# Derivative of Cosine Function/Proof 1

## Theorem

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

## Proof

From the definition of the cosine function, we have:

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

Then:

 $\displaystyle D_x \left({\cos x}\right)$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^n 2n \frac {x^{2n - 1} }{\left({2n}\right)!}$ Power Series is Differentiable on Interval of Convergence $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n - 1} }{\left({2n - 1}\right)!}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^{n+1} \frac {x^{2n + 1} }{\left({2n + 1}\right)!}$ changing summation index $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n + 1} }{\left({2n + 1}\right)!}$

The result follows from the definition of the sine function.

$\blacksquare$