Linear approximation of Tan(44)

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Homework Help Overview

The discussion revolves around finding the linear approximation of the tangent function at 44 degrees, specifically using the function f(x) = tan(44 + x). Participants are exploring the implications of using degrees versus radians in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the linear approximation formula but questions their approach regarding the choice of x and the use of degrees. Other participants suggest considering the expansion around tan(45+x) and express concerns about the derivative calculations in degrees.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the use of radians over degrees. There is acknowledgment of the challenges posed by using degrees, and one participant indicates a successful resolution after reconsidering their approach.

Contextual Notes

There is a noted confusion regarding the conversion between degrees and radians, which affects the derivative calculations. The original poster's assumption about the value of x is also under scrutiny.

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Homework Statement



Find the linear approximation of tan(44)

Homework Equations



f(x) = tan(44 + x), let x = a = 1
f`(x) = sec2(44+x)

The Attempt at a Solution



L(x) = f(a) + f`(a)(x-a)
L(x) = 1 + 2(x-1)
L(0) = 1 + 2(0-1) = -1 (Answer should be approx. 0.965)
Where am I going wrong? X should be zero, because tan(44+0) is what I'm trying to find the approximation of, right?
 
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I think you want to find tan(44 degrees) by expanding tan(45+x) where x=(-1). But be careful about the derivatives when you are working in degrees. I think you are better off expanding tan(pi/4+x) (since pi/4=45 degrees) and then putting x=(-1) degrees, expressed in radians, of course.
 
Well, that'll teach me never to use degrees again. It worked!

Thanks!
 
Yeah, degrees aren't fun. They just represent extra work.
 

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