factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$ [closed]

trigonometry complex-numbers

Factor $z^7-1$ into linear and quadratic factors and prove that $$ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$$

I have been able to prove it using the value of $\cos(\pi/7)$. Given here http://mathworld.wolfram.com/TrigonometryAnglesPi7.html


Solution 1:

Let $z=e^{\frac{i\pi}{7}}$. Then $\cos (\frac{\pi}{7})=(z+z^{-1})/2$, $\cos (\frac{2\pi}{7})=(z^2+z^{-2})/2$, $\cos (\frac{3\pi}{7})=(z^3+z^{-3})/2$. This should get you started.

Solution 2:

Factor $x^7-1$ in $\Bbb C$ and obtain its factorization in $\Bbb R$ by pairing off conjugate roots.

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