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Question 1

Question 2

Решите задачу:

$sin(\frac{15\pi }{8}-2 \alpha )=sin(\frac{16\pi -\pi }{8}-2 \alpha )=sin(2\pi -\frac{\pi}{8}-2 \alpha )=\\\\=sin(-\frac{\pi}{8}-2 \alpha )=-sin(\frac{\pi}{8}+2 \alpha )\\\\cos(\frac{17\pi }{8}-2 \alpha )=cos(\frac{16\pi +\pi }{8}-2 \alpha )=cos(2\pi +\frac{\pi}{8}-2 \alpha )=cos(\frac{\pi}{8}-2 \alpha )$

$sin^2(\frac{15\pi}{8}-2 \alpha )-cos^2(\frac{17\pi}{8}-2 \alpha )=sin^2(\frac{\pi}{8}+2 \alpha )-cos^2(\frac{\pi}{8}-2 \alpha )=\\\\=(sin (\frac{\pi}{8}+2 \alpha )-cos(\frac{\pi}{8}-2 \alpha ))\cdot (sin(\frac{\pi}{8}+ \alpha )+cos(\frac{\pi}{8}-2 \alpha ))=\\\\=[\, \frac{\pi}{8}-2 \alpha =\frac{\pi}{2}-(\frac{3\pi}{8}+2 \alpha )\, ]=\\\\=(sin(\frac{\pi}{8}+2 \alpha )-sin(\frac{3\pi}{8}+2 \alpha ))\cdot (sin(\frac{\pi}{8}+2 \alpha )+sin(\frac{3\pi}{8}+2 \alpha ))=$

$=2sin(-\frac{\pi}{8})\cdot cos(\frac{\pi}{4}+2 \alpha )\cdot 2sin(\frac{\pi}{4}+2 \alpha )cos(\frac{-\pi}{8})=\\\\=(-2sin\frac{\pi}{8}\cdot cos\frac{\pi}{8})\cdot (2sin(\frac{\pi}{4}+2 \alpha )\cdot cos(\frac{\pi}{4}+2 \alpha ))=\\\\=-sin\frac{\pi}{4}\cdot sin(\frac{\pi}{2}+4 \alpha )=-\frac{\sqrt2}{2}\cdot (-cos4 \alpha )= \frac{cos4 \alpha }{\sqrt2}$