Since x - 1 can be calculated precisely, we need to approximate $\displaystyle (\sin x)/2$ on $\displaystyle [0,\pi]$ up to 0.015. We have $\displaystyle \sin x=S_n(x)+R_n(x)$ where $\displaystyle S_n(x)$ is a partial sum of the Taylor series that goes up to $\displaystyle (x-a)^n$ (here $\displaystyle a=\pi/2$) and $\displaystyle R_n(x)=\sin(x)-S_n(x)$. Then $\displaystyle (\sin x)/2=S_n(x)/2+R_n(x)/2$, and we need to ensure that $\displaystyle \sup_{x\in[0,\pi]}|R_n(x)/2|\le 0.015$, i.e., $\displaystyle \sup_{x\in[0,\pi]}|R_n(x)|\le0.03$.

The Lagrange form of the remainder is $\displaystyle \displaystyle R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}$ where $\displaystyle \xi$ is some point between $\displaystyle a$ and $\displaystyle x$. Since $\displaystyle |\sin^{(n)}x|\le 1$ for all n and x, we have $\displaystyle \displaystyle |R_n(x)|\le \frac{(\pi/2)^{n+1}}{(n+1)!}$. According to my calculations, the least n such that $\displaystyle \displaystyle\frac{(\pi/2)^{n+1}}{(n+1)!}\le0.03$ is 5. However, since $\displaystyle \sin^{(5)}(\pi/2)=0$, it is enough to approximate $\displaystyle \sin(x)$ by $\displaystyle S_4(x)$, which has only three terms.

I did this check: $\displaystyle S_4(\pi)=0.019968958<0.03$.