Interesting relation derived from the identity $\sin^2 x + \cos^2 x \equiv 1$

Note that $(2c)!\sigma(c)=\sum_{n=0}^{c-1}\binom{2c}{2n+1}$ while $(2c)!\gamma(c)=\sum_{m=0}^c\binom{2c}{2m}$, so it reduces to the famous result that even-sized subsets of a given nonempty set of finite size are exactly as numerous as the odd-sized subsets.

Incidentally, this mysterious-looking connection of a trigonometry problem to a combinatorics problem makes much more sense if we bring in complex numbers. The Pythagorean identity is then the claim $\exp(z)\exp(-z)=1$ with $z:=ix$, i.e. $\sum_{k+l=n}\frac{(-1)^k}{k!l!}=\delta_{n0}$ if we equate $z^n$ coefficients. Multiplying by $n!$ restates the problem as $\sum_{k+l=n}(-1)^k\binom{n}{k}=0$ for $n>0$, which is just the even-minus-odd calculation.

Here I provide explicit reasoning to make the identity in question obvious (note that I use the notation ${N \choose k} \equiv C(N,k)$).

One can easily note that
$$(2c)! \sigma(c) = \sum_{n = 0}^{c-1} \frac{(2c)!}{(2n+1)! (2c-(2n+1))!} = \sum_{n = 0}^{c-1} \frac{r!} {b_{odd}! (r -b_{odd})!} = \sum_{n = 0}^{c-1} C(r,b_{odd}), $$ where $r= 2c,$ $b_{odd} = 2n+1.$

$$(2c)! \gamma(c) = \sum_{m = 0}^{c} \frac{(2c)!}{(2m)! (2c-2m)!} = \sum_{m = 0}^{c} \frac{r!} {b_{even}! (r - b_{even})!} = \sum_{m = 0}^{c} C(r,b_{even}),$$ where $r = 2c,$ $b_{even} = 2m.$

From the trivial identity $0 \equiv (1 - 1)^{2c}$ one can derive the relation $$0 \equiv (1 - 1)^{2c} = \sum_{d = 0}^{2c} C(2c, d) 1^{2c - d} (-1)^{d} = \sum_{o = 0}^{c-1} C(2c,2o+1) 1^{2c-(2o+1)} (-1)^{2o+1} + \sum_{e = 0}^{c} C(2c,2e) 1^{2c-2e} (-1)^{2e} = -\sum_{o = 0}^{c-1} C(2c,2o+1) + \sum_{e = 0}^{c} C(2c,2e).$$

So $$\sum_{o = 0}^{c-1} C(2c,2o+1) \equiv \sum_{e = 0}^{c} C(2c,2e),$$

and hence

$$\sigma(c) \equiv \gamma(c).$$