Nulling filter coefficients I'm wondering if I have calculated a nulling filter coefficients correctly:

$\ 1, -2\cos(0.44\pi), 1$ 

I have to make a nulling filter that filters out the frequencies $\theta = 0.44\pi$. I've been told that the formula for nulling filters is $y[n] = x[n] - 2\cos(\theta)*x[n-1]+x[n-2]$. 

If I'm correct the two 1s come from the fact that $\ x[n]$ and $\ x[n-2]$ both have a coefficient of 1 and no other variable and the $\ -2\cos(0.44\pi)$ comes from the $\ - 2\cos(\theta)*x[n-1]$ and the fact that $\theta$ = 0.44$\pi$.

If the coefficients are correct, is all I have to do to find the output signal just multiply the input signal by each coefficient and add them together (convolution sum)?
You are looking at an FIR filter whose transfer function is 

$$
\begin{align}
H(z)&amp;=(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})\\[10pt]
&amp;=1-(e^{-j\omega_0}+e^{j\omega_0})z^{-1}+z^{-2}\\[10pt]
&amp;=1-2\cos(\omega_0)z^{-1}+z^{-2}
\end{align}$$

So as you can also see in the comment, there are two zeros at $e^{\pm j\omega_0}$ that null the frequencies $\pm\omega_0$.  

Obviously, here we have $\omega_0=0.44\pi$. The filter's response is 



The impulse response is $$h[n]=\delta[n]-2\cos(0.44\pi)\delta[n-1]+\delta[n-2]$$

Assume an arbitrary input $x[n]$. The output $y[n]$ is given by:

$$y[n]=x[n]*h[n]$$ Therefore $$y[n]=x[n]-2\cos(0.44\pi)x[n-1]+x[n-2]$$