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.

The equation $ \ \cos(5 \theta) \ = \ 0 \ $ has the set of angle solutions $ \ 5 \theta \ = \ \frac{\pi}{2} \ + \ k·\pi \ \Rightarrow \ \theta \ = \ \frac{(2k + 1) \ · \ \pi}{10} \ \ , \ $ arranged on the unit circle as presented in the graph below.  Upon solving the polyomial equation in cosine which is equivalent to $ \ \cos(5 \theta) \ = \ 0 \ \ , \ $ we obtain
$$ 16·\cos^5 \theta \ - \ 20·\cos^3 \theta \ + \ 5·\cos \theta \ \ = \ \ \cos \theta \ · \ ( \ 16·\cos^4 \theta \ - \ 20·\cos^2 \theta \ + \ 5 \ ) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \cos \theta \ \ = \ \ 0 \ \ \ , $$
$$   16·\cos^4 \theta \ - \ 20·\cos^2 \theta \ + \ 5  \ \ = \ \ 0 \ \ \Rightarrow \ \ \cos^2 \theta \ \ = \ \ \frac{20 \ \pm \ \sqrt{80}}{32} \ \ = \ \  \frac{5 \ \pm \ \sqrt5}{8} \ \ . $$
[Thus far, these results have also been shown by earlier posters.]
Because the angle solutions are symmetrical about the $ \ y-$axis (as they are odd multiples of $ \ \frac{\pi}{10} \ ) \ , \ $ they are divided into three "families":
$$ \mathbf{\cos \theta \ \ = \ \ 0 \ \ : }     \quad \quad    \theta \ = \ \frac{5 \pi}{10} \ = \ \frac{\pi}{2} \ \ , \ \ \frac{15 \pi}{10} \ = \ \frac{3 \pi}{2}  \ \ \ \text{[points in blue]} \ \  ; $$
$$ \mathbf{\cos^2 \theta \ \ = \ \ \frac{5 \ - \ \sqrt5}{8} \ \ : }     \quad \quad    \theta \ = \ \frac{3 \pi}{10}   \ \ , \ \ \frac{7 \pi}{10}   \ \ , \ \ \frac{13 \pi}{10}   \ \ , \ \ \frac{17 \pi}{10}  \ \ \ \text{[points in green]} \ \  ; $$
$$ \mathbf{\cos^2 \theta \ \ = \ \ \frac{5 \ + \ \sqrt5}{8} \ \ : }     \quad \quad    \theta \ = \ \frac{ \pi}{10}   \ \ , \ \ \frac{9 \pi}{10}   \ \ , \ \ \frac{11 \pi}{10}   \ \ , \ \ \frac{19 \pi}{10}  \ \ \ \text{[points in red]} \ \  . $$
The cosine function has the property $ \ \cos \theta \ = \ \cos  (2 \pi    -   \theta)   \ = \ -\cos  (  \pi  - \theta  ) \ = \ -\cos  (  \pi  + \theta  ) \ \ , \  $ so four angles will have a common value of cosine-squared (the exception, of course, being $ \ \cos^2 \theta \ = \ 0 \ $ for which $ \ \theta \ = \ \frac{ \pi}{2}   \ = \    \left(  \pi  - \frac{ \pi}{2}  \right) \ \  $ and $ \ \theta \ = \ \frac{ 3 \pi}{2}   \ = \    \left(  \pi  + \frac{ \pi}{2}  \right) \ \ . \ ) $
Applying the "triple-angle formula" for cosine (as Eric Towers suggests), $ \ \cos (3 \theta) \ = \ 4 · \cos^3 \theta \ - \ 3 · \cos \theta \ \ , \ $ we can write
$$ \cos \theta \ · \ \cos (3 \theta) \ \ = \ \ 4 · \cos^4 \theta \ - \ 3 · \cos^2 \theta \ \ . \ $$
The foregoing discussion makes it clear which value of $ \ \cos^2 \theta \ $ should be used among the solutions to $ \ \cos (5 \theta) \ = \ 0 \ \ , \ $ so we may calculate
$$ \cos \left( \frac{\pi}{10} \right) \ · \ \cos \left( \frac{3 \pi}{10} \right) \ \ = \ \ 4 · \cos^4 \left( \frac{\pi}{10} \right) \ - \ 3 · \cos^2 \left( \frac{\pi}{10} \right)   $$
$$ = \ \ 4 ·   \left( \ \frac{5 \ + \ \sqrt5}{8}   \ \right)^2 \ - \ 3 ·   \left( \ \frac{5 \ + \ \sqrt5}{8}  \ \right) $$ $$ = \ \ 4 ·   \left( \ \frac{25 \ + \ 5 \ + \ 10·\sqrt5}{64}   \ \right)  \ - \    \left( \ \frac{15 \ + \ 3·\sqrt5}{8}  \ \right)  $$
$$ = \ \     \frac{30 \ + \ 10·\sqrt5 \ - \ 30 \ - \ 6·\sqrt5}{16} = \ \     \frac{\sqrt5 }{4} \ \ . $$

