Вычислите: а) cos(-3π/4) + cos(π/4) + cos(π/6)sin(π/6) + cos(0)sin(π/2)

Для решения этих примеров воспользуемся значениями тригонометрических функций для основных углов и формулами приведения. а) $$\cos\left(-\frac{3\pi}{4}\right) + \cos\frac{\pi}{4} + \cos\frac{\pi}{6} \cdot \sin\frac{\pi}{6} + \cos 0 \cdot \sin\frac{\pi}{2}$$ **Ответ: $$\frac{\sqrt{3}}{4} + 1$$** 1. $$\cos\left(-\frac{3\pi}{4}\right) = \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ 2. $$\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ 3. $$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin\frac{\pi}{6} = \frac{1}{2}$$ 4. $$\cos 0 = 1, \sin\frac{\pi}{2} = 1$$ Вычисляем: $$-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} + 1 \cdot 1 = 0 + \frac{\sqrt{3}}{4} + 1 = \frac{\sqrt{3}}{4} + 1$$ б) $$\sin\left(-\frac{2\pi}{3}\right) - \cos\frac{3\pi}{4} + \cos\frac{\pi}{3} \cdot \sin\frac{\pi}{6} + \cos\frac{\pi}{2} \cdot \sin\frac{\pi}{2}$$ **Ответ: $$\frac{\sqrt{2} - 2\sqrt{3} + 1}{4}$$** 1. $$\sin\left(-\frac{2\pi}{3}\right) = -\sin\frac{2\pi}{3} = -\frac{\sqrt{3}}{2}$$ 2. $$\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ 3. $$\cos\frac{\pi}{3} = \frac{1}{2}, \sin\frac{\pi}{6} = \frac{1}{2}$$ 4. $$\cos\frac{\pi}{2} = 0, \sin\frac{\pi}{2} = 1$$ Вычисляем: $$-\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{2}}{2}\right) + \frac{1}{2} \cdot \frac{1}{2} + 0 \cdot 1 = -\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} + \frac{1}{4} = \frac{-2\sqrt{3} + 2\sqrt{2} + 1}{4}$$ в) $$\cos\left(-\frac{\pi}{4}\right) + \sin^2\frac{\pi}{3} + \sin\frac{\pi}{6} \cdot \sin\frac{\pi}{2} - \cos\frac{5\pi}{4}$$ **Ответ: $$\sqrt{2} + 1,25$$** 1. $$\cos\left(-\frac{\pi}{4}\right) = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ 2. $$\sin^2\frac{\pi}{3} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$ 3. $$\sin\frac{\pi}{6} = \frac{1}{2}, \sin\frac{\pi}{2} = 1$$ 4. $$\cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$ Вычисляем: $$\frac{\sqrt{2}}{2} + \frac{3}{4} + \frac{1}{2} \cdot 1 - \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} + \frac{3}{4} + \frac{1}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} + \frac{5}{4} = \sqrt{2} + 1,25$$ г) $$\sin\frac{5\pi}{6} + \cos\frac{3\pi}{4} + \cos\frac{\pi}{4} \cdot \sin\frac{3\pi}{4} + \cos\frac{\pi}{2} \cdot \sin\pi$$ **Ответ: $$1 - \frac{\sqrt{2}}{2}$$** 1. $$\sin\frac{5\pi}{6} = \frac{1}{2}$$ 2. $$\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ 3. $$\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$$ 4. $$\cos\frac{\pi}{2} = 0, \sin\pi = 0$$ Вычисляем: $$\frac{1}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} + 0 \cdot 0 = \frac{1}{2} - \frac{\sqrt{2}}{2} + \frac{2}{4} = \frac{1}{2} - \frac{\sqrt{2}}{2} + \frac{1}{2} = 1 - \frac{\sqrt{2}}{2}$$