# Taylor series question

1. Feb 11, 2009

### transgalactic

how to find the taylor series for
$$y(x)=\sin^2 x$$
i need to develop a general series which reaches to the n'th member
so i cant keep doing derivatives on this function till the n'th member

how to solve this??

2. Feb 12, 2009

### Staff: Mentor

Multiply the two Taylor series together (or Maclaurin series if that's what you're using).
$$sin^2(x) = (x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)((x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)$$

The first term will be x^2

3. Feb 12, 2009

### HallsofIvy

Staff Emeritus
Why can't you "keep doing derivatives on this function till the n'th member"?

y= sin^2(x)
y'= 2sin(x)cos(x)= sin(2x)
and the rest is easy.

4. Feb 14, 2009

### transgalactic

what about the remainder ob the multiplication??