3. Вычислите sin (13π/12) + sin (7π/12) - cos (5π/12) + cos (π/12).

Для вычисления этого выражения, воспользуемся формулами суммы синусов и разности косинусов. $\sin a + \sin b = 2 \sin(\frac{a+b}{2}) \cos(\frac{a-b}{2})$ и $\cos a - \cos b = -2 \sin(\frac{a+b}{2}) \sin(\frac{a-b}{2})$. Преобразуем выражение: $\sin(\frac{13\pi}{12}) + \sin(\frac{7\pi}{12}) - \cos(\frac{5\pi}{12}) + \cos(\frac{\pi}{12}) = [\sin(\frac{13\pi}{12}) + \sin(\frac{7\pi}{12})] + [\cos(\frac{\pi}{12}) - \cos(\frac{5\pi}{12})]$. Применим формулы: $[2 \sin(\frac{\frac{13\pi}{12} + \frac{7\pi}{12}}{2}) \cos(\frac{\frac{13\pi}{12} - \frac{7\pi}{12}}{2})] + [-2 \sin(\frac{\frac{\pi}{12} + \frac{5\pi}{12}}{2}) \sin(\frac{\frac{\pi}{12} - \frac{5\pi}{12}}{2})] = [2 \sin(\frac{20\pi}{24}) \cos(\frac{6\pi}{24})] + [-2 \sin(\frac{6\pi}{24}) \sin(\frac{-4\pi}{24})] = 2 \sin(\frac{5\pi}{6}) \cos(\frac{\pi}{4}) + 2 \sin(\frac{\pi}{4}) \sin(\frac{\pi}{6}) = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$.