Could we show $1-(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots)^2=(1-\frac{x^2}{2!}+\frac{x^4}{4!}- \dots)^2$ if we didn't know about Taylor Expansion? Suppose that humanity haven't discovered Taylor Series Expansion of trigonometric functions or of any function that would help us on this. Which means we are not allowed to replace the given infinite series sums with the corresponding $\sin x$ and $\cos x$ functions.
Could we still show that the following identity is true for all real $x$?
$$ 1 - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots \right)^2 = \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right)^2 $$
 A: This boils down to showing $$\tag1\exp(x+y)=\exp(x)\exp(y),$$ where $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$, which is more or less straightforward: On the left hand side, the coefficient of $x^ny^m$ is ${n+m\choose n}\cdot\frac1{(n+m)!}$ and on the right it is $\frac1{n!}\cdot\frac1{m!}$, and these expressions are equal.
Now we find that for real (or in fact arbitrary) $x$ we have 
$$\tag2\exp(ix)\exp(-ix)=\exp(0)=1.$$ If we denote the expression in parentheses on the left of your equation with $s(x)$ and on the right with $c(x)$, we see that $\exp(\pm ix)=c(x)\pm is(x)$, so that $(2)$ becomes
$$c^2(x)+s^2(x)=1.$$
A: A note on mercio's solution:
The proof depends on showing that
$\sum_{i=0}^k \binom {2k}{2i} = \sum_{j=0}^{k-1} \binom {2k}{2j+1}$.
This can be rewritten as
$\sum_{i=0}^{2k} (-1)^i \binom {2k}{i}
=0
$
and this is the expansion of
$0 = (1-1)^{2k}$
by the binomial theorem.
As mercio says,
this is true for any $k$,
not just even $k$,
since
$0 = (1-1)^{k}$.
A: As Andres says in his comment, you can compare the coefficients of $x^{2k}$ in both quantities.
On the left-hand side, this coefficient is $(-1)^k \sum_{i+j=k-1} \frac 1 {(2i+1)!(2j+1)!}$, while on the right-hand side, it is $(-1)^k \sum_{i+j=k} \frac 1{(2i)!(2j)!}$.
Multiplying both coefficients by $(2k)!$, you are left to prove $\sum \binom {2k}{2i} = \sum \binom {2k}{2j+1}$, i.e. that the number of even subsets of a set of size $2k$ is the number of odd subsets.
This is in fact true for any nonempty set, not only those of even size.
To show this combinatorially, pick a distinguished element $x$ of a set $S$ of size $n$. Then given a subset $X \subset S$, map it either to $X \setminus \{ x \}$ or to $X \cup \{x \}$ according to wether $x \in S$ or not. Then you can check that this is a bijection between odd-sized subsets of $S$ and even-sized subsets of $S$.
