cos(pi/17)*cos(2 pi/17)*cos(4pi/17)*cos(8pi/17)=? 27/10/2021 Bởi Josephine cos(pi/17)*cos(2 pi/17)*cos(4pi/17)*cos(8pi/17)=?
Đáp án: \[A = \dfrac{1}{{16}}\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\sin 2x = 2\sin x.\cos x\\A = \cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\left( {2\sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}} \right).\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\sin \dfrac{{2\pi }}{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{4\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.sin\dfrac{{8\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{16\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{{16\pi }}{{17}}\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \left( {\pi – \dfrac{{16\pi }}{{17}}} \right)\\ \Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{\pi }{{17}}\\ \Leftrightarrow A = \dfrac{1}{{16}}\end{array}\) Bình luận
Đáp án:
\[A = \dfrac{1}{{16}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 2x = 2\sin x.\cos x\\
A = \cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\left( {2\sin \dfrac{\pi }{{17}}.\cos \dfrac{\pi }{{17}}} \right).\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\sin \dfrac{{2\pi }}{{17}}.\cos \dfrac{{2\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{4\pi }}{{17}}.\cos \dfrac{{4\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.sin\dfrac{{8\pi }}{{17}}.\cos \dfrac{{8\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\sin \dfrac{{16\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{{16\pi }}{{17}}\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \left( {\pi – \dfrac{{16\pi }}{{17}}} \right)\\
\Leftrightarrow \sin \dfrac{\pi }{{17}}.A = \dfrac{1}{{16}}.\sin \dfrac{\pi }{{17}}\\
\Leftrightarrow A = \dfrac{1}{{16}}
\end{array}\)
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