Given sin(x)=∑n=0∞(−1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}sin(x)=∑n=0∞(2n+1)!(−1)nx2n+1, find the series for sin(2x)\sin(2x)sin(2x).
∑n=0∞(−1)n(2x)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n(2x)2n+1
∑n=0∞(−1)n22n+1x2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1} x^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n22n+1x2n+1
2∑n=0∞(−1)nx2n+1(2n+1)!2\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}2∑n=0∞(2n+1)!(−1)nx2n+1
Both (a) and (b)