I'm trying to show that $\mathbb{R}[\sin t, \cos t]$ consists of all $f$ which has an expression of the form $f(t)=a_0+\sum_{m=1}^n(a_m \cos(mt)+b_m \sin(mt))$ where $a_0,a_m,b_m\in \mathbb{R}$.
Since $e^{it}=\cos t+ i\sin t$, we have $\cos t=\frac{1}{2}(e^{it}+e^{-it})$ and $\sin t=\frac{1}{2i}(e^{it}-e^{-it})$, so $\mathbb{R}[$sin $t$, cos $t]$ is a subring of $\mathbb{C}[e^{it},e^{-it}]$. Can I further conclude that $\mathbb{R}[\sin t, \cos t]$ is a subring of $\mathbb{C}[e^{it}]$? In the polynomial ring we only allow nonnegative powers, so I wasn't sure how to deal with the terms involving $e^{-it}$ to rewrite them as only involving powers of $e^{it}$.
$\sin(t)$ is not a member of the ring $\mathbb{C}[e^{it}]$
Imagine that we could write $\sin(t)$ in the form $\sin(t)=a_0+a_1e^{it}+a_2e^{2it}+...$. By differentiating twice, we would deduce that $-\sin(t)=-a_1e^{it}-4a_2e^{2it}-9a_3e^{3it}-...$. Any attempt to equate coefficients would force all of the coefficients except $a_1$ to be zero. It is not true that $\sin(t)$ can be written in the form $\sin(t)=a_1e^{it}$.
For your original statement, I would suggest that you start by proving that any member of $\mathbb{C}[e^{it},e^{-it}]$ can be written in the form $a_0+\sum_{m=1}^n a_m\cos(mt)+b_m\sin(mt)$ for $a_0,a_m,b_m\in \mathbb{C}$.