## Galois group of C(t,u) over C(t^n,u^n) and application to cosnx

$\mathbf{Problem:}$ Let $E$ be an extension of $\mathbb{C}$ such that $E=\mathbb{C}(t,u)$ where $t$ is transcedental over $\mathbb{C}$ and $u$ satisfies the equation $u^2+t^2=1$ over $\mathbb{C}(t)$. Find the Galois group of $\mathbb{C}(t,u)$ over $\mathbb{C}(t^n,u^n)$ for any $n\in \mathbb{N}$ Show that

$\displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n], i=\sqrt{-1}$

is contained in $\mathbb{C}(t^n,u^n)$. Use this to prove that the function cos$nx$ is expressible rationally with complex coefficients in terms of $\cos^nx$ and $\sin^n x$. Does this hold for sin $nx$.
$\mathbf{Proof:}$ $\forall\, \eta\in \text{Gal }\mathbb{C}(t,u)/\mathbb{C}(t^n,u^n)$, we must have

$\displaystyle\eta(t^n)=\eta^n(t)=t^n$    (1)
$\displaystyle\eta(u^n)=\eta^n(u)=u^n$    (2)
$\displaystyle\eta^2(t)+\eta^2(u)=1$      (3)

This means $(\eta(t)/t)^n=1$. So there exist $w_1$, a n-th root of 1, such that $\eta(t)=w_1t$. Similarly $\exists\, w_2$ such that $\eta(u)=w_2u$. (3) becomes

$w^2_1t^2+w^2_2u^2=1$. Combing with $t^2+u^2=1$, we get $w^2_1=w^2_2=1$. Since $w^n_1=w^n_2=1$, we know that when $n$ is even $w_1=w_2=1$ and when $n$ is odd $w_1=\pm 1$ and $w_2=\pm 1$.

In conclusion, the Galois group is trivial when $n$ is odd and Klein group when $n$ is even.

No matter $n$ is even and odd, $\displaystyle u_n=\frac{1}{2} [(t+iu)^n+(t-iu)^n]$ is invariant, so $u_n$ is in $\mathbb{C}(t^n,u^n)$.

If $n$ is odd, $\displaystyle v_n=\frac{1}{2i}[(t+iu)^n-(t-iu)^n]$ is in $\mathbb{C}(t^n,u^n)$.  If $n$ is even, $v_n$ is not in the ground field, because $\eta(t)=t, \eta(u)=-u$ can not fix $v_n$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ At first I am surprised by the result I got. I can not believe that the galois is trivial when $n$ is odd.  So I conculted to Zhuohui. Thank him for his kind help. I drew the following picture after discussion with him.

case n=4

case n=3