Interesting relation derived from the identity $\sin^2 x + \cos^2 x \equiv 1$ Every one knows that
$$\sin^2 x + \cos^2 x \equiv  1.$$
It is also well known that
$$\sin x \equiv  \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n + 1}}{(2n+1)!},\quad\cos x \equiv  \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.$$
One can rewrite the above mentioned identity as
$$\sum_{n, k = 0}^{\infty} \frac{(-1)^{n+k} x^{2(n+k+1)}}{(2n+1)! (2k+1)!} + \sum_{m, l = 0}^{\infty} \frac{(-1)^{m+l} x^{2(m+l)}}{(2m)! (2l)!} \equiv  1.$$
Terms of the $\sin^2 x$ series cancel out all but one (which is equal to 1)  terms of the $\cos^2 x$ series. This fact can be expressed as
$$(-1)^{c-1} x^{2c} \sigma(c) + (-1)^c x^{2c} \gamma(c) = 0,$$
where $n+k+1 = m+l= c \geq 1$ and
$$\sigma(c) = \sum_{n=0}^{c-1} \frac{1}{(2n+1)! (2(c-n)-1)!},\quad\gamma(c) = \sum_{m=0}^{c} \frac{1}{(2m)! (2(c-m))!}.$$
At the moment it is not obvious for me that $$\sigma(c) \equiv \gamma(c).$$
Does any one have an idea how last identity can be proved?
 A: 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.
A: 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
$$\sigma(c) \equiv \gamma(c).$$
