Is there a way to simplify $ \sum_{i=0}^k \cos^2\left(\frac{\pi}{2}+\left\lfloor\frac{n}{b^i}\right\rfloor \frac{2 \pi}{b}\right) $?

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Sorry for my poor English language, but I really need help. I would like to simplify the sum down in terms of: $n$, $k$, and $b$. $$ \sum_{i=0}^k \cos^2\left(\frac{\pi}{2}+\left\lfloor\frac{n}{b^i}\right\rfloor \frac{2 \pi}{b}\right) $$ Another way to write the same sum is, if $n$ in radix $ b$ is: $$ n=\sum_{j=0}^k a_i\cdot b^i $$

So I can write same cosines sum as:

$$ \sum_{i=0}^k \cos^2{\left(\frac{\pi}{2}+a_i \frac{2 \pi}{b}\right)} $$

The way I would like to write the first summation is as a function of a constant times a cosine of some angle.