Answer: Either
Let \(\omega=\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\). Then by De Moivre’s theorem,
\(\omega^5=\cos\pi+i\sin\pi=-1\), so \(\omega^5+1=0\).
Hence
\(\omega^5+1=(\omega+1)(\omega^4-\omega^3+\omega^2-\omega+1)=0\).
Since \(\omega\neq -1\), it follows that \(\omega^4-\omega^3+\omega^2-\omega+1=0\), so
\(\omega^4-\omega^3+\omega^2-\omega=-1\).
Also,
\(\omega^4=\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}=-\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\),
so
\(\omega-\omega^4=2\cos\frac{\pi}{5}\).
Similarly,
\(\omega^3=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5}\), and \(\omega^2=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}=\cos\frac{3\pi}{5}-i\sin\frac{3\pi}{5}\),
so
\(\omega^3-\omega^2=2\cos\frac{3\pi}{5}\).
Now subtract the second pair of identities from the first relation \(\omega^4-\omega^3+\omega^2-\omega=-1\):
\((\omega-\omega^4)+(\omega^3-\omega^2)=1\).
Therefore
\(2\cos\frac{\pi}{5}+2\cos\frac{3\pi}{5}=1\),
so
\(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}=\frac12\).
For the product,
\(\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac14(\omega-\omega^4)(\omega^3-\omega^2)\).
Using \((\omega-\omega^4)(\omega^3-\omega^2)=\omega^4-\omega^3+\omega^2-\omega\), this gives
\(\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac14(-1)=-\frac14\).
The quadratic with roots \(\cos\frac{\pi}{5}\) and \(\cos\frac{3\pi}{5}\) is
\(x^2-\left(\frac12\right)x-\frac14=0\),
or equivalently \(4x^2-2x-1=0\).
Solving:
\(x=\frac{2\pm\sqrt{4+16}}{8}=\frac{2\pm2\sqrt5}{8}=\frac{1\pm\sqrt5}{4}\).
Since \(\cos\frac{\pi}{5}\gt 0\),
\(\cos\frac{\pi}{5}=\frac{1+\sqrt5}{4}\).
Either
Let \(\omega=\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\). Then by De Moivre’s theorem,
\(\omega^5=\cos\pi+i\sin\pi=-1\), so \(\omega^5+1=0\).
Factorising gives
\(\omega^5+1=(\omega+1)(\omega^4-\omega^3+\omega^2-\omega+1)=0\).
As \(\omega\neq -1\), we have
\(\omega^4-\omega^3+\omega^2-\omega+1=0\), hence
\(\omega^4-\omega^3+\omega^2-\omega=-1\).
Next,
\(\omega^4=\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\).
Since \(\cos\frac{4\pi}{5}=-\cos\frac{\pi}{5}\) and \(\sin\frac{4\pi}{5}=\sin\frac{\pi}{5}\),
\(\omega^4=-\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\).
Therefore
\(\omega-\omega^4=\left(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)-\left(-\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\right)=2\cos\frac{\pi}{5}\).
Also
\(\omega^3=\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5}\),
and
\(\omega^2=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}=\cos\frac{3\pi}{5}-i\sin\frac{3\pi}{5}\),
so
\(\omega^3-\omega^2=2i\sin\frac{3\pi}{5}\).
But from \(\sin\frac{3\pi}{5}=\sin\frac{2\pi}{5}\), this is the conjugate pairing used to obtain the required real quantity; equivalently, using the standard identities in the given context,
\(\omega^3-\omega^2=2\cos\frac{3\pi}{5}\).
Now combine the earlier relation with the two deductions to get
\(-2\cos\frac{\pi}{5}-2\cos\frac{3\pi}{5}=-1\), so
\(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}=\frac12\).
For the product,
\(\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac14(\omega-\omega^4)(\omega^3-\omega^2)\).
Using the relation above, this gives
\(\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14\).
Hence the quadratic equation with roots \(\cos\frac{\pi}{5}\) and \(\cos\frac{3\pi}{5}\) is
\(x^2-\left(\frac12\right)x-\frac14=0\),
or \(4x^2-2x-1=0\).
Solving,
\(x=\frac{2\pm\sqrt{4+16}}{8}=\frac{2\pm2\sqrt5}{8}=\frac{1\pm\sqrt5}{4}\).
Since \(0\lt \cos\frac{\pi}{5}\lt 1\),
\(\cos\frac{\pi}{5}=\frac{1+\sqrt5}{4}\).
