$\left[\Bbb{Q}\left(\cos\left({2\pi\over n}\right)+i\sin\left({2\pi\over n}\right)\right):\Bbb{Q}\right]=\phi(n)$ How to prove that 

$$\left[\Bbb{Q}\left(\cos\left({2\pi\over n}\right)+i\sin\left({2\pi\over n}\right)\right):\Bbb{Q}\right]=\phi(n)$$

Here $[F(a):F] $ is basically the index of field $F(a)$ over $F$ and $\phi(n)$ is the Euler totient function i.e. $\phi(n)=$ number of positive integers that are less than $n$ and are coprime to $n$.
I am not getting how to bring the coprime thing into play.
 A: Edit: I'll try to make this proof work for OP, who hadn't learned Galois theory yet. This may be over their head at the moment, but soon enough OP will encounter the content of this answer anyway. Hopefully, this will be helpful sneak peek into Galois theory.

Preliminaries:
For convenience, let the characteristic be $0$, which ensures that minimal polynomial doesn't have multiple roots in algebraic closure.
Proposition 1. Let $K$ be finite extension of $k$ . Then $[K:k]$ is equal to the number of $k$-linear embeddings of $K$ into algebraic closure $\overline K$.
Sketch of proof. Basically, you can do induction on the number of generators. Assume, thus, that $K = k(\alpha)$ and $f$ its minimal polynomial. Then any embedding $\sigma\colon K\to\overline K$ is uniquely determined by sending $\alpha$ to its conjugate, and these are all the embeddings. Verify this by noting that $k(\alpha)\cong k[x]/(f)$. Thus, both the number of $k$-linear embeddings and the degree of $K/k$ is equal to $\deg f$, which finishes the base of induction. I will let you reflect on the step of induction yourself, hint being $[K:k]=[K:E][E:k]$. $\tag*{$\square$}$
Proposition 2. Let $K$ be algebraic extension of $k$. It is equivalent:
$1)$ If $\alpha\in K$, then all of its conjugates over $k$ are in $K$.
$2)$ Every $k$-linear embedding of $K$ into its algebraic closure if automorphism of $K$.
Proof. Assume $1)$. If $\sigma\colon K\to \overline K$ is $k$-linear embedding, it sends any $\alpha\in K$ to its conjugate, but by $(1)$, it follows that $\sigma(\alpha)\in K$, i.e. $\sigma(K)\subseteq K$. We still have to prove surjectivity. Let $\alpha\in K$ and $f$ its minimal polynomial over $k$. Let $E$ be extension of $k$ generated by roots of $f$. We have inclusions $k\subseteq E\subseteq K$, by $(1)$. Restrict $\sigma$ on $E$, which is finite dimensional over $k$. Notice that $\sigma$ permutes roots of $f$, and thus $\sigma(E)\subseteq E$. By finite dimensionality, $\sigma$ is automorphism of $E$, which means that there is $\beta\in E$ such that $\sigma(\beta) = \alpha$. This proves surjectivity.
Now assume $(2)$. Take $\alpha\in K$ and take one of its conjugate $\beta$. Then $\sigma\colon \alpha\mapsto\beta$ defines $k$-linear embedding of $k(\alpha)$ into $\overline K$. Since $K$ is algebraic extension, we can extend $\sigma$ to embedding of $K$ into $\overline K$ (if $K$ were finite, then you could just use induction to prove the existence of extension, and in infinite case the existence of such extension is guaranteed by Zorn's lemma). By $(2)$, $\sigma$ is then automorphism of $K$, and hence $\beta\in K$. $\tag*{$\square$}$
Extension $K$ of $k$ that satisfies the above conditions is called normal extension.
Corollary. If $K$ is finite normal extension of $k$, then $[K:k]$ is equal to the number of $k$-linear automorphisms of $K$.
Denote the group of $k$-linear automorhpisms of finite normal extension $K$ of $k$ by $\operatorname{Gal}(K/k)$.

Solution to the problem:
Let $\zeta = e^{2\pi i/n}$. Then $\Bbb Q(\zeta)$ is obviously normal extension since it contains all the powers of $\zeta$ and consequently, all the roots of the minimal polyomial of $\zeta$ over $\Bbb Q$. So, $[\Bbb Q(\zeta):\Bbb Q] = |\operatorname{Gal}(\Bbb Q(\zeta)/\Bbb Q)\,|$ by the above Corollary. 
Now, take $f\in \operatorname{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ and let $g\in\operatorname{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ be its inverse. They are uniquely determined by their action on $\zeta$ and they must send $\zeta$ to its conjugate. Let $k,m\in\Bbb Z$ such that $$f(\zeta)=\zeta^k,\ g(\zeta) = \zeta^m.$$ Compose $f$ and $g$ to get that $\zeta^{mk} = \zeta$ which happens if and only if $mk\equiv 1\pmod n$. But, that means $k$ is invertible in $\Bbb Z/n\Bbb Z$, which is equivalent to $k$ being relatively prime to $n$.
Conversely, it is just as easy to see that if $k$ is relatively prime to $n$, $f:\zeta\mapsto \zeta^k$ is invertible, inverse given by map $\zeta\mapsto\zeta^m$, where $mk\equiv 1\pmod n$.
Thus, $|\operatorname{Gal}(\Bbb Q(\zeta)/\Bbb Q)\,|=\varphi(n)$. The claim follows.
A: I've outlined some of the ideas surrounding the question using roots of unity and cyclotomic polynomials, while disregarding the Galois theory.
By Euler's formula 
$$e^{i\theta}=\cos\left(\theta\right)+i\sin\left(\theta\right)\tag{1}$$
Define $\zeta_n:=\cos\left({2\pi \over n}\right)+i\sin\left({2\pi\over n}\right)$, and so by (1) have $\zeta_n=e^{2\pi i/n}$, where $\zeta_n$ is one of the $n$th roots of unity, that is one of the $n$ solutions to
$$X^n=1\tag{2}$$
Taking the $n$ powers of $\zeta_n$, i.e., 
$$\zeta_n^k=e^{2k\pi i/n}\quad \text{for $1\leq k\leq n$,}\tag{3}$$
there will be $\phi(n)$ primitive roots of unity in $\Bbb{Q}(\zeta_n)$, where $\phi$ is Euler's totient function which gives the number of non-negative integers less than $n$ that are prime to $n$; here primitive means these roots of unity have not appeared as any $m$th root of unity having $m<n$, and for this we require $\gcd(n,k)=1$ in (3). This last part follows since if $\zeta_n^{a}=e^{2a\pi i/n}=1$, then $n\mid a$.
For example $\zeta_8^2=e^{2\cdot2\pi i/8}=e^{\pi i/2}=i=\sqrt{-1}$, and this is not a primitive $8$th root of unity since $\gcd(2,8)=2$, and $i$ has already appeared as a primitive $4$th root of unity, $\zeta_4=e^{2\pi i/4}=e^{\pi i/2}=i$, and this is what primitive means. Hence when we adjoin $\zeta_n$  to $\Bbb{Q}$, there are always $\phi(n)$ primitive roots of unity in the field extension $\mathbb{Q}(\zeta_n)/\Bbb{Q}$, and so the degree of this field extension is
$$[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\phi(n)\tag{4}$$

Another way to think of this is that the degree of the $n$th cyclotomic polynomial $\Phi_n(X)$ has degree $\phi(n)$, where $\Phi_n(X)$ is the monic polynomial whose roots are the primitive roots of unity given by
$$\Phi_n(X)=\prod_{\substack{0\leq a<n \\ \gcd(a,n)=1}}(X-\zeta_n^a)\tag{5}$$
Note we call $\mathbb{Q}(\zeta_n)$ a cyclotomic field as the $n$ roots appear periodically apart by $2\pi i/n$ around the unit circle, cyclotomic meaning roughly 'circle-cutting'. Now the $n$th cyclotomic field $\mathbb{Q}(\zeta_n)$ is formed by adjoining to $\Bbb{Q}$ all the $n$th roots of unity.
Now out of all the $n$th roots of unity we can either have 'new' primitive $n$th roots, otherwise the rest will be primitive $d$th roots for some divisor $d$ of $n$, and we can form
$$X^n-1= \prod_{d\mid n}\Phi_d(X)\tag{6}$$
where the product is taken over all divisors of $n$. It can be shown that 
$$\sum_{d\mid n}\phi(d)=n\tag{7}$$
Each of the $\Phi_d(X)$ are monic, and so since $X^n-1$ is monic with integer coefficients then so are all the $\Phi_d(X)$.
Now it can be shown that if $\zeta_n\in\Bbb{C}$ is a solution to (2), then if prime $p\nmid n$ then $\zeta_n$, and $\zeta_n^p$ share the same minimum polynomial over $\Bbb{Q}$. Hence for each integer $a$ prime to $n$, $a=p_1p_2\dotsm p_m$, where each $p_i\nmid n$. Hence $\zeta_n$, $\zeta_n^{p_1}$, $\zeta_n^{p_1p_2},\dotsc,\zeta_n^{p_1p_2\dotsm p_m}$ share the same minimum polynomial over $\Bbb{Q}$, and this is $\Phi_n(X)$. Hence by adjoining $\zeta_n$ to $\Bbb{Q}$ we have 
$$[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\deg(\Phi_n(X))=\phi(n)\tag{8}$$
The first eight cyclotomic polynomials are:
\begin{align*}
\Phi_1(X)&= X-1\\
\Phi_2(X)&= X + 1\\
\Phi_3(X)&= (X-\zeta_3)(X-\zeta_3^2)=X^2+X+1\\
\Phi_4(X)&=  (X-\zeta_4)(X+\zeta_4^3)=(X-i)(X+i)=X^2+1\\
\Phi_5(X)&=X^4+X^3+X^2+X+1\\
\Phi_6(X)&= (X-\zeta_6)(X-\zeta_6^5)=X^2-X+1\\
\Phi_7(X)&= X^6+X^5+X^4+X^3+X^2+X+1\\
\Phi_8(X)&= (X-\zeta_8)(X-\zeta_8^3)(X-\zeta_8^5)(X-\zeta_8^7)=X^4+1\\
\end{align*}
Hence $\zeta_1=e^{2\pi i/1}=1$ and $\zeta_2=e^{2\pi i/2}=-1$, and so on. Now on using (8) we find the degrees of the first eight cyclotomic field extensions over $\Bbb{Q}$ are:
\begin{align*}
[\mathbb{Q}(\zeta_1):\mathbb{Q}]&=\deg(\Phi_1(X))=\phi(1)=1\\
[\mathbb{Q}(\zeta_2):\mathbb{Q}]&=\deg(\Phi_2(X))=\phi(2)=1\\
[\mathbb{Q}(\zeta_3):\mathbb{Q}]&=\deg(\Phi_3(X))=\phi(3)=2\\
[\mathbb{Q}(\zeta_4):\mathbb{Q}]&=\deg(\Phi_4(X))=\phi(4)=2\\
[\mathbb{Q}(\zeta_5):\mathbb{Q}]&=\deg(\Phi_5(X))=\phi(5)=4\\
[\mathbb{Q}(\zeta_6):\mathbb{Q}]&=\deg(\Phi_6(X))=\phi(6)=2\\
[\mathbb{Q}(\zeta_7):\mathbb{Q}]&=\deg(\Phi_7(X))=\phi(7)=6\\
[\mathbb{Q}(\zeta_8):\mathbb{Q}]&=\deg(\Phi_8(X))=\phi(8)=4
\end{align*} 
To finish check (6) with an example:
\begin{align*}
\Phi_1(X)\Phi_2(X)\Phi_4(X)&= (X-1)(X+1)(X^2+1)\\
&=X^4-1
\end{align*}
Note if $p$ is prime the $p$th cyclotomic polynomial can be written as
$$\Phi_p(X)=\frac{X^p-1}{X-1}=X^{p-1}+X^{p-2}+\dotsb+X+1$$
and this can be shown irreducible in $\Bbb{Q}[X]$ by expanding $\Phi_p(X+1)=\frac{(X+1)^p-1}{X}$ and using Eisenstein's criterion with the prime $p$. 
