1102 Найдите tg α, если: а) cos α = 1; б) cos α = -\(\frac{\sqrt{3}}{2}\); в) sin α = \(\frac{\sqrt{2}}{2}\) и 0° < α < 90°; г) sin α = \(\frac{3}{5}\) и 90° < α < 180°.

Используем формулу \(\tan α = \frac{\sin α}{\cos α}\).

а) Если \(\cos α = 1\), то \(\sin α = \sqrt{1 - \cos^2 α} = \sqrt{1 - 1^2} = 0\). Тогда \(\tan α = \frac{0}{1} = 0\).

б) Если \(\cos α = -\frac{\sqrt{3}}{2}\), то \(\sin α = \pm \sqrt{1 - \cos^2 α} = \pm \sqrt{1 - (-\frac{\sqrt{3}}{2})^2} = \pm \sqrt{1 - \frac{3}{4}} = \pm \frac{1}{2}\). Тогда \(\tan α = \frac{\pm \frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \mp \frac{1}{\sqrt{3}} = \mp \frac{\sqrt{3}}{3}\).

в) Если \(\sin α = \frac{\sqrt{2}}{2}\) и 0° < α < 90°, то \(\cos α = \sqrt{1 - \sin^2 α} = \sqrt{1 - (\frac{\sqrt{2}}{2})^2} = \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2}\). Тогда \(\tan α = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1\).

г) Если \(\sin α = \frac{3}{5}\) и 90° < α < 180°, то \(\cos α = -\sqrt{1 - \sin^2 α} = -\sqrt{1 - (\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}\). Тогда \(\tan α = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}\).

Ответ: а) 0; б) \(\mp \frac{\sqrt{3}}{3}\); в) 1; г) -\(\frac{3}{4}\).