Упр.475 ГДЗ Алимов 10-11 класс (Алгебра)

Вычислить:

1)

$$\cos\left(-\frac{\pi}{6}\right)\sin\left(-\frac{\pi}{3}\right)+\tg\left(-\frac{\pi}{4}\right) = \cos\frac{\pi}{6}\cdot\left(-\sin\frac{\pi}{3}\right)-\tg\frac{\pi}{4}$$
$$=\frac{\sqrt{3}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)-1 =-\frac{3}{4}-1 =-\frac{7}{4}$$

Ответ: $$-\frac{7}{4}$$

2)

$$\frac{1+\tg^2\left(-\frac{\pi}{6}\right)}{1+\ctg^2\left(-\frac{\pi}{6}\right)} = \frac{1+\tg^2\frac{\pi}{6}}{1+\ctg^2\frac{\pi}{6}} = \frac{1+\frac{1}{3}}{1+3} = \frac{\frac{4}{3}}{4} = \frac{1}{3}$$

Ответ: $$\frac{1}{3}$$

3)

$$2\sin\left(-\frac{\pi}{6}\right)\cos\left(-\frac{\pi}{6}\right)+\tg\left(-\frac{\pi}{3}\right)+\sin^2\left(-\frac{\pi}{4}\right)$$
$$=2\cdot\left(-\frac{1}{2}\right)\cdot\frac{\sqrt{3}}{2}-\sqrt{3}+\left(\frac{\sqrt{2}}{2}\right)^2$$
$$=-\frac{\sqrt{3}}{2}-\sqrt{3}+\frac{1}{2} =\frac{1-3\sqrt{3}}{2}$$

Ответ: $$\frac{1-3\sqrt{3}}{2}$$

4)

$$\cos(-\pi)+\ctg\left(-\frac{\pi}{2}\right)-\sin\left(-\frac{3\pi}{2}\right)+\ctg\left(-\frac{\pi}{4}\right)$$
$$=\cos\pi-\ctg\frac{\pi}{2}+\sin\frac{3\pi}{2}-\ctg\frac{\pi}{4} =-1-0-1-1=-3$$

Ответ: $$-3$$

5)

$$\frac{3-\sin^2\left(-\frac{\pi}{3}\right)-\cos^2\left(-\frac{\pi}{3}\right)}{2\cos\left(-\frac{\pi}{4}\right)} = \frac{3-\sin^2\frac{\pi}{3}-\cos^2\frac{\pi}{3}}{2\cos\frac{\pi}{4}}$$
$$= \frac{3-\frac{3}{4}-\frac{1}{4}}{2\cdot\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Ответ: $$\sqrt{2}$$

6)

$$2\sin\left(-\frac{\pi}{6}\right)+3+7{,}5\,\tg(-\pi)+\frac{1}{8}\cos\frac{3\pi}{2}$$
$$=2\cdot\left(-\frac{1}{2}\right)+3+7{,}5\cdot 0+\frac{1}{8}\cdot 0 =-1+3=2$$

Ответ: $$2$$