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I'm trying to figure out if for any polynomial $f$, $f(\cos(t),\sin(t))=\sum_{i,j}a_{i,j}\cos(t)^i\sin(t)^j=0$ implies $x^2+y^2-1$ divides $f$.

Trigonometric identities are not exactly central to the class I am taking, so I believe the question simply wants me to assume this, but I'm still curious as to why it is true.

In trying to prove this, I know that through polynomial division one can get $f=qg+r$ where the degree of $x$ in $r$ is 1 or 0 and $r(\cos(t),\sin(t))=0$.

If the degree of $x$ in $r$ is zero we simply have a polynomial in $\sin(t)$, and a polynomial only has finitely many roots whereas $\sin(t)$ can take infinitely many values, so $r=0$.

If the degree of $x$ in $r$ is one, and there is only an $x$ term, one can argue that we can choose $t$ to make all the $\sin$ terms arbitrarily small and the $\cos$ term close to 1, and if the constant term is equal to -1 just stop close to 1, and if it is close to -1 stop at 1.

However, I don't know how to proceed when there are terms like $xy$ in $r$. I was thinking of looking at it as $x$ times a polynomial in $y$ plus a polynomial in $y$, but I'm not sure if that helps.

I'm sure there is a more simple way to do this that I'm not seeing, any hints?

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marked as duplicate by André 3000, Theo Bendit, Eric Wofsey, Travis, Lord Shark the Unknown Mar 12 at 4:51

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Actually I just solved it. If we have a polynomial of degree 1 in $x$ and arbitrary degree in $y$, and split it into $x$ times a polynomial in $y$ plus a polynomial in $y$, looking at $x$ as $\cos(t)$ and $y$ as $\sin(t)$ we see that $\cos(0)=1$ and $\cos(\pi)=-1$ are both roots of the same polynomial, but the polynomial is linear so in $x$ so it can only have 1 root, so $r$ must be zero.

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