Maclaurin polynomial estimating $\sin 15°$

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I would like to know how I would determine the Maclaurin polynomial which will estimate $\sin15°$ with an error less than $10^{-4}$.

Maybe we have to use the formula $$R_n(x) = f^{(n+1)}(c) \, {(x-x_0)^{n+1}\over(n+1)!}$$ where $R_n(x)$ is the remainder of Taylor's formula.

I have no idea how to solve this question. Any help would be appreciated. Thanks!

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The Maclaurin series is $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n\cdot x^{2n+1}}{(2n+1)!}$$

$15°$ corresponds to $x=\frac{15\pi}{180}$. Calculate $n$ such that the remainder $R_n(x)$ has absolute value smaller than $10^{-4}$. This is the point where you can truncate the Maclaurin series. Plug in $x$ into the remaining polynomial.