>>> prove that cos(2*Pi/5)+cos(4*Pi/5)=-1/2
To prove that \(\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}\), we can use
properties of roots of unity and the symmetry of the cosine function.
First, consider the fifth roots of unity. These are the solutions to the equation \(z^5 = 1\). The fifth roots of
unity are given by:
\[1, \omega, \omega^2, \omega^3, \omega^4,\]
where \(\omega = e^{2\pi i / 5}\).
The sum of all the fifth roots of unity is zero:
\[1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.\]
We can rewrite this equation as:
\[\omega + \omega^2 + \omega^3 + \omega^4 = -1.\]
Next, we use the fact that \(\omega = e^{2\pi i / 5}\) and its complex conjugate is \omega^4\). The real part of \(\omega + \omega^4\) is \(2\cos\left(\frac{2\pi}{5}\right)\), and the real
part of \(\omega^2 + \omega^3\) is \(2\cos\left(\frac{4\pi}{5}\right)\).
Thus, we can separate the sum of the roots into their real parts:
\[2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = -1.\]
Dividing both sides by 2, we get:
\[\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}.\]
Therefore, the proof is complete and the answer is:
\[ -\frac{1}{2} \]
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