# Derivative of Sine Function/Proof 1

## Theorem

$\map {D_x} {\sin x} = \cos x$

## Proof

From the definition of the sine function, we have:

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

From Radius of Convergence of Power Series over Factorial, this series converges for all $x$.

 $\displaystyle D_x \left({\sin x}\right)$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \left({2n+1}\right) \frac {x^{2n} } {\left({2n+1}\right)!}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n} } {\left({2n}\right)!}$

The result follows from the definition of the cosine function.

$\blacksquare$