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

Рассмотрим вариант решения задания из учебника Алимов, Колягин, Ткачёва 10 класс, Просвещение: 475 Вычислить:

1. $$\cos\left(-\frac{\pi}{6}\right)\sin\left(-\frac{\pi}{3}\right)+\tg\left(-\frac{\pi}{4}\right)$$

   $$\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}=\frac{\sqrt3}{2},\quad \sin\left(-\frac{\pi}{3}\right)=-\sin\frac{\pi}{3}=-\frac{\sqrt3}{2},\quad \tg\left(-\frac{\pi}{4}\right)=-1$$

   $$ \frac{\sqrt3}{2}\cdot\left(-\frac{\sqrt3}{2}\right)-1 =-\frac34-1 =-\frac74 $$

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

   $$\tg\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt3},\quad \ctg\left(-\frac{\pi}{6}\right)=-\sqrt3$$

   $$ \frac{1+\left(-\frac{1}{\sqrt3}\right)^2}{1+\left(-\sqrt3\right)^2} =\frac{1+\frac13}{1+3} =\frac{\frac43}{4} =\frac13 $$

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

   $$ 2\cdot\left(-\frac12\right)\cdot\frac{\sqrt3}{2} +\left(-\sqrt3\right) +\left(\frac{\sqrt2}{2}\right)^2 $$

   $$ -\frac{\sqrt3}{2}-\sqrt3+\frac12 =-\frac{3\sqrt3}{2}+\frac12 =\frac{1-3\sqrt3}{2} $$

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

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

   $$ -1+0-1-1=-3 $$

5. $$\frac{3-\sin^2\left(-\frac{\pi}{3}\right)-\cos^2\left(-\frac{\pi}{3}\right)}{2\cos\left(-\frac{\pi}{4}\right)}$$

   $$ \sin^2\left(-\frac{\pi}{3}\right)=\left(-\frac{\sqrt3}{2}\right)^2=\frac34,\quad \cos^2\left(-\frac{\pi}{3}\right)=\left(\frac12\right)^2=\frac14 $$

   $$ 3-\frac34-\frac14=2,\qquad 2\cos\left(-\frac{\pi}{4}\right)=2\cdot\frac{\sqrt2}{2}=\sqrt2 $$

   $$ \frac{2}{\sqrt2}=\sqrt2 $$

6. $$2\sin\left(-\frac{\pi}{6}\right)+3+7{,}5\,\tg(-\pi)+\frac18\cos\frac{3\pi}{2}$$

   $$ \sin\left(-\frac{\pi}{6}\right)=-\frac12,\quad \tg(-\pi)=0,\quad \cos\frac{3\pi}{2}=0 $$

   $$ 2\cdot\left(-\frac12\right)+3+7{,}5\cdot0+\frac18\cdot0=-1+3=2 $$

Ответ

1) $$-\frac74$$; 2) $$\frac13$$; 3) $$\frac{1-3\sqrt3}{2}$$; 4) $$-3$$; 5) $$\sqrt2$$; 6) $$2$$.