21

Краткий ответ:

а) \(\displaystyle \frac{5\sqrt2}{2}\);

б) \(\displaystyle \frac{15}{4}\);

в) \(3\);

г) \(\displaystyle \frac{\sqrt3}{6}\).

Сначала подставляем значение \(\alpha\), затем приводим углы к табличным.

а)

\[ 3\sin\left(2\alpha-\frac{\pi}{4}\right)+ 2\cos(3\alpha-\pi),\qquad \alpha=\frac{\pi}{4}. \]

\[ 2\alpha-\frac{\pi}{4} =2\cdot\frac{\pi}{4}-\frac{\pi}{4} =\frac{\pi}{4}, \]

\[ 3\alpha-\pi =3\cdot\frac{\pi}{4}-\pi =-\frac{\pi}{4}. \]

Тогда \[ 3\sin\frac{\pi}{4}+2\cos\left(-\frac{\pi}{4}\right) =3\cdot\frac{\sqrt2}{2} +2\cdot\frac{\sqrt2}{2} =\frac{5\sqrt2}{2}. \]

б)

Во второй дроби читаем \(\frac{3\alpha}{2}\):

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

\[ \alpha-\frac{\pi}{3} =\frac{2\pi}{3}-\frac{\pi}{3} =\frac{\pi}{3}, \]

\[ \frac{5\pi}{4}-\frac{3\alpha}{2} =\frac{5\pi}{4}-\frac{3}{2}\cdot\frac{2\pi}{3} =\frac{5\pi}{4}-\pi =\frac{\pi}{4}. \]

\[ \sin^2\frac{\pi}{3}+3\tg\frac{\pi}{4} =\left(\frac{\sqrt3}{2}\right)^2+3\cdot1 =\frac34+3 =\frac{15}{4}. \]

в)

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

\[ 3\alpha-\frac{\pi}{6} =3\cdot\frac{\pi}{6}-\frac{\pi}{6} =\frac{\pi}{3}, \]

\[ \alpha+\frac{\pi}{12} =\frac{\pi}{6}+\frac{\pi}{12} =\frac{\pi}{4}. \]

\[ 4\cos\frac{\pi}{3}+\ctg\frac{\pi}{4} =4\cdot\frac12+1 =3. \]

г)

\[ \cos\left(\alpha+\frac{\pi}{3}\right) \tg^2\left(2\alpha+\frac{\pi}{2}\right), \qquad \alpha=-\frac{\pi}{6}. \]

\[ \alpha+\frac{\pi}{3} =-\frac{\pi}{6}+\frac{2\pi}{6} =\frac{\pi}{6}, \]

\[ 2\alpha+\frac{\pi}{2} =-\frac{\pi}{3}+\frac{\pi}{2} =\frac{\pi}{6}. \]

\[ \cos\frac{\pi}{6}\cdot\tg^2\frac{\pi}{6} =\frac{\sqrt3}{2}\cdot \left(\frac1{\sqrt3}\right)^2 =\frac{\sqrt3}{2}\cdot\frac13 =\frac{\sqrt3}{6}. \]

Проверка: все полученные углы — табличные, а \(\tg\) и \(\ctg\) определены.