Trigonometry : verify that $\cos \frac{A}{2} +\cos 2A = 0$ for $A=\frac{2\pi}{5}$ Question in trigonometry: verify that $\cos\frac{A}{2} + \cos 2A = 0$.
Let $A = \frac{2\pi}{5}$. 


*

*Verify that $\cos \frac{A}{2} + \cos 2A = 0$.

*Show that $x=\cos A$ satisifies the equation $x=2(4x^4-4x^2+1)-1$.

*Factor $8x^4-8x^2-x+1$ over $\mathbb Z$, and deduce that $\cos A$ is a zero of a quadratic polynomial over $\mathbb Z$.

*Determine $\cos A.$


My solution is as follows :
We know $\cos(A/2)=\sqrt{(1+\cos A)/2}$ 
and $\cos2A=2\cos^2A-1$. 
$\cos2A=2\cos^2A-1$
It is sufficient if we prove that $\cos2A=-\cos(A/2)$   for the first question.
$2\cos^2A-1=\pm\sqrt{(1+\cos A)/2}$
squaring on both sides and simplifying:
$8\cos4A+2-8\cos^2A=1+\cos A$
subsitituting $\cos A$ with $x$ :
$8x^4- 8x^2+2-1 = x$
$2(4x^4-4x^2+1)-1=x$
I believe that if I do the same thing backwards, I will get the answer for question (2)
I have factorized the equation and ended up with this:
$(x-1)[8x^3+8x-1]$
Now, the third question asks me to deduce that cosA is a zero of quadratic polynomial, but I have only linear and cubic factors, how do I proceed?
Regarding the first question, I’m clueless, I have simplified it as much as possible, but its not going anywhere.
 A: $1$.
$\cos \frac A2=\cos \frac \pi5$ and 
as $\cos(\pi-x)=-\cos x,$ 
$\cos2A=\cos\frac{4\pi}5=\cos\left(\pi-\frac\pi5\right)=-\cos\frac\pi5 $
$2$.
As $5A=2\pi,A=2\pi-4A$
So using $\cos2x=2\cos^2x-1$ and $\cos(2\pi-x)=\cos x$,
$\cos A=\cos(2\pi-4A)=\cos 4A=2(\cos2A)^2-1=2(2\cos^2A-1)^2-1=8\cos^4A-8\cos^2A+1$
$3$. 
But $\cos4A=\cos A$  
$\implies 4A=2n\pi\pm A$ where $n$ is any integer
Taking $'+'$ sign, $A=\frac{2n\pi}5$ where $n=0,1,2,3,4$
So, the roots are $\cos0=1,$
$\cos\frac{2\pi}5, \cos\frac{4\pi}5=\cos\left(\pi-\frac{\pi}5\right)=-\cos\frac\pi5<0,$
$\cos\frac{6\pi}5=\cos\left(\pi+\frac{\pi}5\right)=-\cos\frac\pi5,$
$\cos\frac{8\pi}5=\cos\left(2\pi-\frac{8\pi}5\right)=\cos\frac{2\pi}5$
Taking $'-'$ sign, $A=\frac{2n\pi}3$ where $n=0,1,2$
So, the roots are $\cos0=1,$
$\cos\frac{2\pi}3=\cos\left(\pi-\frac{\pi}3\right)=-\cos\frac\pi3=-\frac12,$
$\cos\frac{4\pi}3=\cos\left(\pi+\frac{\pi}3\right)=-\cos\frac\pi3=-\frac12,$
So, $8x^4-8x^2-x+1=(x-1)\{x-(-\frac12)\}\left(x-\cos\frac{2\pi}5\right)\{x-(-\cos\frac{\pi}5)\}$
So, $\cos\frac{2\pi}5>0,-\cos\frac{\pi}5<0$ are the roots of the qaudratic eqaution $$\frac{8x^4-8x^2-x+1}{(x-1)(x+\frac12)}=0\ \ \ \ (1)$$
$4$.
As $A=\frac{2\pi}5,0<\frac{2\pi}5<\frac\pi2,$ the value of $\cos\frac{2\pi}5$ will be the positive root of $(1)$
