Question

how do you figure out the maclaurin series for sin^2 (x)? can I use the mac series for plain sin x?

Answer 1

its just an algorithm, the \(n'th\) term is given by \(\frac{f^{(n)}(0)}{n!}x^n\)

Answer 2

Recall the identity:
\[\sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x\]
You can use the Maclaurin series for \(\cos x\) if you replace the \(x\) with \(2x\).