Finding the Taylor series for sin(x)^2 by squaring the sin(x) series

[myusernameis]

Homework Statement

sin(x)= sum((-1)^k* (x^(2k+1)/(2k+1)!))k=0 to infinity

Homework Equations

so if i want to find sin(x)^2, (not sin(x^2), that would be easier though...)

The Attempt at a Solution

then...
do i square the whole thing, like this?

sum(((-1)^k* (x^(2k+1)/(2k+1)!))^2)k=0 to infinity

thanks a bunch!

 

[Dick]

You have to square the whole series (x-x^3/3!+x^5/5!-...)*(x-x^3/3!+x^5/5!-...). It's not just the sum of the squares of each term. It's a double sum. There are cross terms. It's easy enough to find the first few terms that way.

 

[myusernameis]

You have to square ... that way.

ahhh thanks for answering both of my questions!

good night!