Finding the Taylor Series for y(x)=sin^2x

[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 can't keep doing derivatives on this function till the n'th member

how to solve this??

 

[Mark44]

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

 

[HallsofIvy]

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.

 

[transgalactic]

what about the remainder ob the multiplication??