# Is $\sin^2+\cos^2-1=0$ the only relation for $\sin$ and $\cos$? [duplicate]

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?

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.