Show that $ 2 \sinh(z)=\exp(z)-\exp(-z)$

Question

$ 2 \sinh(z)=\exp(z)-\exp(-z)$;

$ 2 \cosh(z)=\exp(z)+\exp(-z)$

where $z \in \mathbb{C} $

$$\sin(z) := \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1!} z^{2k+1}$$

$$\cos(z) := \sum_{k=0}^{\infty}\frac{(-1)^k}{2k!} z^{2k}$$

I guess that I have to use a trigonometrical identity, but I don't know which one and is that step equivalent to use it for hyperbolic functions? Thank in advance for your help.

Answer 1

It is hard to answer this question without knowing what definitions you are using for your functions. However, if you use the series expansions, then

$$\exp(z)-\exp(-z)=\sum_{n=0}^\infty \left[\frac{z^n}{n!}(1-(-1)^n)\right]$$

Note that at even $n$ the expression in the brackets is $0$ and at odd $n$ it is $2\frac{z^n}{n!}$. Then the expression simplifies to

$$=2\sum_{n\text{ odd}}^\infty \frac{z^{n}}{n!}=2\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}=2\sinh(z)$$

You can do a similar manipulation to get the identity involving $\cosh(z)$. Again, I want to emphasize that this method only works if you have previously shown that these functions have these series expansions. Without knowing how you define the functions, this is one possible way to show the identities in question.