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Trig substitutions

  1. Mar 3, 2006 #1
    Yeah I know this is pretty basic, but I just can't think of it right now.

    The slope of the wire is tan(theta) = y' = dy/dx
    which yields sin(theta) = y' / sqrt (1 + y'^2)

    How do you get from the first line to the second one?



    this is what I I have done:
    x^2 + y^2 = 1
    cos^2(theta) + sin^2(theta) = 1
    so cos^2(theta) = 1 - sin^2(theta)
    so cos(theta) = sqrt (1 - y^2).

    And
    tan(theta) = sin(theta) / cos (theta).
    So using tan(theta) = y' = dy/dx
    sin(theta) / cos (theta) = y'.
    sin(theta) = y' cos(theta)
    plugging in what we got for cos earlier
    sin(theta) = y' [sqrt (1 - y^2)]
    now I am stuck.


    I am just using this example to try and figure out how to do it. I actually want it in the form cos(theta) = something * x'.
     
  2. jcsd
  3. Mar 3, 2006 #2
    Recall that sec(theta)=1/cos(theta). What's the relationship between tan^2 and sec^2?

    -Dan
     
  4. Mar 3, 2006 #3
    BAM. I think I got it.

    1 + tan^2 = sec^2
    sec = sqrt (1 + tan^2)
    sec = 1/cos
    so sin * sqrt (1 + tan^2) = y'
    so sin = y' / sqrt (1 + tan^2) , and tan = y'
    so sin = y' / sqrt (1 + y'^2)

    Thx for your help.
     
    Last edited: Mar 3, 2006
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