Answer
$0.61566$
Work Step by Step
Given: $f(x)=\sin x; a=\dfrac{\pi}{6}$
$(38^{\circ}) \times (\dfrac{\pi}{180^{\circ}}) \approx \dfrac{19\pi}{90}$
Here, we have $|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$
$\implies |R_4(x)|\leq \dfrac{1}{5!}|\dfrac{19\pi}{90}-\dfrac{\pi}{6}|^{5}\approx 0.00000041$
and $ 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 \approx T_4(\dfrac{19\pi}{90})$
Thus, $ T_4(38^{\circ}) \approx 0.61566$
