# Power Series Expansion for Sine Function

## Theorem

The sine function has the power series expansion:

 $\displaystyle \sin x$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!}$ $\displaystyle$ $=$ $\displaystyle x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots$

valid for all $x \in \R$.

## Proof

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

Hence:

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

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

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

where $k \in \Z$.

This leads to the Maclaurin series expansion:

 $\displaystyle \sin x$ $=$ $\displaystyle \sum_{r \mathop = 0}^\infty \left({\frac {x^{4 k} } {\left({4 k}\right)!} \sin \left({0}\right) + \frac {x^{4 k + 1} } {\left({4 k + 1}\right)!} \cos \left({0}\right) - \frac {x^{4 k + 2} } {\left({4 k + 2}\right)!} \sin \left({0}\right) - \frac {x^{4 k + 3} } {\left({4 k + 3}\right)!} \cos \left({0}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{r \mathop = 0}^\infty \left({\frac {x^{4 k + 1} } {\left({4 k + 1}\right)!} - \frac {x^{4 k + 3} } {\left({4 k + 3}\right)!} }\right)$ Sine of Zero is Zero, Cosine of Zero is One $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!}$ setting $n = 2 k$

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

$\blacksquare$