Find the Maclaurin series for f(x)=sin2(x)f(x) = \sin^2(x)f(x)=sin2(x) using trigonometric identities.
∑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=0∞(−1)n22nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!}∑n=0∞(2n)!(−1)n22nx2n
∑n=1∞(−1)n−12nx2n(2n)!\sum_{n=1}^{\infty} \frac{(-1)^{n-1} 2^{n} x^{2n}}{(2n)!}∑n=1∞(2n)!(−1)n−12nx2n
∑n=1∞(−1)n22n−1x2n(2n)!\sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n-1} x^{2n}}{(2n)!}∑n=1∞(2n)!(−1)n22n−1x2n