Answer
(a) $f\left( x \right) \approx 1 - \dfrac{1}{2}{x^2}$
(b) $\cos 2^\circ = \cos \dfrac{\pi }{{90}} \approx 0.999391$
Calculator: $\cos 2^\circ \approx 0.999391$
Work Step by Step
(a) Let $f\left( x \right) = \cos x$. Taking the derivatives gives
$f'\left( x \right) = - \sin x$, ${\ \ \ \ \ }$ $f{\rm{''}}\left( x \right) = - \cos x$
The local quadratic approximation of $f$ at ${x_0} = 0$ is
$f\left( x \right) \approx f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right) + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{\left( {x - 0} \right)^2}$
$f\left( x \right) \approx 1 - \dfrac{1}{2}{x^2}$
(b) We convert $2^\circ $ to the unit in radians:
$2^\circ = \dfrac{2}{{180}}\pi = \dfrac{\pi }{{90}}$
Thus,
$f\left( {\dfrac{\pi }{{90}}} \right) = \cos \dfrac{\pi }{{90}} \approx 1 - \dfrac{1}{2}\cdot{\left( {\dfrac{\pi }{{90}}} \right)^2}$
$\cos 2^\circ = \cos \dfrac{\pi }{{90}} \approx 0.999391$
Using a calculating facility, we get $\cos 2^\circ \approx 0.999391$.
