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Ответ:

$\frac{1}{2}$

Объяснение:

$1)cos(\frac{\pi}{11} )+cos(\frac{3\pi}{11} )+cos(\frac{5\pi}{11} )+cos(\frac{7\pi}{11} )+cos(\frac{9\pi}{11} )=\\=\frac{2sin(\frac{\pi}{11})cos(\frac{3\pi}{11} )+sin(\frac{2}{11} )2sin(\frac{\pi}{11} )cos(\frac{5\pi}{11} )+2sin(\frac{\pi}{11} )cos(\frac{7\pi}{11} )+2sin(\frac{\pi}{11} ) cos(\frac{9\pi}{11} )}{2cos(\frac{\pi}{11} ) } =\\=2sin(\frac{\pi}{11} )cos(\frac{9\pi}{11} )=-sin(\frac{8\pi}{11} )+sin(\frac{10\pi}{11} )=\frac{sin(\frac{10\pi}{11} )}{2sin(\frac{2\pi}{11} )} =\frac{1}{2}\\$

H=\frac{1}{2}\\" alt="2)H=cos(\frac{\pi}{11} )+cos(\frac{3\pi}{11} )+cos(\frac{5\pi}{11} )+cos(\frac{7\pi}{11} )+cos(\frac{9\pi}{11} )\\H=cos(-\frac{\pi}{11} )+cos(-\frac{3\pi}{11} )+cos(-\frac{5\pi}{11} )+cos(-\frac{7\pi}{11} )+cos(-\frac{9\pi}{11} )\\-1=cos(-\frac{11\pi}{11} )\\2H-1=>H=\frac{1}{2}\\" align="absmiddle" class="latex-formula">

Посчитать, ** фотографии​

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