我们知道\[\begin{split}\cos\dfrac{\pi}7\cdot\cos\dfrac{2\pi}7\cdot\cos\dfrac{4\pi}7&=\dfrac {\sin\dfrac{\pi}7\cdot\cos\dfrac{\pi}7\cdot\cos\dfrac{2\pi}7\cdot\cos\dfrac{4\pi}7}{\sin\dfrac{\pi}7}\\&=\dfrac {\dfrac 18\sin\dfrac{8\pi}7}{\sin\dfrac{\pi}7}\\&=-\dfrac 18.\end{split}\] 那么如何求三角代数式\[\sin\dfrac{\pi}7\cdot\sin\dfrac{2\pi}7\cdot\sin\dfrac{4\pi}7\]的值呢？
- 法一
考虑方程\(\sin 3\theta=\sin 4\theta\)的根为\[\dfrac{\pi}7,\dfrac{3\pi}7,\dfrac{5\pi}7,\pi,\dfrac{9\pi}7,\dfrac{11\pi}7,\dfrac{13\pi}7.\] 化简上述方程，并去掉根\(\pi\)，有\[\left(3-4\sin^2\theta\right)^2=16\left(1-\sin^2\theta\right)\left(-1+2\sin^2\theta\right)^2.\] 其六根之积为\(\dfrac 7{64}\)，因此原式的值为\(\dfrac{\sqrt 7}8\)．
- 法二
原式的平方为\[\begin{split}\qquad&\dfrac{1-\cos\dfrac{2\pi}7}2\cdot\dfrac{1-\cos\dfrac{4\pi}7}2\cdot\dfrac{1-\cos\dfrac{6\pi}7}2\\&=\dfrac 18\left[1-\left(\cos\dfrac{2\pi}7+\cos\dfrac{4\pi}7+\cos\dfrac{6\pi}7\right)+\cos\dfrac{2\pi}7\cos\dfrac{4\pi}7+\cos\dfrac{4\pi}7\cos\dfrac{6\pi}7+\cos\dfrac{6\pi}7\cos\dfrac{2\pi}7-\cos\dfrac{2\pi}7\cos\dfrac{4\pi}7\cos\dfrac{6\pi}7\right]\\&=\dfrac 18\left[1-\left(-\dfrac 12\right)+\dfrac 12\left(\cos\dfrac{6\pi}7+\cos\dfrac{12\pi}7+\cos\dfrac{10\pi}7+\cos\dfrac{2\pi}7+\cos\dfrac{8\pi}7+\cos\dfrac{4\pi}7\right)-\dfrac 18\right]\\&=\dfrac 7{64}.\end{split}\] 于是原式的值为\(\dfrac{\sqrt 7}8\)．
- 法三
记\(z^n=1\)的根分别为 \[\omega,\omega^2,\omega^3,\cdots,\omega^n,\] 则由 \[z^n-1=0\] 得 \[(z-1)\left(z^{n-1}+z^{n-2}+\cdots+1\right)=0,\] 因此 \[(z-\omega)\left(z-\omega^2\right)\cdots\left(z-\omega^{n-1}\right)=z^{n-1}+z^{n-2}+\cdots+1.\] 令\(z=1\)，两边取模，得 \[\prod_{k=1}^{n-1}\left|\left(1-\cos\dfrac{2k\pi}n\right)+\mathrm i\sin\dfrac{2\pi}n\right|=n.\] 于是可得 \[\left|\prod_{k=1}^{n-1}\sin\dfrac{k\pi}n\right|=\dfrac n{2^{n-1}}.\] 取\(n=7\)，即可得原式的值为\(\dfrac{\sqrt 7}8\)． 想想看，如果令\(z=-1\)，两边取模可以得到什么呢？