I will prove the identity $\cos(x+y)=\cos x\cos y-\sin x\sin y$, using with the following definitions of sine and cosine:
$ \sin x:= \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} \ \ \ \ ;\ \ \ \cos x:= \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$
Proof:
$\cos (x+y)= \sum_{n=0}^{\infty}(-1)^n\frac{(x+y)^{2n}}{(2n)!}$
Using the Binomial theorem, we will have
$\sum_{n=0}^{\infty}(-1)^n\sum^{2n}_{k=0}\binom{2n}{k}\frac{x^ky^{2n-k}}{(2n)!}=$ $=\sum_{n=0}^{\infty}(-1)^n\sum^{2n}_{k=0}\frac{x^ky^{2n-k}}{k!(2n-k)!}$
Now, separating the inner sum into two, for even $k$ and for odd $k$:
$=\sum_{n=0}^{\infty}(-1)^n\sum^{n}_{k=0}\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}+\sum_{n=1}^{\infty}(-1)^n\sum^{n-1}_{k=0}\frac{x^{2k+1}y^{2n-2k-1}}{(2k+1)!(2n-2k-1)!}$
Now, let us look on the first sum,
$\sum_{n=0}^{\infty}(-1)^n\sum^{n}_{k=0}\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}=$ $=\sum_{n=0}^{\infty}\sum^{n}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}(-1)^{n-k}\frac{y^{2(n-k)}}{(2(n-k))!}=$
By Cauchy product, we have:
$=\sum_{n=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}\sum_{n=0}^{\infty}(-1)^k\frac{y^{2k}}{(2k)!}=$
$=\cos x\cos y$
For the second sum,
$\sum_{n=1}^{\infty}(-1)^n\sum^{n-1}_{k=0}\frac{x^{2k+1}y^{2n-2k-1}}{(2k+1)!(2n-2k-1)!}=$
By Cauchy product, we have:
$\sum_{n=1}^{\infty}\sum^{n-1}_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}(-1)^{n-k}\frac{y^{2((n-1)-k)+1}}{(2((n-1)-k)+1)!}$
And by substituting $t=n-1$, we will have:
$\sum_{t=0}^{\infty}\sum^{t}_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}(-1)^{t+1-k}\frac{y^{2(t-k)+1}}{(2(t-k)+1)!}$
$=-[ \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!} ][\sum_{k=0}^{\infty}(-1)^k\frac{y^{2k+1}}{(2k+1)!}] $
$=-\sin x\sin y$
Q.E.D