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Решите задачу:

$sin^2 \alpha +cos^2 \alpha =1\\cos^2 \alpha =1-sin^2 \alpha \\cos \alpha =б\sqrt{1-sin^2 \alpha }\\\\1.sin \alpha =\frac{\sqrt{91}}{10}, \ a\in(0;\frac{\pi}2)\\cos \alpha =б\sqrt{1-\frac{91}{100}}=б\sqrt{\frac{100-91}{100}}=б\sqrt{\frac{9}{100}}=б\frac{3}{10}=\frac{3}{10}(a\in(0;\frac{\pi}2))\\\\2.sin \alpha =-\frac{\sqrt{51}}{10}, \ a\in(\frac{3\pi}2;2\pi)\\cos \alpha =б\sqrt{1-\frac{51}{100}}=б\sqrt{\frac{100-51}{100}}=б\sqrt{\frac{49}{100}}=б\frac{7}{10}=\frac{7}{10}(a\in(\frac{3\pi}2;2\pi))$