Question

How do you find the Maclaurin Series for `sin(3x)`?

Answer 1

`sin3x= sum_(n=0)^oo (-1)^n ((3x)^(2n+1))/((2n+1)!)`

Explanation:

You can use the development in MacLaurin series of `sint`

`sint= sum_(n=0)^oo (-1)^n (t^(2n+1))/((2n+1)!)`

and substitute `t=3x`

`sin3x= sum_(n=0)^oo (-1)^n ((3x)^(2n+1))/((2n+1)!)`