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Linear approximation of Tan(44)

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the linear approximation of tan(44)

    2. Relevant equations

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

    3. 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?
  2. jcsd
  3. Dec 10, 2008 #2


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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.
  4. Dec 10, 2008 #3
    Well, that'll teach me never to use degrees again. It worked!

  5. Dec 10, 2008 #4


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