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Trig step in a chain rule question

  1. Feb 21, 2006 #1
    I need to show that two equations equal one another. It's too complicated to display fully on here but I'm stuck on a step:

    dF/dr = df/dx cos2(h) + df/dy sin(h)

    (dF/dr)^2 = (df/dx)^2 cos^2(h) + (df/dy)^2 sin^2(h)

    Does anybody know how to get rid of the cos squared and sin squared?

    P.S. By (dF/dr)^2 I mean the first derivative squared not the second derivative
  2. jcsd
  3. Feb 21, 2006 #2


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    [tex]\frac{dF}{dr} = \frac{df}{dx} \cos ^2(h) + \frac{df}{dy} \sin (h) ,[/tex]


    [tex]\left( \frac{dF}{dr}\right) ^2 =\left( \frac{df}{dx} \cos ^2(h) + \frac{df}{dy} \sin (h) \right) ^2 = \left( \frac{df}{dx}\right) ^2 \cos ^4(h) +2\frac{df}{dx}\cdot\frac{df}{dy} \cos ^2(h)\sin (h)+ \left( \frac{df}{dy}\right) ^2 \sin ^2(h) [/tex]
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