Упр.17.15 ГДЗ Мордкович 10-11 класс (Алгебра)

а) sin (arctg 3/4);
б) cos (arcctg 12/5);
в) sin (arcctg (-4/3));
г) cos (arctg (-5/12)).

Используем формулы:

$$\sin(\arctg x)=\frac{x}{\sqrt{1+x^2}}, \qquad \cos(\arctg x)=\frac{1}{\sqrt{1+x^2}}.$$

Для арккотангенса:

$$\sin(\arcctg x)=\frac{1}{\sqrt{1+x^2}}, \qquad \cos(\arcctg x)=\frac{x}{\sqrt{1+x^2}}.$$

1. $$\sin\left(\arctg \frac{3}{4}\right)=\frac{\frac{3}{4}}{\sqrt{1+\left(\frac{3}{4}\right)^2}}=\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{16}}}=\frac{\frac{3}{4}}{\sqrt{\frac{25}{16}}}=\frac{3}{5}.$$

2. $$\cos\left(\arcctg \frac{12}{5}\right)=\frac{\frac{12}{5}}{\sqrt{1+\left(\frac{12}{5}\right)^2}}=\frac{\frac{12}{5}}{\sqrt{1+\frac{144}{25}}}=\frac{\frac{12}{5}}{\sqrt{\frac{169}{25}}}=\frac{12}{13}.$$

3. $$\sin\left(\arcctg\left(-\frac{4}{3}\right)\right)=\frac{1}{\sqrt{1+\left(-\frac{4}{3}\right)^2}}=\frac{1}{\sqrt{1+\frac{16}{9}}}=\frac{1}{\sqrt{\frac{25}{9}}}=\frac{3}{5}.$$

4. $$\cos\left(\arctg\left(-\frac{5}{12}\right)\right)=\frac{1}{\sqrt{1+\left(-\frac{5}{12}\right)^2}}=\frac{1}{\sqrt{1+\frac{25}{144}}}=\frac{1}{\sqrt{\frac{169}{144}}}=\frac{12}{13}.$$

Ответ

$$\frac{3}{5}, \ \frac{12}{13}, \ \frac{3}{5}, \ \frac{12}{13}.$$