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Решите задачу:

$\sqrt{32}-\sqrt{128}sin^2\frac{9\pi}{8}=[\, sin(\frac{9\pi}{8})=sin(\pi +\frac{\pi}{8})=-sin\frac{\pi}{8}\, ]=\\\\=\sqrt{2^5}-\sqrt{2^7}sin^2\frac{\pi}{8}=[\, sin^2 \alpha =\frac{1-cos2 \alpha }{2}\; \to \; sin^2\frac{\pi}{8}=\frac{1-cos\frac{\pi}{4}}{2}\, ]=\\\\=\sqrt{2^4\cdot 2}-\sqrt{2^6\cdot 2}\cdot \frac{1-cos\frac{\pi}{4}}{2}=2^2\sqrt2-2^3\sqrt3\cdot \frac{1-\frac{\sqrt2}{2}}{2}=\\\\=4\sqrt2-4\sqrt2\cdot (1-\frac{\sqrt2}{2})=4\sqrt2-4\sqrt2+\frac{4\sqrt2\cdot \sqrt2}{2}=4$