Prove that $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ using Cauchy product. Show the correctness of the equality by making use of the Cauchy product of the series $\cos(x)=\sum_{n=0}^{\infty}(-1)^n.\frac{x^{2n}}{(2n)!}$ and $\sin(x)=\sum_{n=0}^{\infty}(-1)^n.\frac{x^{2n+1}}{(2n+1)!}$. I've reached $$\cos(x).\cos(y)=\sum_{n=0}^{\infty}(-1)^n.\sum_{k=0}^{n}\frac{1}{(2k)!(2(n-k))!}.x^{2k}.y^{2(n-k)}$$ and $$\sin(x).\sin(y)=\sum_{n=0}^{\infty}(-1)^n.\sum_{k=0}^{n}\frac{1}{(2k+1)!(2(n-k)+1)!}.x^{2k+1}.y^{2(n-k)+1}$$
and $$\cos(x+y)=\sum_{n=0}^{\infty}(-1)^n.\sum_{k=0}^{2n}\frac{1}{k!(2n-k)!}.x^{k}.y^{2n-k}$$
but I guess I'm missing some steps or got something wrong.
 A: Everything is fine so far. We can start from
\begin{align*}
\cos(x)\cos(y)&=\sum_{n=0}^\infty (-1)^n\sum_{k=0}^n\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}\tag{1}\\
\sin(x)\sin(y)&=\sum_{n=0}^\infty (-1)^n\sum_{k=0}^n\frac{x^{2k+1}y^{2n-2k+1}}{(2k+1)!(2n-2k+1)!}\tag{2}\\
\cos(x+y)&=\sum_{n=0}^\infty (-1)^n\sum_{k=0}^{2n}\frac{x^{k}y^{2n-k}}{k!(2n-k)!}\\
\end{align*}

We obtain
  \begin{align*}
\cos(x+y)&=\sum_{n=0}^\infty (-1)^n\sum_{k=0}^{2n}\frac{x^{k}y^{2n-k}}{k!(2n-k)!}\tag{3}\\
&=\sum_{n=0}^\infty (-1)^n\sum_{k=0}^{n}\frac{x^{2k}y^{2n-2k}}{(2k)!(2n-2k)!}\\
&\qquad\qquad +\sum_{n=1}^\infty (-1)^n\sum_{k=0}^{n-1}\frac{x^{2k+1}y^{2n-2k-1}}{(2k+1)!(2n-2k-1)!}\tag{4}\\
&=\cos(x)\cos(y)+\sum_{n=0}^\infty (-1)^{n+1}\sum_{k=0}^{n}\frac{x^{2k+1}y^{2n-2k+1}}{(2k+1)!(2n-2k+1)!}\tag{5}\\
&=\cos(x)\cos(y)-\sin(x)\sin(y)
\end{align*}
  and the claim follows.

Comment:


*

*In (3) we split the inner sum in even and odd parts. We do so by replacing $k\rightarrow 2k$ and $k\rightarrow 2k+1$.

*In (4) we see the left-hand series is already $\cos(x)\cos(y)$ according to (1). We shift the index $n$ of the right-hand series by one to start from $n=0$.

*In (5) we see the right-hand series is $-\sin(x)\sin(y)$ according to (2).
