Đáp án:
\[A = \dfrac{1}{{16}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 2x = 2\sin x.\cos x\\
A = \cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\left( {2\sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}} \right).\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\sin \dfrac{{2\pi }}{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{4\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.sin\dfrac{{8\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{16\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{{16\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \left( {\pi - \dfrac{{16\pi }}{{17}}} \right)\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{\pi }{{17}}\\
\Leftrightarrow A = \dfrac{1}{{16}}
\end{array}\)
