# How do you find the maclaurin series expansion of f(x) =sin(3x)?

Jul 31, 2017

sin(3x)=sum_(n=1)^oo ((-1)^n(3x)^(2n-1))/((2n-1)!)

#### Explanation:

First off the Maclaurin series is a special case of the Taylor series where $a = 0$

So a Maclaurin series would be of the form sum_(n=0)^oo ((d^n f)/(dx^n)(0)x^n)/(n!)

Since $\sin \left(x\right)$ is $0$ if $x = 0$, but its derivative $\cos \left(x\right)$ is $1$ when $x = 0$ we need to consider the odd derivatives

So that means we'll have a series like this

sum_(n=1)^oo ((d^(2n-1)f)/(dx^(2n-1))(0)x^(2n-1))/((2n-1)!)

and since the derivatives alternate between positive and negative $1$ we have

sum_(n=1)^oo ((-1)^nx^(2n-1))/((2n-1)!)

Also since we are finding the maclaurin series for $\sin \left(3 x\right)$ instead of $\sin \left(x\right)$ we'll get

sin(3x)=sum_(n=1)^oo ((-1)^n(3x)^(2n-1))/((2n-1)!)