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

Question

as we know $ S\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}x^{2n+1} $

and

$ C\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}x^{2n} $

the power series of sinx and cosx.

I want to prove that : $ S\left(x+y\right)=S\left(x\right)C\left(y\right)+S\left(y\right)C\left(x\right) $

What ive done:

let $ x,y \in \mathbb{R} $ So

$ S\left(x+y\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\left(x+y\right)^{2n+1}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\sum_{k=0}^{2n+1}\binom{2n+1}{k}x^{k}y^{2n-k+1}=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\sum_{k=0}^{2n+1}\frac{\left(2n+1\right)!x^{k}y^{2n-k+1}}{\left(2n+1\right)!\left(k!\right)\left(2n-k+1\right)!}\right)=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\sum_{k=0}^{n}\frac{x^{k}y^{2n-k+1}}{k!\left(2n-k+1\right)!}+\sum_{k=n+1}^{2n+1}\frac{x^{k}y^{2n-k+1}}{k!\left(2n-k+1\right)!}\right) $

and also

$ S\left(x\right)C\left(y\right)+S\left(y\right)C\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}x^{2n+1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}y^{2n}+\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}y^{2n+1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}x^{2n} $

we know that the series absolutely converge, so we can use cauchy's product. so

$ \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}x^{2n+1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}y^{2n}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}x^{2k+1}\frac{\left(-1\right)^{n-k}}{\left(2n-2k\right)!}y^{2n-2k}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{\left(-1\right)^{n}}{\left(2k+1\right)!\left(2n-2k\right)!}x^{2k+1}y^{2n-2k} $

and

$ \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}y^{2n+1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}x^{2n}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{\left(2k\right)!}x^{2k}\frac{\left(-1\right)^{n-k}}{\left(2n-2k+1\right)!}y^{2n-2k+1}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{\left(-1\right)^{n}}{\left(2k\right)!\left(2n-2k+1\right)!}x^{2k}y^{2n-2k+1} $

if we sum the last 2 equations we get:

$ \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)= $

now i want to change the summation index to make it closer to the expression of $ S(x+y) $ so:

$ \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=n+1}^{2n+1}\frac{1}{\left(2\left(k-n-1\right)\right)!\left(2n-2\left(k-n-1\right)+1\right)!}x^{2\left(k-n-1\right)}y^{2n-2\left(k-n-1\right)+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=n+1}^{2n+1}\frac{1}{\left(2k-2n-2\right)!\left(4n-2k+3\right)!}x^{2k-2n-2}y^{4n-2k+3}\right)=\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(\sum_{k=0}^{n}\frac{x^{k}y^{2n-k+1}}{\left(k!\right)\left(2n-k+1\right)!}\cdot\frac{x^{k+1}\left(k!\right)\left(2n-k+1\right)!}{y^{k+1}\left(2k+1\right)!\left(2n-2k\right)!}+\sum_{k=n+1}^{2n+1}\frac{x^{k}y^{2n-k+1}}{\left(k!\right)\left(2n-k+1\right)!}\cdot\frac{x^{k}y^{2n+2}\left(k!\right)\left(2n-k+1\right)!}{x^{2n+2}y^{k}\left(2k-2n-2\right)!\left(4n-2k+3\right)!}\right)$

Now im stuck. i dont know how to continue. Any ideas will help, thanks.

Answer 1

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*}

Answer 2

For any complex $z$ we have $\displaystyle{e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!}}$. As is well-known, this implies $e^{iz}=\cos(z)+i\sin(z)$. Then $$ \cos(x+y)+i \sin(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y)) $$ $$ \cos(x+y)+i \sin(x+y)=\bigg(\cos(x)\cos(y)-\sin(x)\sin(y)\bigg)+i\bigg(\cos(x)\sin(y)+\sin(x)\cos(y)\bigg) $$

Another way using the derivative formulas (hopefully not circular; pun intended) $S'=C$, $C'=-S$. For a sufficiently nice function $f(x)$, we have $$ f(x+y) = \sum_{k=0}^{\infty}\frac{f^{k}(x)}{k!}y^k $$Put $f(x)=\sin(x)$. Grouping into even and odd derivatives, we have $$ \sin(x+y) = \sum_{k=0}^{\infty}\frac{\sin(x) (-1)^k}{(2k)!}y^{2k}+\sum_{k=0}^{\infty}\frac{\cos(x) (-1)^k}{(2k+1)!}y^{2k+1} $$ $$ = \sin(x)\sum_{k=0}^{\infty}\frac{ (-1)^k}{(2k)!}y^{2k}+\cos(x)\sum_{k=0}^{\infty}\frac{ (-1)^k}{(2k+1)!}y^{2k+1} = \sin(x)\cos(y)+\cos(x)\sin(y) $$

Answer 3

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.