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Решите задачу:

$cos^2x=\frac{1+cos2x}{2}\\\\(cos\frac{7\pi}{8})^2=(cos(\pi-\frac{\pi}{8}))^2=(-cos\frac{\pi}{8})^2=cos^2\frac{\pi}{8}=\frac{1+cos(2*\frac{\pi}{8})}{2}=\\=\frac{1}{2}+\frac{cos\frac{\pi}{4}}{2}=\frac{1}{2}+\frac{\sqrt{2}}{4}\\\\...=\sqrt{2}-\sqrt{8}(\frac{1}{2}+\frac{\sqrt{2}}{4})=[\sqrt{8}=\sqrt{4*2}=2\sqrt{2}]=\\=\sqrt{2}-2\sqrt{2}(\frac{1}{2}+\frac{\sqrt{2}}{4})=\sqrt{2}-\sqrt{2}-\frac{2\sqrt{2}*\sqrt{2}}{4}=-\frac{4}{4}=-1$

$sin2x=2sinx*cosx\\\\2sin(\frac{5\pi}{6})cos(\frac{5\pi}{6})=sin(2*\frac{5\pi}{6})=sin\frac{5\pi}{3}=sin(2\pi-\frac{\pi}{3})=\\=-sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}$