Linear Approximations and Differentials

[loadsy]

Alright I just did the following question, and was hoping I did it right:

Use differentials(or, equivalently, a linear approximation) to estimate the given number)

cos 31.5* (* meaning degrees)

f(x) = cosx
f(31.5*) = ?

a is chosen to be the closest number to the number evaluating in the function of, such that the equation can be easily evaluated.
a= 30*, then dx = 31.5*-30*= 1.5*

delta y is approximately equal to dy
delta y = f(31.5*)-f(30*)
= cos31.5* - cos30*
= cos31.5* - root3/2

dy = f'(30*)dx

f'(x) = -sinx -> f'(30*) = -sin(30*) = -0.5

dy = (-1/2)(1.5*) = -1/2(pi/120) = -(pi)/240

Hence: cos31.5* - root3/2 is approximately equal to -(pi)/240 or:
cos(31.5*) = root3/2 - (pi)/240

I'm just checking to see if I followed the correct steps in solving this question. Thanks guys.

 

[Steve4Physics]

Hence: cos31.5* - root3/2 is approximately equal to -(pi)/240 or:
cos(31.5*) = root3/2 - (pi)/240

At the time of writing, this is a 15+ year old question. However the following may help someone who comes across it.

The question specifically asks for an estimate. Giving the answer in terms of √3 and π (which are exact) is not appropriate. A numerical answer to 3 significant figures is needed. (Be guided by the fact that ‘31.5º' has 3 significant figures.)

The answer is correct though working is a bit untidy. The answer can easily be checked: find cos(31.5º) and then evaluate the expression ##\frac {\sqrt 3}{2} - \frac {\pi}{240}##. They agree - both give 0.853

 

[WWGD]

Yes, to add some detail, this is the tangent line approximation, which follows from the differentiability of cosx. For those interested in the Math side.