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 cosnx 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

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