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Simple differentiation problem, brainfart

  1. Jul 26, 2009 #1

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


    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}

    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


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    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:

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


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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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