1015 Найдите tg а, если: a) cos α = 1; б) cos α = √3 2 ; в) sina = √2 2 . г) sin α = 3 5 и 90°< α <180°.

a) cos α = 1

$$\sin^2 α + \cos^2 α = 1$$

$$\sin α = \pm \sqrt{1 - \cos^2 α}$$

$$\sin α = \pm \sqrt{1 - 1^2} = 0$$

$$tg α = \frac{\sin α}{\cos α} = \frac{0}{1} = 0$$

б) cos α = $$\frac{\sqrt{3}}{2}$$

$$\sin α = \pm \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \pm \sqrt{1 - \frac{3}{4}} = \pm \frac{1}{2}$$

$$tg α = \frac{\sin α}{\cos α} = \frac{\pm \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$$

в) sin α = $$\frac{\sqrt{2}}{2}$$

$$\cos α = \pm \sqrt{1 - (\frac{\sqrt{2}}{2})^2} = \pm \sqrt{1 - \frac{2}{4}} = \pm \sqrt{\frac{2}{4}} = \pm \frac{\sqrt{2}}{2}$$

$$tg α = \frac{\sin α}{\cos α} = \frac{\frac{\sqrt{2}}{2}}{\pm \frac{\sqrt{2}}{2}} = \pm 1$$

г) sin α = $$\frac{3}{5}$$ и $$90°< α <180°$$

$$\cos α = -\sqrt{1 - (\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$

$$tg α = \frac{\sin α}{\cos α} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$

Ответ: a) $$0$$; б) $$\pm \frac{\sqrt{3}}{3}$$; в) $$\pm 1$$; г) $$- \frac{3}{4}$$