= Solution
The trigonometric addition formula gives
$$
\sin((n+1)\theta)+\sin((n-1)\theta)=2\cos\theta\sin(n\theta).
$$
Dividing by $\sin\theta$ and writing $x=\cos\theta$ gives the <recurrence relation>
$$
\boxed{y_{n+1}=2xy_n-y_{n-1}.}
$$
The first two members are $y_1=1$ and $y_2=2x$. If $y_n$ has degree $n-1$ and leading coefficient $2^{n-1}$, then $2xy_n$ has degree $n$ and leading coefficient $2^n$, whereas $y_{n-1}$ has lower degree and cannot cancel that term. <Mathematical induction> proves
$$
\boxed{\deg y_n=n-1,\qquad\text{leading coefficient of }y_n=2^{n-1}.}
$$
For example the sequence begins $1,2x,4x^2-1,8x^3-4x$, agreeing with the <Chebyshev polynomial of the second kind>.
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