#### Answer

$\approx 0.17365 $

#### Work Step by Step

We are given that $f(x)=\cos x$ and $a=\dfrac{\pi}{2}$
Here, we have $80^{\circ} \times \dfrac{\pi}{180^{\circ}} \approx \dfrac{4\pi}{9}$
Check the Taylor inequality for the interval $\dfrac{4\pi}{9} \leq x\leq \dfrac{\pi}{2}$
Now, $|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}$
and $|R_3(x)|\leq \dfrac{1}{5!}|\dfrac{4\pi}{9}-\dfrac{\pi}{2}|^{5}\approx 0.00000135$
and $f(x) \approx T_3(x)\approx -(x-\dfrac{\pi}{2}) +\dfrac{1}{6}-(x-\dfrac{\pi}{2})^3 $
Thus, we have $f(x) \approx 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 $