Usbrth

Using $$z^2n+1 = prod_k=1^nleft(z^2-2zcosleft(fracleft(2k-1right)pi2nright)+1right)$$
and $$z^2n+1 = left(1+z^2right)left(1-z^2+z^4-cdots+z^2n-2right)$$
prove, for $n$ odd, $$cos^2left(fracpi2nright) cos^2left(frac3pi2nright) cos^2left(frac5pi2nright) cdots cos^2left(fracleft(n-2right)pi2nright) = n2^1-n$$

The provided solution (from the original source of this problem) involves noting that in the first result, the product will contain a factor of $left(1+z^2right)$, where $k=fracn+12$ (which is an integer since $n$ is odd). Thus, substituting in the second result, they cancel the factor of $left(1+z^2right)$ on both sides, and then set $z=i$ to deduce the required result.

But my question is, since $z=i$ gives $1+z^2=0$, are we allowed to make this cancellation and then still set $z=i$, since we seem to be dividing by zero? If there is in fact a problem, can their proof be easily repaired?

There is no problem; you can view the identity as an equality of polynomials in $BbbC[z]$. This is an integral domain, meaning that you can cancel non-zero factors; the factor $1+z^2$ is a non-zero polynomial, so you can cancel it.

This gives you a simpler equality of polynomials. You can then evaluate these two polynomials in $z=i$ to obtain the given identity.

More formally; combining the given expressions for $z^2n+1$ shows that
begineqnarray*
0&=&left(1+z^2right)left(1-z^2+z^4-cdots+z^2n-2right)
-prod_k=1^nleft(z^2-2zcosleft(fracleft(2k-1right)pi2nright)+1right)\
&=&(1+z^2)left(1-z^2+z^4-cdots+z^2n-2
-prod_k=1^n-1left(z^2-2zcosleft(fracleft(2k-1right)pi2nright)+1right)right),\
endeqnarray*
as polynomials. This means one of the two factors is the zero polynomial. Clearly this is not $1+z^2$, so
$$1-z^2+z^4-cdots+z^2n-2
-prod_k=1^n-1left(z^2-2zcosleft(fracleft(2k-1right)pi2nright)+1right)=0,$$
which shows that indeed
$$1-z^2+z^4-cdots+z^2n-2
=prod_k=1^n-1left(z^2-2zcosleft(fracleft(2k-1right)pi2nright)+1right).$$
Now plugging in $z=i$ yields the desired identity.