What is the Maclaurin series for f(x)=sin2(x)f(x) = \sin^2(x)f(x)=sin2(x) using sin2(x)=1−cos(2x)2\sin^2(x) = \frac{1-\cos(2x)}{2}sin2(x)=21−cos(2x)?
∑n=1∞(−1)n−122n−1x2n(2n)!\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{2n-1} x^{2n}}{(2n)!}∑n=1∞(2n)!(−1)n−122n−1x2n
∑n=1∞(−1)n22nx2n(2n)!\sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!}∑n=1∞(2n)!(−1)n22nx2n
∑n=1∞22n−1x2n(2n)!\sum_{n=1}^{\infty} \frac{2^{2n-1} x^{2n}}{(2n)!}∑n=1∞(2n)!22n−1x2n
∑n=1∞(−1)n−1x2n2n\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{2n}∑n=1∞2n(−1)n−1x2n