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Derivative with a range constraint - mystified!

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    x=sin(a) / cos(b)

    a+b < pi/2
    a>0, b>0
    0<x<1

    show that

    da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b)


    3. The attempt at a solution

    dx/da = cos(a) / cos(b) therefore
    da/dx = cos(b) / cos(a)

    =cos^3(b)cos(a) / cos^2(b)cos^2(a)

    However the denominator of the desired format = cos(a+b)cos(a-b) = cos^2(a)cos^2(b)-sin^2(a)sin^2(b)

    Not sure how to get rid of the sin^2(a)sin^2(b) term.

    Is the question wrong or is there something special that needs to be done to take into account the range of a, b and x?

    Very much appreciate help as Ive been totally stumped on this and the equality is required further on
     
  2. jcsd
  3. Jun 11, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you add and subtract cos2(a)sin2(b) in the denominator your equality comes out:

    [tex]\frac{\cos(b)}{\cos(a)}=\frac{\cos^3(b)\cos(a)}{\cos^2(a)-\sin^2(b)}[/tex]

    At least I think it does; you can check it. The problem is if you let a = b = π/6, you get the left side = 0.9742785794, the right side = 1, and x = 0.5773502695.

    So it would appear something is amiss. :frown:
     
  4. Jun 12, 2010 #3

    Mark44

    Staff: Mentor

    This seems to be the same question you posted in this thread.
    This equation is not an identity
    [tex]\frac{cos(b)}{cos(a)}=\frac{cos^3(b)cos(a)}{cos^2(a)-sin^2(b)}[/tex]
    LCKurtz points out a counterexample. Another is a = π/6 and b = π/4. Using these numbers the left side value is sqrt(6)/3 and the right side value is sqrt(6)/2.
     
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