Упр.21 ГДЗ Колмогоров 10-11 класс (Алгебра)

Подставим заданные значения угла и вычислим выражения.

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

   $$3\sin\left(2\cdot\frac{\pi}{4}-\frac{\pi}{4}\right)+2\cos\left(3\cdot\frac{\pi}{4}-\pi\right) =3\sin\frac{\pi}{4}+2\cos\left(-\frac{\pi}{4}\right)$$
   $$=3\cdot\frac{\sqrt{2}}{2}+2\cdot\frac{\sqrt{2}}{2} =\frac{5\sqrt{2}}{2}.$$

2. $$\sin^2\left(\alpha-\frac{\pi}{3}\right)+3\tg\left(\frac{5\pi}{4}-\frac{3\pi}{2}\right), \quad \alpha=\frac{2\pi}{3}$$

   $$\sin^2\left(\frac{2\pi}{3}-\frac{\pi}{3}\right)+3\tg\left(\frac{5\pi}{4}-\frac{3\pi}{2}\right) =\sin^2\frac{\pi}{3}+3\tg\left(-\frac{\pi}{4}\right)$$
   $$=\left(\frac{\sqrt{3}}{2}\right)^2+3\cdot(-1) =\frac{3}{4}-3 =-\frac{9}{4}.$$

3. $$4\cos\left(3\alpha-\frac{\pi}{6}\right)+\ctg\left(\alpha+\frac{\pi}{12}\right), \quad \alpha=\frac{\pi}{6}$$

   $$4\cos\left(3\cdot\frac{\pi}{6}-\frac{\pi}{6}\right)+\ctg\left(\frac{\pi}{6}+\frac{\pi}{12}\right) =4\cos\frac{\pi}{3}+\ctg\frac{\pi}{4}$$
   $$=4\cdot\frac{1}{2}+1=3.$$

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

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

Ответ: а) $$\frac{5\sqrt{2}}{2}$$; б) $$-\frac{9}{4}$$; в) $$3$$; г) $$\frac{\sqrt{3}}{6}$$.