Let $\omega= \cos \frac{2\pi}{10}+i\sin \frac{2\pi}{10}$, let $K=\mathbb{Q}(\omega^2)$ and $L=\mathbb{Q}(\omega)$. Let $\omega= \cos \frac{2\pi}{10}+i\sin \frac{2\pi}{10}$. Let $K=\mathbb{Q}(\omega^2)$ and $L=\mathbb{Q}(\omega)$. Then
A. $[L,\mathbb{Q}]=10$
B. $ [L,K]=2$
C. $[K,\mathbb{Q}]=4$
D. $L=K$ 
 A: Note that $\omega$ is a primitive $10^{th}$ root of unity and $\omega^2$ is a primitive $5^{th}$ root of unity. 
The Cyclotomic Polynomials are the minimal polynomials of primitive roots of unity and are easy to compute with the formulas on the wikipedia page.
$$\Phi_5(x) = x^4 + x^3 + x^2 + x + 1 \\
\Phi_{10}(x)= x^4 - x^3 + x^2 - x + 1.\\$$
This tells you that $$[\mathbb{Q}(\omega): \mathbb{Q}] = 4 = [\mathbb{Q}(\omega^2):\mathbb{Q}].$$
Therefore since $\mathbb{Q}(\omega^2) \subseteq \mathbb{Q}(\omega)$ and they have the same degree as an extension of $\mathbb{Q}$, $\mathbb{Q}(\omega^2) = \mathbb{Q}(\omega).$
More generally, if $n$ is an odd number greater than 1, $$\Phi_{2n}(x) = \Phi_{n}(-x).$$
Now if $\omega$ is a primitive $(2n)^{th}$ root of unity, then $\omega = e^{\frac{2\pi i}{2n}k}$ for some $k$ is relatively prime to $2n$. Then $\omega^2 = e^{\frac{2\pi i}{n}k}$ is a primitive $n^{th}$ root of unity. Using the fact above, $$[\mathbb{Q}(\omega): \mathbb{Q}] = \Phi(n) = [\mathbb{Q}(\omega^2):\mathbb{Q}]$$
Where $\Phi(n)$ is the number of numbers less than $n$ and coprime to it.
Therefore since $\mathbb{Q}(\omega^2) \subseteq \mathbb{Q}(\omega)$ and they have the same degree as an extension of $\mathbb{Q}$, $\mathbb{Q}(\omega^2) = \mathbb{Q}(\omega)$.
A: You have that $\omega ^2=e^{\frac{2 i\pi}{5}}$, and thus $$X^5-1=(X-1)(X^4+X^3+X^2+X+1).$$
Is an annihilator polynomial. The fact that $$X^4+X^3+X^2+X+1$$
is irreducible is a famous result.
Moreover, $\mathbb Q(\omega ^2)\subset \mathbb Q(\omega )$. By the way $X^5+1$ is an annihilator polynomial of $\omega $. With all these remarks, you will have no problem to show that $L=K$ and $[K:\mathbb Q]=4$.
