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

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- Sep 23rd 2008, 08:48 PMthecountlinear approximation
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 - Sep 23rd 2008, 10:08 PMChop Suey
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. - Sep 23rd 2008, 10:18 PMChris L T521
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:

http://img.photobucket.com/albums/v4.../coslinear.jpg

Here's a graph of the function with its linear approximation:

http://img.photobucket.com/albums/v4...coslinear2.jpg

(the blue line is $\displaystyle \cos(x)$ and the purple is $\displaystyle L(x)$

So I would say its**true**.

--Chris - Sep 23rd 2008, 10:23 PMChop Suey