Question

How do you find the maclaurin series expansion of `f(x) = sin(pi*x/2)`?

Answer 1

`sin((pix)/2) = sum_(n=0)^oo (-1)^n ((pi)/2)^(2n+1)x^(2n+1)/((2n+1)!)`

Explanation:

You can just start from the MacLaurin series for `sint`:

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

and calculate it for `t= (pix)/2`:

`sin((pix)/2) = sum_(n=0)^oo (-1)^n ((pix)/2)^(2n+1)/((2n+1)!)=sum_(n=0)^oo (-1)^n ((pi)/2)^(2n+1)x^(2n+1)/((2n+1)!)`