\frac{1}{4}} = \frac{1}{2};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{3}}{2} \bullet \frac{2}{1} = \sqrt{3}.\]
\[\textbf{г)}\sin\alpha = \frac{1}{4}:\]
\[\cos\alpha = \sqrt{1 - \sin^{2}\alpha} =\]
\[= \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4};\]
\[tg\ \alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1}{4} \bullet \frac{4}{\sqrt{15}} = \frac{1}{\sqrt{15}} =\]
\[= \frac{\sqrt{15}}{15}.\]