how to compute the signal passing from the low pass filter? I am currently trying to solve this question.

Let $x[n]=\cos(\frac{\pi}{2}n)$ and $h[n]=\frac{1}{5}\text{sinc}(\frac{n}{5})$.
Compute the convolution $y[n]=x[n]∗h[n],$ and write the value of $y[5].$
Hint - Given for the question is to compute the convolution in the frequency domain first.

when I calculated the Fourier transform of $y[n]$, I got
$$Y(e^{jw}) = X(e^{jw}) \times H(e^{jw})$$
$$X(e^{jw}) = (1/2) * [\delta(\omega-\pi/2)+\delta(\omega+\pi/2)]$$
$$H(e^{jw}) = 
     \begin{cases}
       \text{1,} &amp;\quad \text{if } |\omega| &lt;= \pi/5\\
       \text{0,} &amp;\quad \text{otherwise} \\ 
     \end{cases}$$
But what I don't understand is, since the range of $X$ is beyond the range of $H$, how do we multiply the two FT's? I must be misunderstanding the concept somewhere. Could anyone help explain to me how do we find the output?
Low pass filters get the name because they pass signals at low frequencies and attenuate signals at higher frequencies. The definition of low and high depend on the the cut-off frequency of the filter. In your case, the cut-off frequency is $\frac{\pi}{5}$. Your filter is the ideal low pass filter where signals above the cut-off frequency are completely zeroed out. The input signal is a perfect cosine signal so all of its energy is concentrated at one specific frequency. It happens to be at a frequency above the cut-off frequency, so what should happen?