Given that the $0$ degree Taylor polynomial centered at $x=\frac{1}{2}$ for $f(x)=\arcsin(x)$ is given by $T_{0,\frac{1}{2}}(x)=\frac{\pi}{6}$ how would I use this to show that $1.2(x-\frac{1}{2})\leq R_{0,\frac{1}{2}}(x) \leq(x-\frac{1}{2})$ on $[0.4,0.5]$
I know that by Taylor's theorem $R_{0,\frac{1}{2}}(x)=\frac{1}{\sqrt{1-c^2}}(x-\frac{1}{2}$) and since $\frac{1}{\sqrt{1-x^2}}$ is an increasing function on $[0.4,0.5]$ that $R_{0,\frac{1}{2}}(x) \leq \frac{1}{\sqrt{1-0.5^2}}(x-\frac{1}{2})$
Since as you notice $$ \frac{1}{\sqrt{1-0.4^2}}\le\frac{1}{\sqrt{1-c^2}}\le\frac{1}{\sqrt{1-0.5^2}}, $$ all you need to prove is that $$ \color{red}{1\le\frac{1}{\sqrt{1-0.4^2}}}\le\frac{1}{\sqrt{1-c^2}}\le\color{blue}{\frac{1}{\sqrt{1-0.5^2}}\le 1.2} $$ and then multiply by $x-0.5$, which is negative for $0.4\le x\le 0.5$, so the inequalities become reversed.