Power Series Expansion for Sine Function

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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

From Derivative of Sine Function:

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

From Derivative of Cosine Function:

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


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 $x$.

$\blacksquare$


Also see


Sources