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Help me to clear my confusion, this is regarding nth derivative of e^ax cos(bx+c)

  1. Mar 1, 2012 #1
    My question is about the nth derivative of e^ax cos(bx+c). Though i can calculate it easily but i am confused at one point.
    When we calculate the first derivative we put a = r.cos(theta), b = r.sin(theta) (every thing is ok till here)
    My confusion starts when we use (theta) = tan^-1(b/a) [tan inverse]
    the reason for my confusion can be understood by:
    suppose we have a = -1, b = 1
    we put a = sqrt{2}*cos(3*pi/4)
    b = sqrt{2}*sin(3*pi/4)
    but the tan^-1(b/a) = tan^-1(-1) = -pi/4
    but our theta is 3*pi/4
    according to this theta our a will be 1 and b will be -1 which is different from our values of a and b
     
  2. jcsd
  3. Mar 2, 2012 #2

    mathman

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    Science Advisor
    Gold Member

    tan is periodic with period π, so arctan (-1) = -π/4 + kπ, for any integer k.
     
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