<s>[INST] Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By considering the equation $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$
I have completed part (a). I am stuck on part (b) however.
My Workings:
By Factor formulae, $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{1}{2} (\cos(\frac{2\pi}{5}) + \cos(\frac{\pi}{5}))$
Considering roots of $z^5 = 1$. The sum of the roots equals 0.
So $\cos(\frac{2\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{8\pi}{5}) + \cos(\frac{10\pi}{5})=0$
$\cos(\frac{2\pi}{5}) = \cos(\frac{8\pi}{5})$, $\cos(\frac{4\pi}{5}) = \cos(\frac{6\pi}{5}) = -\cos(\frac{\pi}{5})$
So $2\cos(\frac{2\pi}{5})-2\cos(\frac{\pi}{5}) = -1$
$\cos(\frac{2\pi}{5})-\cos(\frac{\pi}{5}) = -\frac{1}{2}$
Been on stuck on this problem for an hour. Please help me.
 [/INST] I don't know if this is what is intended, but here is where I went with this problem.
let $u = \cos \theta$
$u(16 u^4 - 20 u^2 + 5) = 0$
From the binomial theorem:
$u^2 = \frac {20 \pm \sqrt {80}}{32}\\
u^2 = \frac {5 \pm \sqrt {5}}{8}$
Yes, $u = 0$ is a solution associated with $\theta = \frac {\pi}{2}$ but that isn't particularly interesting.
$u^2 = \cos^2 \theta = \frac 12 (1+\cos 2\theta)$
$\cos 2\theta = \frac {5 \pm \sqrt {5}}{4} - 1\\
\cos 2\theta = \frac {1 \pm \sqrt {5}}{4}$
There are multiple values of $\theta$ such that $\cos 5\theta = 0$ the smallest of which is $\theta = \frac {\pi}{10}.$  This will be associated with the largest possible value of $\cos \theta$
$\cos \frac {\pi}{5}  = \frac {1 + \sqrt {5}}{4}\\
\cos \frac {2\pi}{5} = 2\cos^2 \frac {\pi}{5} - 1 = 2 \frac {3 + \sqrt 5}{8} - 1\\
\cos \frac {2\pi}{5} = \frac {-1 + \sqrt 5}{4}$
$\frac 12 (\cos \frac {\pi}{5} + \cos \frac {2\pi}{5}) = \frac 12(\frac {1 + \sqrt {5}}{4} + \frac {-1 + \sqrt 5}{4}) = \frac {\sqrt 5}{4}$
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