This is one of my favourite exercises from Jacobson so far. Let $t$ be transcendental over $\C$, and $u$ be an element such that $t^2+u^2=1$.
Theorem. $\C(t^n,u^n)/\C(t,u)$ is normal. The Galois group is trivial if $n$ is odd and $V_4$ if $n$ is even.
Proof. The extension is the splitting field of the polynomial $(x^n-t^n)(x^n-u^n)$ since $\C$ contains all $n$th roots of unity, thus it is normal. For $n$ odd, consider the following lattice:
$$\xymatrix{ & \C(t,u) & \\ & \C(t,u^n)\ar@{-}[u] & \\ \C(t^n,u^n)\ar@{-}[ur] & & \C(t)\ar@{-}[ul] \\ \C(t^n,u^{2n})\ar@{-}[u]\ar@{-}[urr] & & \\ & & \C(t^n)\ar@{-}[ull]\ar@{-}[uu] }$$
We have $\C(t,u^n)=\C(t,u)$ since $u^n/u$ is a power of $u^2\in\C(t)$. And $[\C(t,u):\C(t)]=2$, since $u$ satisfies the quadratic $x^2+(t^2-1)$ which has no roots in $\C(t)$ and is hence irreducible. Thus $[\C(t,u^n):\C(t)]=2$ as well.
Next, I show that $[\C(t):\C(t^n)]=n$, which implies $[\C(t,u^n):\C(t^n)]=2n$. Clearly the degree is at most $n$. Suppose $p(x)\in\C(t^n)[x]$ has degree $<n$ and $p(t)=0$. Multiplying common denominators, we obtain $q(t)=0$ for some $q(x)\in\C[t^n][x]$. Expanding this out gives
$$q_{n-1}(t^n)t^{n-1} + q_{n-2}(t^n)t^{n-2} + \ldots + q_0(t^n) = 0,\quad q_i(x)\in\C[x].$$
The reader can convince himself that all the $q_i(x)$'s are forced to be 0, so $p(x)=q(x)=0$.
Alternatively, we can factor the extension into a tower of subextensions with prime degrees $\C(t)/\C(t^{p_1})/\C(t^{p_1p_2})/\ldots$, where $n=\prod_{i=1}^n p_i$. The degree $[\C(t):\C(t^{p_1})]$ is $p_1$, since $x^{p_1}-t^{p_1}$ either has a root in $\C(t^{p_1})$ or is irreducible, and the former cannot be true. Likewise for $[\C(t^{p_1}):\C(t^{p_1p_2})]$ and so on.
For the reader's sake, here is the lattice with the aforementioned degrees filled in:
$$\xymatrix{ & \C(t,u) & \\ & \C(t,u^n)\ar@{-}[u]_1 & \\ \C(t^n,u^n)\ar@{-}[ur] & & \C(t)\ar@{-}[ul]_2 \\ \C(t^n,u^{2n})\ar@{-}[u]\ar@{-}[urr] & & \\ & & \C(t^n)\ar@{-}[ull]\ar@{-}[uu]_n }$$
It remains to determine $[\C(t^n,u^n):\C(t^n,u^{2n})]$ and $[\C(t^n,u^{2n}):\C(t^n)]$, which will allow us to determine the desired degree $[\C(t,u^n):\C(t^n,u^n)]=[\C(t,u):\C(t^n,u^n)]$. The first degree is obviously $\le2$, and it is $>1$ since $\C(t^n,u^{2n})\subset\C(t)$ whereas $\C(t^n,u^n)$ isn't. Thus it is 2.
As for the second degree, observe that $\C(t)/\C(t^n)$ is Galois with automorphisms generated by $t\mapsto \zeta t$, where $\zeta$ is a primitive $n$th root of unity. And $\C(t^n,u^{2n})/\C(t^n)$ is a subextension since $u^{2n}=(1-t^2)^n\subset \C(t^n,t^2)=\C(t)$. Observe that $(1-\zeta^2 t^2)^n \neq (1-t^2)^n$, since by virtue of $\C[t]$ being a UFD this implies $1-\zeta^2t^2 = 1-t^2$, a contradiction. Therefore, $u^{2n}$ is fixed only by the trivial automorphism, and so $[\C(t^n,u^{2n}):\C(t^n)]=n$.
Here is the completed lattice diagram:
$$\xymatrix{ & \C(t,u) & \\ & \C(t,u^n)\ar@{-}[u]_1 & \\ \C(t^n,u^n)\ar@{-}[ur]^1 & & \C(t)\ar@{-}[ul]_2 \\ \C(t^n,u^{2n})\ar@{-}[u]_2 \ar@{-}[urr]_1 & & \\ & & \C(t^n)\ar@{-}[ull]_n \ar@{-}[uu]_n }$$
The arguments for the even case are quite similar so I won't go through them. The lattice is
$$\xymatrix{ & \C(t,u) \\ & \C(t,u^n)\ar@{-}[u]_2 \\ & \C(t)\ar@{-}[u]_1 \\ & \C(t^2)\ar@{-}[u]_2 \\ \C(t^n,u^n)\ar@{-}[ur]^1 & \\ & \C(t^2)\ar@{-}[ul]_{n/2}\ar@{-}[uu]_{n/2} \\ }$$
Thus $[\C(t,u):\C(t^n,u^n)]=4$. There are degree 2 subextensions of $\C(t,u)/\C(t^n,u^n)$, namely $\C(t)$ and $\C(u)$, so the Galois group is $V_4$. The proof is done. $\qed$
As a corollary, the element
$${(t+iu)^n+(t-iu)^n\over 2}$$
lies in $\C(t^n,u^n)$ for all $n$. If $n$ is odd, this is trivial. If $n$ is even, it is fixed by the automorphisms $t\mapsto -t, u\mapsto u$ and $t\mapsto t, u\mapsto -u$ of $\C(t,u)/\C(t^n,u^n)$. In particular,
$$\cos nx = {(\cos x+i\sin x)^n+(\cos x-i\sin x)^n\over 2}$$
can be expressed in terms of $\cos^n x$ and $\sin^n x$. A little bit of experimentation shows this is not obvious at all! Courtesy of Ariana, here's an expression for $n=3$:
$$\cos3x={\cos^3x\left(-2+5\cos^6x-7\sin^6x\right) \over 1+2\cos^6x-\sin^6x}.$$