177. Вычислите, не применяя таблицы значения синуса и косинуса для чисел \(\frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{7\pi}{12},\ \frac{11\pi}{12}\): \(\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}\), \(\frac{5\pi}{12}=\frac{\pi}{6}+\frac{\pi}{4}\), \(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\), \(\frac{11\pi}{12}=\pi-\frac{\pi}{12}\).

Используем формулы синуса и косинуса суммы и разности, а также точные значения для углов \(\frac{\pi}{6}\), \(\frac{\pi}{4}\) и \(\frac{\pi}{3}\).

1. \(\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}\):

   \(\sin\frac{\pi}{12}=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt6-\sqrt2}{4}\).

   \(\cos\frac{\pi}{12}=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt6+\sqrt2}{4}\).

2. \(\frac{5\pi}{12}=\frac{\pi}{6}+\frac{\pi}{4}\):

   \(\sin\frac{5\pi}{12}=\sin\frac{\pi}{6}\cos\frac{\pi}{4}+\cos\frac{\pi}{6}\sin\frac{\pi}{4}=\frac{\sqrt6+\sqrt2}{4}\).

   \(\cos\frac{5\pi}{12}=\cos\frac{\pi}{6}\cos\frac{\pi}{4}-\sin\frac{\pi}{6}\sin\frac{\pi}{4}=\frac{\sqrt6-\sqrt2}{4}\).

3. \(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\):

   \(\sin\frac{7\pi}{12}=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt6+\sqrt2}{4}\).

   \(\cos\frac{7\pi}{12}=\cos\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt2-\sqrt6}{4}\).

4. \(\frac{11\pi}{12}=\pi-\frac{\pi}{12}\). По формулам приведения:

   \(\sin\frac{11\pi}{12}=\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}\), \(\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}=-\frac{\sqrt6+\sqrt2}{4}\).