699. Найдите: а) \( \sin \alpha \) и \( \tan \alpha \), если \( \cos \alpha = \frac{1}{2} \); б) \( \sin \alpha \) и \( \tan \alpha \), если \( \cos \alpha = \frac{\sqrt{2}}{2} \); в) \( \cos \alpha \) и \( \tan \alpha \), если \( \sin \alpha = \frac{\sqrt{3}}{2} \); г) \( \cos \alpha \) и \( \tan \alpha \), если \( \sin \alpha = \frac{1}{3} \).

a) Если \( \cos \alpha = \frac{1}{2} \), то \( \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). Тогда \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \). б) Если \( \cos \alpha = \frac{\sqrt{2}}{2} \), то \( \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (\frac{\sqrt{2}}{2})^2} = \sqrt{1 - \frac{2}{4}} = \sqrt{\frac{2}{4}} = \frac{\sqrt{2}}{2} \). Тогда \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 \). в) Если \( \sin \alpha = \frac{\sqrt{3}}{2} \), то \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \). Тогда \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \). г) Если \( \sin \alpha = \frac{1}{3} \), то \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{1}{3})^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \). Тогда \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \).