Đáp án:
\[\cos \frac{{2\pi }}{{31}}.\cos \frac{{4\pi }}{{31}}.\cos \frac{{8\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}} = \frac{1}{{32}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 2x = 2\sin x.\cos x\\
\cos \frac{{2\pi }}{{31}}.\cos \frac{{4\pi }}{{31}}.\cos \frac{{8\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{2}.\frac{1}{{\sin \frac{{2\pi }}{{31}}}}.\left( {2\sin \frac{{2\pi }}{{31}}.\cos \frac{{2\pi }}{{31}}} \right).\cos \frac{{4\pi }}{{31}}.\cos \frac{{8\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{{2\sin \frac{{2\pi }}{{31}}}}.\sin \frac{{4\pi }}{{31}}.\cos \frac{{4\pi }}{{31}}.\cos \frac{{8\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{{4.\sin \frac{{2\pi }}{{31}}}}.sin\frac{{8\pi }}{{31}}.\cos \frac{{8\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{{8.\sin \frac{{2\pi }}{{31}}}}.sin\frac{{16\pi }}{{31}}.\cos \frac{{16\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{{16.\sin \frac{{2\pi }}{{31}}}}.sin\frac{{32\pi }}{{31}}.\cos \frac{{32\pi }}{{31}}\\
= \frac{1}{{32.\sin \frac{{2\pi }}{{31}}}}.\sin \frac{{64\pi }}{{31}}\\
= \frac{1}{{32.\sin \frac{{2\pi }}{{31}}}}.\sin \left( {2\pi + \frac{{2\pi }}{{31}}} \right)\\
= \frac{1}{{32.\sin \frac{{2\pi }}{{31}}}}.\sin \frac{{2\pi }}{{31}}\\
= \frac{1}{{32}}
\end{array}\)
