Найди sin a, если cos a = 1/2

1100. a) Если $\cos \alpha = \frac{1}{2}$, то $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (\frac{1}{2})^2} = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$. б) Если $\cos \alpha = -\frac{2}{3}$, то $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (-\frac{2}{3})^2} = \pm \sqrt{1 - \frac{4}{9}} = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}$. в) Если $\cos \alpha = -1$, то $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (-1)^2} = \pm \sqrt{1 - 1} = 0$. *Перевод:* 1100. a) If $\cos \alpha = \frac{1}{2}$, then $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (\frac{1}{2})^2} = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$. b) If $\cos \alpha = -\frac{2}{3}$, then $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (-\frac{2}{3})^2} = \pm \sqrt{1 - \frac{4}{9}} = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}$. c) If $\cos \alpha = -1$, then $\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha} = \pm \sqrt{1 - (-1)^2} = \pm \sqrt{1 - 1} = 0$.