Cannot expand $\sin(2x^2-4x+3)$ at $x_0 = 1$

Question

Trying to expand $\sin(2x^2 - 4x+3)$ at $x_0 = 1$ to the $O(x-x_0)^n$.

After substitution $t = x - 1 $, the problem becames $$\sin(2t^2+1) \text{ at } t_0 = 0$$ While we know that $$\sin(s) = \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}s^{2n+1}}{(2n+1)!},$$ can we substitute $2t^2+1$ instead of $s$? It looks like no, because when you expand $\sin(2t^2+1)$ at $t_0 = 0$, obviously, you would get $\sin(1), \cos(1)$ terms. Could someone help me?

Answer 1

You could start by writing $\sin(1+t^2) = \sin(1)\cos(t^2)+\cos(1)\sin(t^2)$, then use the Taylor expansions of $\sin$,$\cos$ at $0$.

Answer 2

Just simplify that $$\sin(2x^2-4x)=\sin(2(x-2)x+3)$$

$$\sin(2(x-2)x+3)=\sin(2(x-2)x)\cos(3)+cos((2(x-2)x)\sin(3)$$

then use the taylor expansion..