Complex Numbers and Binomial Expansion 
I was able to show the above by equating the solution of $cos \ 5\theta=0$ (which gives $\pi/10$) and the solution of ${16cos}^5\theta - {20cos}^3\theta+5cos\theta = 0$ (which is what you get when you use De Moivre's Theorem with binomial expansion). However, I do not understand how the second part of the question has anything to do with the first, and thus am lost as to where to begin. 
 A: What you have to realize about the equation $\cos5\theta=0$ is that $\frac{\pi}{10}$ is a solution because $\cos\frac{\pi}{2}=0$, but also that $\frac{7\pi}{10}$ is a solution because $\cos\frac{35\pi}{10}=\cos\frac{5\pi}2=\cos\left(2\pi+\frac{\pi}2\right)=\cos\frac\pi 2=0$. Therefore, we can use this equation to find both $\cos\frac\pi {10}$ and $\cos\frac{7\pi}{10}$.
We have the following:
$$16\cos^5\theta-20\cos^3\theta+5\cos\theta=0$$
We want to solve this equation to find $\cos \frac{\pi}{10}$ and $\cos \frac{7\pi}{10}$, which we know is not $0$. Therefore, we can divide by $\cos \theta$:
$$16\cos^4\theta-20\cos^2\theta+5=0$$
Now, we can use the quadratic formula to find $\cos^2\theta$:
$$\cos^2\theta=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2(16)}$$
$$\cos^2\theta=\frac{20\pm4\sqrt 5}{32}$$
$$\cos^2\theta=\frac{5\pm\sqrt5}{8}$$
Now, since $\frac{\pi}{10}$ is very close to $0$, it will have a bigger $\cos$, so we want to take the bigger solution. Therefore, the $\pm$ becomes a $+$.
$$\cos^2\frac{\pi}{10}=\frac{5+\sqrt5}{8}$$
$\frac\pi{10}$ is in Quadrant I, so we take the positive square root of both sides:
$$\cos\frac\pi{10}=\sqrt{\frac{5+\sqrt5}{8}}$$
Now, for $\frac{7\pi}{10}$, we have:
$$\cos^2\frac{7\pi}{10}=\frac{5\pm\sqrt5}{8}$$
For this, you need to ask yourself two questions:


*

*Does the $\pm$ become a $+$ or a $-$? (Think bigger vs smaller $\cos$.)

*Do you take the positive or negative square root? (Think sign of $\cos$ in this quadrant.)


Good luck finding $\cos\frac{7\pi}{10}$!
