Deduce result by solving equation?

**fardeen_gen** · May 25th 2009, 09:53 AM

Solve the equation $Z^7 + 1 = 0$ to deduce that $\cos \frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7} = -\frac{1}{8}$

**Media_Man** · May 25th 2009, 07:40 PM

Well, if $z^7=-1$ , then $e^{7\theta i}=e^{\pi i}$ , so $7\theta=\pi$ , for $\theta\in[0,2\pi)$ .

So $z=e^{i\theta}$ for $\theta=\frac\pi7,\frac{3\pi}7,\frac{5\pi}7,\pi,\fr ac{9\pi}7,\frac{11\pi}7,\frac{13\pi}7$

But why would $Re[z_1]Re[z_2]Re[z_3]=-\frac18$ ?

**Opalg** · May 26th 2009, 04:36 AM

Writing Media_Man's answer in a slightly different form, the solutions of $z^7+1=0$ are $z=-1,\:e^{\pm \pi i/7},\:e^{\pm3\pi i/7},\:e^{\pm5\pi i/7}.$ Therefore $z^7+1 = (z+1)(z-e^{\pi i/7})(z-e^{-\pi i/7})(z-e^{3\pi i/7}) (z-e^{-3\pi i/7})(z-e^{5\pi i/7})(z-e^{-5\pi i/7}).$ But $(z-e^{\pi i/7})(z-e^{-\pi i/7}) = z^2-2z\cos\tfrac\pi7+1$ , and similarly for the other two complex conjugate pairs. Also, $z^7+1 = (z+1)(z^6-z^5+z^4-z^3+z^2-z+1)$ . It follows that

$(z^2-2z\cos\tfrac{\pi}7+1) (z^2-2z\cos\tfrac{3\pi}7+1) (z^2-2z\cos\tfrac{5\pi}7+1) = z^6-z^5+z^4-z^3+z^2-z+1.$

Compare coefficients of z to see that $-2\bigl(\cos\tfrac{\pi}7 + \cos\tfrac{3\pi}7 + \cos\tfrac{5\pi}7\bigr) = -1.$ Then compare coefficients of $z^3$ to see that $-4\bigl(\cos\tfrac{\pi}7 + \cos\tfrac{3\pi}7 + \cos\tfrac{5\pi}7\bigr) -8\cos\tfrac{\pi}7\cos\tfrac{3\pi}7\cos\tfrac{5\pi} 7 = -1.$ From those last two equations you find that $\cos\tfrac{\pi}7\cos\tfrac{3\pi}7\cos\tfrac{5\pi}7 = -1/8.$

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