prove that $ \sin(x+y)=\sin{x} \cdot \cos{y} +\sin{y} \cdot \cos{x} $ using power series (without trigonometric identities)

Starting from the line just before "Now I want to change...", \begin{align*} \sum_{n=0}^{\infty}\sum_{k=0}^{n}&\left(\frac{\left(-1\right)^{n}}{\left(2k+1\right)!\left(2n-2k\right)!}x^{2k+1}y^{2n-2k}+\frac{\left(-1\right)^{n}}{\left(2k\right)!\left(2n-2k+1\right)!}x^{2k}y^{2n-2k+1}\right)\\ &=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\sum_{k=0}^{n}\frac{1}{\left(2k+1\right)!\left(2n-2k\right)!}x^{2k+1}y^{2n-2k}+\sum_{k=0}^{n}\frac{1}{\left(2k\right)!\left(2n-2k+1\right)!}x^{2k}y^{2n-2k+1}\right)\\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\sum_{k=0}^{n}\frac{(2n+1)!}{\left(2k+1\right)!\left(2n-2k\right)!}x^{2k+1}y^{2n-2k}+\sum_{k=0}^{n}\frac{(2n+1)!}{\left(2k\right)!\left(2n-2k+1\right)!}x^{2k}y^{2n-2k+1}\right)\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\sum_{k=0}^{2n+1} \binom{2n+1}{k}x^ky^{2n-k+1}\\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(x+y)^{2n+1}. \end{align*}

Here is another approach based on the

Lemma: Let $f:\mathbb {R} \to\mathbb {R} $ be a twice differentiable function satisfying $$f''(x) =-f(x), f(0)=0,f'(0)=0$$ then $f(x) =0$ for all $x$.

The function $f(x) =\operatorname{S} (x+a) - \operatorname {S} (x) \operatorname {C} (a) -\operatorname{C} (x) \operatorname{S} (a) $ satisfies the hypotheses of the lemma above and hence is zero for all $x$.

It remains to prove the lemma without using any idea about circular functions (including the series definitions given above).

The proof is not difficult and can be given by noting that the function $$g(x) =\{f(x) \} ^2 +\{f'(x) \} ^2 $$ is such that $g'(x) =0$ for all $x$ and therefore $g(x) =g(0)=0$ for all $x$. It follows that $f$ vanishes everywhere.

The above proof uses the fact that we can differentiate power series term by term in the interior of region of convergence.