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Simple differentiation problem, brainfart

  1. Jul 26, 2009 #1
    Hai,

    I have the easiest problem but im stuck at the last step, simplification.

    [tex]\frac{cos3x+sin3x}{cos3x-sin3x}[/tex]

    [tex]\begin{aligned}
    F(x) = cos(3x)+sin(3x)\\
    F'(x) = -3sin(3x)+3cos(3x) \\
    G(x) = cos(3x)-sin(3x) \\
    G'(x) = -3sin(3x)-3cos(3x)\end{aligned}
    [/tex]

    This gives [tex]\frac{(-3sin(3x)+3cos(3x))*(cos(3x)-sin(3x))+(cos(3x)+sin(3x))*(cos(3x)-sin(3x))}{(cos(3x)-sin(3x))^2}[/tex]

    But how would i simplify this?
     
  2. jcsd
  3. Jul 26, 2009 #2

    djeitnstine

    User Avatar
    Gold Member

    Expand the quadratics and look for identities. Already I can find sin^2 + cos^2 identities on both the top and bottom.
     
  4. Jul 26, 2009 #3
    Also, it will help to remember that sin(u)cos(u) = (1/2)sin(2u)
     
  5. Jul 26, 2009 #4
    Okay, after some more trying:

    I was able to reduce the first part[tex](-3sin(3x)+3cos(3x))(cos(3x)-sin(3x))[/tex]
    Down to:
    [tex]-3sin(3x)*cos(3x) + 3 -3cos(3x)*sin(3x)[/tex]

    And the second part:[tex](cos(3x)+sin(3x))*(cos(3x)-sin(3x))[/tex]
    Down to:
    [tex]cos(3x)^{2}-sin(3x)^{2}-2sin(3x)cos(3x)[/tex]

    Is that any better?
     
  6. Jul 26, 2009 #5

    djeitnstine

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    Gold Member

    Take note of AUMathtutor's comment
     
  7. Jul 27, 2009 #6
    Also, naturally,

    cos(u)^{2} - sin(u)^{2} = cos(2u)
     
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