The function $f(x)=\sin (9 x)$ has a Maclaurin series. Find the first 4 nonzero terms in the series, that is write down the Taylor polynomial with 4 nonzero terms.

Step 1 ：The Maclaurin series for \(\sin x\) is given by \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\).

Step 2 ：Substitute \(9x\) into the series for \(x\), we get \(\sin (9x) = 9x - \frac{(9x)^3}{3!} + \frac{(9x)^5}{5!} - \frac{(9x)^7}{7!} + \ldots\).

Step 3 ：The first four nonzero terms of the series are \(9x, -\frac{(9x)^3}{3!}, \frac{(9x)^5}{5!}, -\frac{(9x)^7}{7!}\).

Step 4 ：Simplify these terms, we get \(9x, -\frac{729x^3}{6}, \frac{59049x^5}{120}, -\frac{4782969x^7}{5040}\).

Step 5 ：So, the first four nonzero terms in the Maclaurin series for \(\sin (9x)\) are \(9x, -\frac{729x^3}{6}, \frac{59049x^5}{120}, -\frac{4782969x^7}{5040}\).

Step 6 ：Therefore, the Taylor polynomial with 4 nonzero terms is \(\boxed{9x - \frac{729x^3}{6} + \frac{59049x^5}{120} - \frac{4782969x^7}{5040}}\).