Question: I'm trying to find the area of the region bounded by $r =|\sin \theta | $ in the $xy$-plane using the formula: 
\begin{align\*}
A &= \frac{1}{2}\int\_{0}^{\pi} (|\sin \theta|)^2 d\theta \\
&=\frac{1}{2}\int\_{0}^{\pi} (1-\cos2\theta)^2 d\theta \\
&= \frac{1}{2}\int\_{0}^{\pi} (1-2\cos\theta + \cos^22\theta)^2 d\theta \\
\end{align\*}

which I get $\frac{\pi}{4}$ as the answer but I got it wrong on a test.

The other options are:

1. $\frac{\pi}{2}$
2. $\frac{\pi}{4}$
3. $\pi$
4. $1$
5. $2$

Could there have been a mistake on the test or did I miss something?

Answer: 
Below I'll show how to find 

(1) the minimal polynomial for $\omega+\omega^{-1}$,

(2) the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$, and

(3) the minimal polynomial for $\sin\left(\dfrac{2\pi}{5}\right)$.

(1) First note that $\omega$ is a root of $x^5-1$, but not of $x-1$. Hence $\omega$ is a root of $\dfrac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$. So we have that

$$\omega^4+\omega^3+\omega^2+\omega+1=0.$$

Since $\omega\ne0$, we can divide through by $\omega^2$ to obtain

$$\omega^2+\omega+1+\omega^{-1}+\omega^{-2}=0.$$

Which we can rewrite as

$$\left(\omega^2+2+\omega^{-2}\right)+\left(\omega+\omega^{-1}\right)+1-2=0.$$

So that 

$$\left(\omega+\omega^{-1}\right)^2+\left(\omega+\omega^{-1}\right)-1=0.$$

Hence $\omega+\omega^{-1}$ is a root of $x^2+x-1$. Since the roots of this polynomial are not rational, this polynomial is irreducible. Hence $x^2+x-1$ is the minimal polynomial of $\omega+\omega^{-1}$.

(2) $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\omega+\omega^{-1}}{2}$, and $4\left(\dfrac{\omega+\omega^{-1}}{2}\right)^2+2\left(\dfrac{\omega+\omega^{-1}}{2}\right)-1=0$.

So the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$ is $4x^2+2x-1$.

(3) The roots of $4x^2+2x-1$ are $\dfrac{-1\pm\sqrt{5}}{4}$, and $\cos\left(\dfrac{2\pi}{5}\right)>0$, so $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\sqrt{5}-1}{4}$.

It follows that $\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$.

For simplicity, let's let $\alpha=\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$. So we have that

$$\alpha^2=\dfrac{\sqrt{5}+5}{8}.$$

$$8\alpha^2-5=\sqrt{5}.$$

$$64\alpha^4-80\alpha^2+25=5.$$

It follows that $\alpha$ is a root $16x^4-20x^2+5$, which is irreducible because of Eisenstein's criterion. So the minimal polynomial of $\sin\left(\dfrac{2\pi}{5}\right)$ is $16x^4-20x^2+5$.
Hint: square it, then make use of the fact that the 5 fifth roots of unity sum to 0.