Answer
$0.17365 $
Work Step by Step
Given: $f(x)=\cos (x)$
Also, $ T_3(x)\approx -(x-\dfrac{\pi}{2}) +(\dfrac{1}{6})-(x-\dfrac{\pi}{2})^3 $
Now, we have $(80^{\circ}) \times (\dfrac{\pi}{180^{\circ}}) \approx \dfrac{4\pi}{9}$
$|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$
For $n=3$, we have $|R_3(x)|\leq \dfrac{1}{5!}|\dfrac{4\pi}{9}-\dfrac{\pi}{2}|^{5}\approx 0.00000135$
Hence, $ T_3(\dfrac{4\pi}{9})\approx -(\dfrac{4\pi}{9}-\dfrac{\pi}{2}) +\dfrac{1}{6}-(\dfrac{4\pi}{9}-\dfrac{\pi}{2})^3\approx 0.17365 $
