Using the linear approximation to f(x) = cos x at the point x = 1 to estimate
the value of cos(1.03) works well because cos x looks like a line when you zoom
in on it.
true/false
I say it's false since you don't know an exact value for cos(1), and since there is a closer exact value near 1.03, such as $\displaystyle \frac{\pi}{3}$.
I may have answered this in a different way than the question expects you to answer.
The linear approximation takes on the form $\displaystyle L(x)=-\sin(1)(x-1)+\cos(1)\implies L(x)=-.841x+1.382$
So, is $\displaystyle L(1.03)$ a good approximation to $\displaystyle \cos(1.03)$?
$\displaystyle L(1.03)=-.841(1.03)+1.382\approx .515$
Now, $\displaystyle \cos(1.03)\approx .5148$
This is a pretty good approximation, with an error of about $\displaystyle 10^{-3}$.
If you were to zoom in on the cosine function at about 1, it appears linear:
Here's a graph of the function with its linear approximation:
(the blue line is $\displaystyle \cos(x)$ and the purple is $\displaystyle L(x)$
So I would say its true.
--Chris