Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}}&{f\left( {{x_n}} \right)}&{f'\left( {{x_n}} \right)}&{\frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}}&{{x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}} \\
1&{1.6000}&{ - 0.0292}&{ - 0.9996}&{0.0292}&{1.5708} \\
2&{1.5708}&{0.0000}&{ - 1.0000}&{0.0000}&{1.5708}
\end{array}}\]
Work Step by Step
$$\eqalign{
& f\left( x \right) = \cos x,{\text{ }}{x_1} = 1.6 \cr
& f\left( {{x_1}} \right) = \cos \left( {1.6} \right) \approx - 0.0292 \cr
& {\text{Use Newton's Method}} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }}\left( {\bf{1}} \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos x} \right] = - \sin x \cr
& f'\left( {{x_1}} \right) = - \sin \left( {1.6} \right) \approx - 0.9996 \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{1 + 1}} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}} \cr
& {x_2} = 1.6 - \frac{{ - 0.0292}}{{ - 0.9996}} \approx 1.5708 \cr
& {\text{The calculations for two iterations are shown in the table}} \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}}&{f\left( {{x_n}} \right)}&{f'\left( {{x_n}} \right)}&{\frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}}&{{x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}} \\
1&{1.6000}&{ - 0.0292}&{ - 0.9996}&{0.0292}&{1.5708} \\
2&{1.5708}&{0.0000}&{ - 1.0000}&{0.0000}&{1.5708}
\end{array}}\]
