Question

помогите с косинусами

Answer 1

$cos\frac{2\pi}{7}cos\frac{4\pi}{7}+cos\frac{4\pi}{7}cos\frac{6\pi}{7}+cos\frac{6\pi}{7}\frac{2\pi}{7}\\\frac{1}{2}\left [ cos\frac{6\pi}{7}+cos\frac{2\pi}{7}+cos\frac{10\pi}{7}+\frac{2\pi}{7}+cos\frac{8\pi}{7}+\frac{4\pi}{7} \right ]\\\frac{1}{2}\left [ 2cos\frac{2\pi}{7}+cos\frac{6\pi}{7}+cos\frac{4\pi}{7}+\frac{8\pi}{7}+cos\frac{10\pi}{7} \right ]\\\bigstar  \ \ cosx=-cos\left ( \pi-x \right )\\\frac{1}{2}\left [ 2cos\frac{2\pi}{7}-2cos\frac{\pi}{7}-2cos\frac{3\pi}{7} \right ]=cos\frac{2\pi}{7}-cos\frac{\pi}{7}-cos\frac{3\pi}{7}\\\frac{2sin\frac{\pi}{7}\left ( cos\frac{2\pi}{7}-cos\frac{\pi}{7}-cos\frac{3\pi}{7} \right )}{2sin\frac{\pi}{7}}=\frac{sin\frac{3\pi}{7}-sin\frac{2\pi}{7}-sin\frac{2\pi}{7}-sin\frac{4\pi}{7}+sin\frac{2\pi}{7}}{2sin\frac{\pi}{7}}=-\frac{sin\frac{\pi}{7}}{2sin\frac{\pi}{7}}=-\frac{1}{2}$