Страница 138 номер 459, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
По значению одной из тригонометрических функций (\(\sin \alpha\), \(\cos \alpha\), \(\operatorname{tg} \alpha\), \(\operatorname{ctg} \alpha\)) найти значения остальных трёх: 1) \(\cos \alpha = \frac{5}{13}\) и \(\frac{3\pi}{2} < \alpha < 2\pi\); 2) \(\sin \alpha = 0{,}8\) и \(\frac{\pi}{2} < \alpha < \pi\); 3) \(\operatorname{tg} \alpha = \frac{15}{8}\) и \(\pi < \alpha < \frac{3\pi}{2}\); 4) \(\operatorname{ctg} \alpha = -3\) и \(\frac{3\pi}{2} < \alpha < 2\pi\); 5) \(\cos \alpha = 0{,}8\) и \(0 < \alpha < \frac{\pi}{2}\); 6) \(\sin \alpha = -\frac{5}{13}\) и \(\frac{3\pi}{2} < \alpha < 2\pi\); 7) \(\operatorname{tg} \alpha = -2{,}4\) и \(\frac{\pi}{2} < \alpha < \pi\); 8) \(\operatorname{ctg} \alpha = \frac{7}{24}\) и \(\pi < \alpha < \frac{3\pi}{2}\).
1) \(\cos \alpha = \frac{5}{13}\) и \(\frac{3\pi}{2} < \alpha < 2\pi\)
Четвёртая четверть.
\(\sin \alpha = -\sqrt{1 - \dfrac{25}{169}} = -\dfrac{12}{13};\)
\(\operatorname{tg} \alpha = -\dfrac{12}{13} \cdot \dfrac{13}{5} = -\dfrac{12}{5};\)
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = -\dfrac{5}{12};\)
2) \(\sin \alpha = 0{,}8\) и \(\frac{\pi}{2} < \alpha < \pi\)
Вторая четверть.
\(\cos \alpha = -\sqrt{1 - 0{,}64} = -0{,}6;\)
\(\operatorname{tg} \alpha = -\dfrac{0{,}8}{0{,}6} = -\dfrac{4}{3};\)
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = -\dfrac{3}{4};\)
3) \(\operatorname{tg} \alpha = \frac{15}{8}\) и \(\pi < \alpha < \frac{3\pi}{2}\)
Третья четверть.
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = \dfrac{8}{15};\)
\(\cos \alpha = -\sqrt{\dfrac{1}{1 + \operatorname{tg}^{2} \alpha}} = -\sqrt{\dfrac{1}{1 + \dfrac{225}{64}}} = -\sqrt{\dfrac{64}{289}} = -\dfrac{8}{17};\)
\(\sin \alpha = -\sqrt{1 - \dfrac{64}{289}} = -\sqrt{\dfrac{225}{289}} = -\dfrac{15}{17};\)
4) \(\operatorname{ctg} \alpha = -3\) и \(\frac{3\pi}{2} < \alpha < 2\pi\)
Четвёртая четверть.
\(\operatorname{tg} \alpha = \dfrac{1}{\operatorname{ctg} \alpha} = -\dfrac{1}{3};\)
\(\cos \alpha = \sqrt{\dfrac{1}{1 + \operatorname{tg}^{2} \alpha}} = \sqrt{\dfrac{1}{1 + \dfrac{1}{9}}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}};\)
\(\sin \alpha = -\sqrt{1 - \dfrac{9}{10}} = -\dfrac{1}{\sqrt{10}};\)
5) \(\cos \alpha = 0{,}8\) и \(0 < \alpha < \frac{\pi}{2}\)
Первая четверть.
\(\sin \alpha = \sqrt{1 - 0{,}64} = 0{,}6;\)
\(\operatorname{tg} \alpha = \dfrac{0{,}6}{0{,}8} = \dfrac{3}{4};\)
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = \dfrac{4}{3};\)
6) \(\sin \alpha = -\frac{5}{13}\) и \(\frac{3\pi}{2} < \alpha < 2\pi\)
Четвёртая четверть.
\(\cos \alpha = \sqrt{1 - \dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13};\)
\(\operatorname{tg} \alpha = -\dfrac{5}{13} \cdot \dfrac{13}{12} = -\dfrac{5}{12};\)
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = -\dfrac{12}{5};\)
7) \(\operatorname{tg} \alpha = -2{,}4\) и \(\frac{\pi}{2} < \alpha < \pi\)
Вторая четверть.
\(\cos \alpha = -\sqrt{\dfrac{1}{1 + \operatorname{tg}^{2} \alpha}} = -\sqrt{\dfrac{1}{1 + \dfrac{576}{100}}} = -\dfrac{10}{26} = -\dfrac{5}{13};\)
\(\sin \alpha = \sqrt{1 - \dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13};\)
\(\operatorname{ctg} \alpha = \dfrac{1}{\operatorname{tg} \alpha} = -1 : \dfrac{24}{10} = -\dfrac{5}{12};\)
8) \(\operatorname{ctg} \alpha = \frac{7}{24}\) и \(\pi < \alpha < \frac{3\pi}{2}\)
Третья четверть.
\(\operatorname{tg} \alpha = \dfrac{1}{\operatorname{ctg} \alpha} = \dfrac{24}{7};\)
\(\cos \alpha = -\sqrt{\dfrac{1}{1 + \operatorname{tg}^{2} \alpha}} = -\sqrt{\dfrac{1}{1 + \dfrac{576}{49}}} = -\sqrt{\dfrac{49}{625}} = -\dfrac{7}{25};\)
\(\sin \alpha = -\sqrt{1 - \dfrac{49}{625}} = -\sqrt{\dfrac{576}{625}} = -\dfrac{24}{25}.\)
