# Power Series Expansion for Cosine Function

## Theorem

The cosine function has the power series expansion:

 $\ds \cos x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ $\ds$ $=$ $\ds 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots$

valid for all $x \in \R$.

## Proof

$\dfrac \d {\d x} \cos x = -\sin x$
$\dfrac \d {\d x} \sin x = \cos x$

Hence:

 $\ds \dfrac {\d^2} {\d x^2} \cos x$ $=$ $\ds -\cos x$ $\ds \dfrac {\d^3} {\d x^3} \cos x$ $=$ $\ds \sin x$ $\ds \dfrac {\d^4} {\d x^4} \cos x$ $=$ $\ds \cos x$

and so for all $m \in \N$:

 $\ds m = 4 k: \ \$ $\ds \dfrac {\d^m} {\d x^m} \cos x$ $=$ $\ds \cos x$ $\ds m = 4 k + 1: \ \$ $\ds \dfrac {\d^m} {\d x^m} \cos x$ $=$ $\ds -\sin x$ $\ds m = 4 k + 2: \ \$ $\ds \dfrac {\d^m} {\d x^m} \cos x$ $=$ $\ds -\cos x$ $\ds m = 4 k + 3: \ \$ $\ds \dfrac {\d^m} {\d x^m} \cos x$ $=$ $\ds \sin x$

where $k \in \Z$.

This leads to the Maclaurin series expansion:

 $\ds \sin x$ $=$ $\ds \sum_{k \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} \map \cos 0 - \frac {x^{4 k + 1} } {\paren {4 k + 1}!} \map \sin 0 - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} \map \cos 0 + \frac {x^{4 k + 3} } {\paren {4 k + 3}!} \map \sin 0}$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} }$ Sine of Zero is Zero, Cosine of Zero is One $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ setting $n = 2 k$

From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$.

$\blacksquare$