Упр.3 Повторение ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

а) Так как $$\sin t=\frac{8}{17}$$ и $$t\in\left(\frac{\pi}{2};\pi\right),$$ то $$\cos t<0.$$

$$\cos t=-\sqrt{1-\sin^2 t} =-\sqrt{1-\frac{64}{289}} =-\sqrt{\frac{225}{289}} =-\frac{15}{17}.$$

$$\tg t=\frac{\sin t}{\cos t} =\frac{8/17}{-15/17} =-\frac{8}{15}, \qquad \ctg t=\frac{1}{\tg t} =-\frac{15}{8}.$$

б) Так как $$\sin t=-\frac{7}{25}$$ и $$t\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right),$$ то $$\cos t>0.$$

$$\cos t=\sqrt{1-\sin^2 t} =\sqrt{1-\frac{49}{625}} =\sqrt{\frac{576}{625}} =\frac{24}{25}.$$

$$\tg t=\frac{\sin t}{\cos t} =\frac{-7/25}{24/25} =-\frac{7}{24}, \qquad \ctg t=\frac{1}{\tg t} =-\frac{24}{7}.$$

в) Так как $$\sin t=\frac{9}{41}$$ и $$t\in\left(\frac{\pi}{2};\frac{3\pi}{2}\right),$$ то $$\cos t<0.$$

$$\cos t=-\sqrt{1-\sin^2 t} =-\sqrt{1-\frac{81}{1681}} =-\sqrt{\frac{1600}{1681}} =-\frac{40}{41}.$$

$$\tg t=\frac{\sin t}{\cos t} =\frac{9/41}{-40/41} =-\frac{9}{40}, \qquad \ctg t=\frac{1}{\tg t} =-\frac{40}{9}.$$

г) Так как $$\sin t=-\frac{35}{37}$$ и $$t\in\left(\pi;\frac{3\pi}{2}\right),$$ то $$\cos t<0.$$

$$\cos t=-\sqrt{1-\sin^2 t} =-\sqrt{1-\frac{1225}{1369}} =-\sqrt{\frac{144}{1369}} =-\frac{12}{37}.$$

$$\tg t=\frac{\sin t}{\cos t} =\frac{-35/37}{-12/37} =\frac{35}{12}, \qquad \ctg t=\frac{1}{\tg t} =\frac{12}{35}.$$

Ответ

а) $$\cos t=-\frac{15}{17},\ \tg t=-\frac{8}{15},\ \ctg t=-\frac{15}{8};$$
б) $$\cos t=\frac{24}{25},\ \tg t=-\frac{7}{24},\ \ctg t=-\frac{24}{7};$$
в) $$\cos t=-\frac{40}{41},\ \tg t=-\frac{9}{40},\ \ctg t=-\frac{40}{9};$$
г) $$\cos t=-\frac{12}{37},\ \tg t=\frac{35}{12},\ \ctg t=\frac{12}{35}.$$