maclaurin series for sin(2x): show that it converges to sin2x for all x.

Question

if the function were sinx we can prove that the error term tends to zero as the degree of the polynomial tends to infinity. however, with sin 2x the (n+1)th derivative is $$2^{n+1} (sin x )or (cos x)$$ so that method doesnt work.. how else can we prove that it converges to sin 2x, can i write sin 2x = 2 sinx cosx and say that sin x and cos x both converge?

Answer 1

If the Maclaurin series for $f(x)$ is $S(x)=\sum_{n=0}^\infty a_n x^n$, the Maclaurin series for $f(2x)$ is $$\sum_{n=0}^\infty a_n 2^n x^n = \sum_{n=0}^\infty a_n (2x)^n=S(2x)$$ If you know the series $S(x)$ converges to $f(x)$ for all $x$, then $S(2x)$ converges to $f(2x)$ for all $x$.