Упр.17.4 ГДЗ Мордкович 10-11 класс (Алгебра)

1. а) $$\operatorname{arcctg}(-1)+\operatorname{arctg}(-1)$$;
   б) $$\arcsin\left(-\frac{\sqrt{2}}{2}\right)+\operatorname{arcctg}(-\sqrt{3})$$;
   в) $$\operatorname{arcctg}\left(-\frac{\sqrt{3}}{3}\right)-\operatorname{arcctg}\left(\frac{\sqrt{3}}{3}\right)$$;
   г) $$\arccos\left(-\frac{1}{2}\right)-\operatorname{arcctg}(-\sqrt{3})$$.

Воспользуемся значениями обратных тригонометрических функций:

• $$\operatorname{arcctg}(-1)=\frac{3\pi}{4}, \quad \operatorname{arctg}(-1)=-\frac{\pi}{4}$$
• $$\arcsin\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\pi}{4}, \quad \operatorname{arcctg}(-\sqrt{3})=\frac{5\pi}{6}$$
• $$\operatorname{arcctg}\left(-\frac{\sqrt{3}}{3}\right)=\frac{2\pi}{3}, \quad \operatorname{arctg}\frac{\sqrt{3}}{3}=\frac{\pi}{6}$$
• $$\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3}, \quad \operatorname{arcctg}(-\sqrt{3})=\frac{5\pi}{6}$$

Тогда:

$$\begin{aligned} \text{а)}\;& \operatorname{arcctg}(-1)+\operatorname{arctg}(-1) = \frac{3\pi}{4}-\frac{\pi}{4} = \frac{\pi}{2};\\[4pt] \text{б)}\;& \arcsin\left(-\frac{\sqrt{2}}{2}\right)+\operatorname{arcctg}(-\sqrt{3}) = -\frac{\pi}{4}+\frac{5\pi}{6} = \frac{7\pi}{12};\\[4pt] \text{в)}\;& \operatorname{arcctg}\left(-\frac{\sqrt{3}}{3}\right)-\operatorname{arctg}\frac{\sqrt{3}}{3} = \frac{2\pi}{3}-\frac{\pi}{6} = \frac{\pi}{2};\\[4pt] \text{г)}\;& \arccos\left(-\frac{1}{2}\right)-\operatorname{arcctg}(-\sqrt{3}) = \frac{2\pi}{3}-\frac{5\pi}{6} = -\frac{\pi}{6}. \end{aligned}$$

Ответ: а) $$\frac{\pi}{2}$$; б) $$\frac{7\pi}{12}$$; в) $$\frac{\pi}{2}$$; г) $$-\frac{\pi}{6}$$.