Упр.1268 ГДЗ Алимов 10-11 класс (Алгебра)
(0 < a < пи/2): 1)cosa=0,8; 2) sina=5/13; 3) tga=2,4; 4) ctga= 7/24.
$$\cos \alpha = 0{,}8 = \frac{4}{5}$$
Так как $$0<\alpha<\frac{\pi}{2}$$, то $$\sin \alpha>0$$:
$$ \sin \alpha=\sqrt{1-\cos^2\alpha} =\sqrt{1-\left(\frac45\right)^2} =\sqrt{1-\frac{16}{25}} =\sqrt{\frac{9}{25}} =\frac35 $$
$$ \tg \alpha=\frac{\sin \alpha}{\cos \alpha} =\frac{3/5}{4/5} =\frac34 $$
$$ \ctg \alpha=\frac{1}{\tg \alpha} =\frac43 $$
$$\sin \alpha=\frac{5}{13}$$
Так как $$0<\alpha<\frac{\pi}{2}$$, то $$\cos \alpha>0$$:
$$ \cos \alpha=\sqrt{1-\sin^2\alpha} =\sqrt{1-\left(\frac{5}{13}\right)^2} =\sqrt{1-\frac{25}{169}} =\sqrt{\frac{144}{169}} =\frac{12}{13} $$
$$ \tg \alpha=\frac{\sin \alpha}{\cos \alpha} =\frac{5/13}{12/13} =\frac{5}{12} $$
$$ \ctg \alpha=\frac{1}{\tg \alpha} =\frac{12}{5} $$
$$\tg \alpha=2{,}4=\frac{12}{5}$$
Используем формулу:
$$ 1+\tg^2\alpha=\frac{1}{\cos^2\alpha} $$
Тогда
$$ \cos \alpha=\frac{1}{\sqrt{1+\tg^2\alpha}} =\frac{1}{\sqrt{1+\left(\frac{12}{5}\right)^2}} =\frac{1}{\sqrt{1+\frac{144}{25}}} =\frac{1}{\sqrt{\frac{169}{25}}} =\frac{5}{13} $$
$$ \sin \alpha=\sqrt{1-\cos^2\alpha} =\sqrt{1-\left(\frac{5}{13}\right)^2} =\frac{12}{13} $$
$$ \ctg \alpha=\frac{1}{\tg \alpha} =\frac{5}{12} $$
$$\ctg \alpha=\frac{7}{24}$$
Используем формулу:
$$ 1+\ctg^2\alpha=\frac{1}{\sin^2\alpha} $$
Тогда
$$ \sin \alpha=\frac{1}{\sqrt{1+\ctg^2\alpha}} =\frac{1}{\sqrt{1+\left(\frac{7}{24}\right)^2}} =\frac{1}{\sqrt{1+\frac{49}{576}}} =\frac{1}{\sqrt{\frac{625}{576}}} =\frac{24}{25} $$
$$ \cos \alpha=\sqrt{1-\sin^2\alpha} =\sqrt{1-\left(\frac{24}{25}\right)^2} =\frac{7}{25} $$
$$ \tg \alpha=\frac{1}{\ctg \alpha} =\frac{24}{7} $$
Ответ
1) $$\sin \alpha=\frac35,\ \tg \alpha=\frac34,\ \ctg \alpha=\frac43$$;
2) $$\cos \alpha=\frac{12}{13},\ \tg \alpha=\frac{5}{12},\ \ctg \alpha=\frac{12}{5}$$;
3) $$\cos \alpha=\frac{5}{13},\ \sin \alpha=\frac{12}{13},\ \ctg \alpha=\frac{5}{12}$$;
4) $$\sin \alpha=\frac{24}{25},\ \cos \alpha=\frac{7}{25},\ \tg \alpha=\frac{24}{7}$$.
