Show that $ \frac{{d}}{dx}\sin(x) = \cos(x)$, using summation forms. It is well-known that: $ \frac{{d}}{dx}\sin(x) = \cos(x)$. [statement (1)]
Given the definitions:
$$\sin(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}$$
And:
$$\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} x^{2n}$$
Can you show that statement (1) is valid?

Expanding the first one:
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$$
Calculate its derivative:
$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}...$$
Then calculate the summation form. You're done.
