Answer
$0.61566$
Work Step by Step
Here, we have $f(x)=\sin x$ and $a=\dfrac{\pi}{6}$
and $f(x) \approx T_4(x)\approx |\dfrac{1}{2}+\dfrac{\sqrt 3}{2}(x-\dfrac{\pi}{6})-\dfrac{1}{4}(x-\dfrac{\pi}{6}))^2 +\dfrac{\sqrt 3}{12}-(x-\dfrac{\pi}{2})^6+\dfrac{1}{48}(x-\dfrac{\pi}{6})^4-\sin x |$
and $38^{\circ} \times \dfrac{\pi}{180^{\circ}} \approx \dfrac{19\pi}{90}$
Now, we need to check the Taylor inequality for the interval $\dfrac{\pi}{6}-\dfrac{2\pi}{45} \leq x\leq \dfrac{\pi}{6}+\dfrac{2\pi}{45}$
Now, $|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$
and $|R_4(x)|\leq \dfrac{1}{5!}|\dfrac{19\pi}{90}-\dfrac{\pi}{6}|^{5}\approx 0.00000041$
Thus, $ T_4(38^{\circ})=\dfrac{1}{2}+\dfrac{\sqrt 3}{2}(x-\dfrac{\pi}{6})-\dfrac{1}{4}(x-\dfrac{\pi}{6}))^2 +\dfrac{\sqrt 3}{12}-(x-\dfrac{\pi}{2})^6+\dfrac{1}{48}(x-\dfrac{\pi}{6})^4 $
Hence, $ T_4(38^{\circ})=T_4(\dfrac{19\pi}{90}) \approx 0.61566$
