Trigonometry substitution

Question

Given the expansion of $\cos(5x) = 16\cos^5(x)-20\cos^3(x)+5\cos(x)$. How can I find the find the value of $\cos(\frac{\pi}{10})$ and $\cos(\frac{7\pi}{10})$ using $\cos(5x) = 0$ ? I know that I can change $\cos (5x)$ as a quartic equation and can get $\cos(x) = \pm\sqrt \frac{5 \pm\sqrt 5}{8}$. Which means $\cos(\frac{5\pi}{10}) =\cos(\frac{\pi}{2})= 0$ but how I am going to decide about the $\pm$ signs? and what about $\cos(\frac{7\pi}{10})$ ?

Answer 1

First of all, $$0<\frac\pi{10}<\frac\pi2<\frac{7\pi}{10}$$

So, $\displaystyle\cos\frac\pi{10}>0;\cos\frac{7\pi}{10}<0$

$\displaystyle\implies-\cos\frac{7\pi}{10}=\cos\left(\pi-\frac{7\pi}{10}\right)=\cos\frac{3\pi}{10}>0$

and using Prosthaphaeresis Formulas, $\displaystyle\cos\frac\pi{10}-\cos\frac{3\pi}{10}=2\sin\frac\pi5\sin\frac{2\pi}5>0$

$\displaystyle\implies\cos\frac\pi{10}>\cos\frac{3\pi}{10}$

Can you determine the required values from here?