# Linear Approximation of 3cos^2(3) to estimate the value of |cos(3)

• May 13th 2011, 07:08 PM
RezMan
Linear Approximation of 3cos^2(3) to estimate the value of |cos(3)
hey guys can you help me with this:

Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|

where do I begin? would it look something like this:
y-(3cos^2(3))=$\displaystyle \frac{d(3cos^2(3))}{dy}(x-(3))$
• May 13th 2011, 08:04 PM
TKHunny
Is there a particular linear approximation you had in mind?

The "slope" piece of your line is zero (0). That's not very helpful.
• May 14th 2011, 05:15 AM
HallsofIvy
Frankly your work doesn't make a whole lot of sense to me. You want are trying to get a linear approximation at x= 3. The slope is, of course, the derivative of $\displaystyle 3cos^2(x)$ which is $\displaystyle -6cos(x)sin(x)$, evaluated at x= 3 which requires calculating cos(3) and sin(3) anyway.

TKHunny is suggesting you use a linear approximation at $\displaystyle x= \pi$ rather than at x= 3. There, as TKHunny says, the slope is 0 so your approximation is just the value at $\displaystyle x= \pi$ which is 3.
• May 14th 2011, 06:59 PM
RezMan
Well this question is the third part of the initial question, which is:
Given the function y=xcos^2(x)
a) find the differential of the function:
which I found to be: cos^2(x)-2xcos(x)sin(x)'

b) Give the linearization of the function at x=pi. Which I found to be: y=x

and then there's that question
c) Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|
which I don't really understand how to do it.
I guess I have to find value of x that allow 3cos^2(3) to equal |cos(3)|?
• May 14th 2011, 07:15 PM
TKHunny
Very good.

a) OK
b) OK
c) Well, what is the approximation? You have A(x) = x, the linearization of y(x) close to pi. x = 3 is pretty close to pi. Use this to create an approximation of y(3). Then use algebra to produce the approximation of |cos(3)|.
• May 14th 2011, 08:21 PM
RezMan
(Worried) you know I should understand what that means but I don't. Could you please detail it a little more
• May 14th 2011, 08:37 PM
TheEmptySet
Quote:

Originally Posted by RezMan
(Worried) you know I should understand what that means but I don't. Could you please detail it a little more

So you know the linear approximation is

$\displaystyle y \approx L(x)=x$

So just evaluate this at

$\displaystyle x=3 \implies y(3) \approx L(3)=3$
• May 15th 2011, 07:25 PM
RezMan
this is what my teacher says:
"So in this case, SUPPOSE you had y = 2x - 5 as a linear approximation for y = f(x) near x = 3 (that is not the case, i'm just saying pretend that is your linear approximation).

then we could say f(3)= 3 cos^2(3) is APPROXIMATELY EQUAL to y = 2(3) - 5 = 1

Since 3 cos^2(3) is APPROXIMATELY EQUAL to 1, we conclude that cos^2(3) is APPROXIMATELY EQUAL to 1/3, so by taking square roots we find |cos(3)| is APPROXIMATELY EQUAL to sqrt (1/3)."

So 3cos^2(3) = 3, because y=x. So maybe cos^2(3) =1? so 3(1)=3. So sqrt(1)=|cos(3)| ?
• May 15th 2011, 07:38 PM
RezMan
she says yes. Case closed. Thanks guys!