Which real numbers are contained in $\mathbb{Q}(\cos40+i\sin40)$?
I have a field extension $\mathbb{Q}(\cos40+i\sin40)$. I would like to find all the real numbers in this field and prove that they form a subfield (S) and find $|S:\mathbb{Q}|$. And also calculate the splitting field of S.
I know that $\mathbb{Q}(\cos40) \leq \mathbb{Q}(\cos40+i\sin40) = \mathbb{Q}(\cos40, i\sin40)$. And that $|\mathbb{Q}(\cos40):\mathbb{Q}|=3$. And $\mathbb{Q}(\cos40)$ only contains real numbers.
($\cos40+i\sin40$ is a primitive 9th root of unity)
My assumption is that S = $\mathbb{Q}(\cos40+\sin40)$. I have to show that $\cos40+\sin40$ could be written with the help of elements from $\mathbb{Q}(\cos40+i\sin40)$. I believe I could calculate these in question if I knew $S$. I would be grateful if you could help me.
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$\begingroup$In their series of comments, @diracdeltafunk seems to have led you almost all the way to the full answering of your questions. To cut to the chase, your number, call it $\zeta$, is a primitive ninth root of unity, and thus is a root of the Cyclotomic Polynomial $\Phi(X)=X^6+X^3+1$, known to be irreducible. If we set $\xi=\zeta+\zeta^{-1}=\zeta+\overline\zeta$, then it’s not at all hard to see that the minimal polynomial of $\xi$ is $X^3-3X+1$.
The Galois group of $\Bbb Q(\zeta)$ over $\Bbb Q$ is isomorphic to $(\Bbb Z/9\Bbb Z)^\times$, a cyclic group generated by $5$. Since a cyclic group has just one subgroup of each possible order, there are only the two proper subfields of $\Bbb Q(\zeta)$, the quadratic field generated by the cube roots of unity (it’s imaginary) and the other, generated by $\xi=2\cos(40^\circ)$.
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