Упр.5.7 ГДЗ Мордкович Семенов 10 класс (Алгебра)

Задача

а) $$\tg\left(-\frac{2\pi}{3}\right)-\ctg\left(\frac{2\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)$$

б) $$\tg\left(-\frac{\pi}{4}\right)+\sin^2\left(\frac{\pi}{3}\right)+\ctg\left(\frac{5\pi}{4}\right)$$

в) $$2\tg\left(\frac{\pi}{6}\right)+\sin\left(\frac{\pi}{3}\right)+2\ctg\left(\frac{2\pi}{3}\right)$$

г) $$\tg^2\left(-\frac{\pi}{4}\right)+\cos\left(\frac{2\pi}{3}\right)+\ctg\left(\frac{7\pi}{4}\right)$$

Подробный ответ

а) $$\tg\left(-\frac{2\pi}{3}\right)-\ctg\left(\frac{2\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)$$

$$\tg\left(-\frac{2\pi}{3}\right)=\sqrt{3}, \qquad \ctg\left(\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{3}, \qquad \cos\left(\frac{\pi}{3}\right)=\frac12$$

$$\sqrt{3}-\left(-\frac{\sqrt{3}}{3}\right)+\frac12=\frac{4\sqrt{3}}{3}+\frac12$$

б) $$\tg\left(-\frac{\pi}{4}\right)+\sin^2\left(\frac{\pi}{3}\right)+\ctg\left(\frac{5\pi}{4}\right)$$

$$\tg\left(-\frac{\pi}{4}\right)=-1, \qquad \sin^2\left(\frac{\pi}{3}\right)=\left(\frac{\sqrt{3}}{2}\right)^2=\frac34, \qquad \ctg\left(\frac{5\pi}{4}\right)=1$$

$$-1+\frac34+1=\frac34$$

в) $$2\tg\left(\frac{\pi}{6}\right)+\sin\left(\frac{\pi}{3}\right)+2\ctg\left(\frac{2\pi}{3}\right)$$

$$2\tg\left(\frac{\pi}{6}\right)=2\cdot\frac{1}{\sqrt{3}}=\frac{2\sqrt{3}}{3}, \qquad \sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, \qquad 2\ctg\left(\frac{2\pi}{3}\right)=2\cdot\left(-\frac{1}{\sqrt{3}}\right)=-\frac{2\sqrt{3}}{3}$$

$$\frac{2\sqrt{3}}{3}+\frac{\sqrt{3}}{2}-\frac{2\sqrt{3}}{3}=\frac{\sqrt{3}}{2}$$

г) $$\tg^2\left(-\frac{\pi}{4}\right)+\cos\left(\frac{2\pi}{3}\right)+\ctg\left(\frac{7\pi}{4}\right)$$

$$\tg^2\left(-\frac{\pi}{4}\right)=(-1)^2=1, \qquad \cos\left(\frac{2\pi}{3}\right)=-\frac12, \qquad \ctg\left(\frac{7\pi}{4}\right)=-1$$

$$1-\frac12-1=-\frac12$$