The function
sin(pi x)/(pi x)
is zero at the nonzero integers, so it should be equal to:
sin(pi x)/(pi x) = product over n of (1-x^2/n^2)
The right hand side is a product that converges and it is zero when x isa nonzero integer. The normalization is correct, because for x = 1 it is 1 while the limit for x to 1 of the left hand side is also 1. So, the above indentity looks correct (but you can't rigorously prove that, Euler only conjectured the identity).
The coefficient of x^2 of the left hand side is:
-pi^2/6
And from the right hand side it is minus the sum of 1/n^2 from n = 1 to infinity. To get an x^2 term, you need to take it from one factor of the infinite product, say the nth, and then you need to take the 1 from all other factors. You then get -1/n^2, and all n from 1 to infinity contribute.
Now, you can just as well do this using cos(pi x). The zeroes are at x = (n+1/2), so you would conjecture that:
cos(pi x) = Product over n of [1-x^2/(n+1/2)^2]
Normalization is correct as can be seen from putting x = 1 on bith sides. Extracting the coefficient of x^2 gives:
-pi^2/2 = -Sum over n from n= 0 to infinity of 1/(n+1/2)^2
You can rewrite this as:
sum over n 1/(2n+1)^2 = pi^2/8
You then use the following trick. If we put:
Zeta(2) = sum from n = 1 to infinity of 1/n^2
Then clearly the sum of the inverse squares of the even numbers only is:
sum from n = 1 to infinity of 1/(2n)^2 = 1/4 Zeta(2)
So, the sum over only the inverse squares if the odd numbers must be
3/4 Zeta(2). So we have:
pi^2/8 = 3/4 Zeta(2) -------->
Zeta(2) = pi^2/6.