Упр.21 ГДЗ Колмогоров 10-11 класс (Алгебра)
а) Подставим $$\alpha=\frac{\pi}{4}$$:
$$3\sin\left(2\alpha-\frac{\pi}{4}\right)+2\cos(3\alpha-\pi) =3\sin\left(2\cdot\frac{\pi}{4}-\frac{\pi}{4}\right)+2\cos\left(3\cdot\frac{\pi}{4}-\pi\right)$$
$$=3\sin\frac{\pi}{4}+2\cos\left(-\frac{\pi}{4}\right) =3\cdot\frac{\sqrt2}{2}+2\cdot\frac{\sqrt2}{2} =\frac{5\sqrt2}{2}.$$
б) Подставим $$\alpha=\frac{2\pi}{3}$$:
$$\sin^2\left(\alpha-\frac{\pi}{3}\right)+3\tg\left(\frac{5\pi}{4}-\frac{3\pi}{2}\right) =\sin^2\left(\frac{2\pi}{3}-\frac{\pi}{3}\right)+3\tg\left(\frac{5\pi}{4}-\frac{3\pi}{2}\right)$$
$$=\sin^2\frac{\pi}{3}+3\tg\left(-\frac{\pi}{4}\right) =\left(\frac{\sqrt3}{2}\right)^2+3\cdot(-1) =\frac34-3=-\frac94.$$
в) Подставим $$\alpha=\frac{\pi}{6}$$:
$$4\cos\left(3\alpha-\frac{\pi}{6}\right)+\ctg\left(\alpha+\frac{\pi}{12}\right) =4\cos\left(3\cdot\frac{\pi}{6}-\frac{\pi}{6}\right)+\ctg\left(\frac{\pi}{6}+\frac{\pi}{12}\right)$$
$$=4\cos\frac{\pi}{3}+\ctg\frac{\pi}{4} =4\cdot\frac12+1=3.$$
г) Подставим $$\alpha=-\frac{\pi}{6}$$:
$$\cos\left(\alpha+\frac{\pi}{3}\right)\tg^2\left(2\alpha+\frac{\pi}{2}\right) =\cos\left(-\frac{\pi}{6}+\frac{\pi}{3}\right)\tg^2\left(2\cdot\left(-\frac{\pi}{6}\right)+\frac{\pi}{2}\right)$$
$$=\cos\frac{\pi}{6}\tg^2\frac{\pi}{6} =\frac{\sqrt3}{2}\cdot\left(\frac{1}{\sqrt3}\right)^2 =\frac{\sqrt3}{2}\cdot\frac13 =\frac{\sqrt3}{6}.$$
Ответ
а) $$\frac{5\sqrt2}{2}$$; б) $$-\frac94$$; в) $$3$$; г) $$\frac{\sqrt3}{6}$$.
