Вычислите: 27.4. а) 2 sin π/8 cos π/8; б) sin π/8 cos π/8 + 1/4; в) cos^2 π/8 - sin^2 π/8; г) √2/2 - (cos π/8 + sin π/8)^2.

Для решения этих задач воспользуемся формулами двойного угла: 1) $\sin 2\alpha = 2 \sin \alpha \cos \alpha$ 2) $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ 3) $(\cos \alpha + \sin \alpha)^2 = \cos^2 \alpha + 2\sin \alpha \cos \alpha + \sin^2 \alpha = 1 + \sin 2\alpha$ Решение: а) $2 \sin \frac{\pi}{8} \cos \frac{\pi}{8} = \sin(2 \cdot \frac{\pi}{8}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ б) $\sin \frac{\pi}{8} \cos \frac{\pi}{8} + \frac{1}{4} = \frac{1}{2} \cdot (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}) + \frac{1}{4} = \frac{1}{2} \sin \frac{\pi}{4} + \frac{1}{4} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{4} = \frac{\sqrt{2}}{4} + \frac{1}{4} = \frac{\sqrt{2} + 1}{4}$ в) $\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8} = \cos(2 \cdot \frac{\pi}{8}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ г) $\frac{\sqrt{2}}{2} - (\cos \frac{\pi}{8} + \sin \frac{\pi}{8})^2 = \frac{\sqrt{2}}{2} - (1 + \sin(2 \cdot \frac{\pi}{8})) = \frac{\sqrt{2}}{2} - (1 + \sin \frac{\pi}{4}) = \frac{\sqrt{2}}{2} - (1 + \frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{2}}{2} = -1$