$\left(1+\cos\dfrac {\pi}7\right)\left(1+\cos\dfrac{3\pi}7\right)\left(1+\cos\dfrac{5\pi}7\right)$ 的值为 \((\qquad)\)
【难度】
【出处】
无
【标注】
【答案】
B
【解析】
由题意，根据单位根的推论一有\[\begin{split}m&=\left(1+\cos\dfrac{\pi}7\right)\left(1+\cos\dfrac{3\pi}7\right)\left(1+\cos\dfrac{5\pi}7\right)\\
&=\left(1-\cos\dfrac{2\pi}7\right)\left(1-\cos\dfrac{4\pi}7\right)\left(1-\cos\dfrac{6\pi}7\right)\\
&=8\sin^2\dfrac{\pi}{7}\sin^2\dfrac{2\pi}{7}\sin^2\dfrac{3\pi}{7}\\
&=8\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}\sin\dfrac{4\pi}{7}\sin\dfrac{5\pi}{7}\sin\dfrac{6\pi}{7}\\
&=\dfrac{7}{8}.
\end{split}\]
&=\left(1-\cos\dfrac{2\pi}7\right)\left(1-\cos\dfrac{4\pi}7\right)\left(1-\cos\dfrac{6\pi}7\right)\\
&=8\sin^2\dfrac{\pi}{7}\sin^2\dfrac{2\pi}{7}\sin^2\dfrac{3\pi}{7}\\
&=8\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}\sin\dfrac{4\pi}{7}\sin\dfrac{5\pi}{7}\sin\dfrac{6\pi}{7}\\
&=\dfrac{7}{8}.
\end{split}\]
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