TITLE: 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?
QUESTION [12 upvotes]: 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 $x$?
$$ 1 - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots \right)^2 = \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right)^2 $$

REPLY [2 votes]: 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
$0 = (1-1)^{k}$.