Реши пример 1 под цифрой 11 и пример 1 под цифрой 12

Задание 11. 1) Если $sin \alpha = \frac{3}{5}$ и $0^\circ \le \alpha \le 90^\circ$, то $cos \alpha = \sqrt{1 - sin^2 \alpha} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. 2) Если $sin \alpha = \frac{1}{3}$ и $90^\circ < \alpha < 180^\circ$, то $cos \alpha = -\sqrt{1 - sin^2 \alpha} = -\sqrt{1 - (\frac{1}{3})^2} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$. 3) Если $sin \alpha = \frac{\sqrt{3}}{4}$, то $cos \alpha = \sqrt{1 - sin^2 \alpha} = \sqrt{1 - (\frac{\sqrt{3}}{4})^2} = \sqrt{1 - \frac{3}{16}} = \sqrt{\frac{13}{16}} = \frac{\sqrt{13}}{4}$. 4) Если $cos \alpha = -0.8$, то $sin \alpha = \sqrt{1 - cos^2 \alpha} = \sqrt{1 - (-0.8)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$. 5) Если $sin \alpha = \frac{4}{5}$ и $90^\circ < \alpha < 180^\circ$, то $tg \alpha = \frac{sin \alpha}{cos \alpha} = \frac{\frac{4}{5}}{-\sqrt{1 - (\frac{4}{5})^2}} = \frac{\frac{4}{5}}{-\sqrt{1 - \frac{16}{25}}} = \frac{\frac{4}{5}}{-\sqrt{\frac{9}{25}}} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}$. 6) Если $cos \alpha = \frac{12}{13}$ и $0^\circ < \alpha < 90^\circ$, то $ctg \alpha = \frac{cos \alpha}{sin \alpha} = \frac{\frac{12}{13}}{\sqrt{1 - (\frac{12}{13})^2}} = \frac{\frac{12}{13}}{\sqrt{1 - \frac{144}{169}}} = \frac{\frac{12}{13}}{\sqrt{\frac{25}{169}}} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}$. Задание 12. 1) Если $sin \alpha = \frac{5}{13}$, то $cos \alpha = \sqrt{1 - sin^2 \alpha} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. 3) Если $sin \alpha = \frac{5}{13}$, то $tg \alpha = \frac{sin \alpha}{cos \alpha} = \frac{\frac{5}{13}}{\sqrt{1 - (\frac{5}{13})^2}} = \frac{\frac{5}{13}}{\sqrt{1 - \frac{25}{169}}} = \frac{\frac{5}{13}}{\sqrt{\frac{144}{169}}} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$.