# Definitions of sin and cos using the exponent

I follows these steps:

1. Define $$e^z:=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$.
2. Show thatthe series is absolutely convergent.
3. Define $$\sin(z):=\frac{e^{iz}-e^{-iz}}{2i}$$, and $$\cos(z):=\frac{e^{iz}+e^{-iz}}{2}.$$
4. Derive $$\sin(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}z^{2n+1}, \quad \cos(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}.$$ just by sum manipulation.

Can I now trust that $$\frac{e^{iz}+e^{-iz}}{2}=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}z^{2n+1}$$? Or: is this question equivalent to the following lemma?

Lemma If $$f(x)=\sum a_k$$ and $$g(x)=\sum b_k$$ are abs. conv. then (a) $$\sum ca_k$$ is abs. conv. and $$cf=\sum ca_k$$; (b) $$\sum (a_k+b_k)$$ is abs. conv. and $$f+g=\sum (a_k+b_k)$$.

Is the given lemma treated in a standard textbook or is assumed as evident?

(a) If $$\sum_{n=0}^\infty\lvert a_k\rvert$$ converges, then $$\sum_{n=0}^\infty\lvert ca_k\rvert$$ converges too, since it is equal to $$\sum_{n=0}^\infty\lvert c\rvert\lvert a_k\rvert$$.
(b) If both series $$\sum_{n=0}^\infty\lvert a_k\rvert$$ and $$\sum_{n=0}^\infty\lvert b_k\rvert$$ converge, then $$\sum_{n=0}^\infty\lvert a_k+b_k\rvert$$ converges too, since$$(\forall k\in\mathbb Z^+):\lvert a_k+b_k\rvert\leqslant\lvert a_k\rvert+\lvert b_k\rvert$$and $$\sum_{n=0}^\infty\bigl(\lvert a_k\rvert+\lvert b_k\rvert\bigr)$$ converges.