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Math Help - Trigonometry again derivative

  1. #1
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    Trigonometry again derivative

    I have this:
    y=3cosx-cos3x

    I need to solve it.

    I got this:
    y'=3sin(3x)-3sin(x)

    I simplified it to get:
    y'=3(sin(3x)-sin(x))

    I equate it to 0 and I get:
    sin(3x)-sin(x)=0

    How do I solve this?
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  2. #2
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    Hello,

    You don't have to derivate, since it'll only (the things that could interest us) give you variations.

    Transform cos(3x) into a form with cos(x) and sin(x).


    cos(3x)=cos(2x+x)

    So use the following formulas :

    \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)

    \cos(2x)=2 \cos^2(x)-1

    \sin(2x)=2 \sin(x)\cos(x)

    Try to get cos(3x) first...
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  3. #3
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    Well I need to find all the maximum and minimum points so that is why I derived it.

    I guess I could do what you did and then derive it. Let me try.
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  4. #4
    Moo
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    I need to solve it.
    This was the only thing I knew you had to do



    Well, for solving the derivative, do the same :

    \sin(3x)=\sin(2x+x)


    Formulas you need :

    \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)

    \sin(2x)=2\sin(x)\cos(x)

    \cos(2x)= hmmm, you'll see which one of the three below would be useful :

    \begin{array}{ccc} \cos(2x) & = & \cos^2(x)-\sin^2(x) \\ & = & 2 \cos^2(x)-1 \\ & = & 1-2\sin^2(x) \end{array}
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  5. #5
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    How do I solve:
    cos^2(x)=1/4

    Nevermind I am dumb I got it...
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  6. #6
    Moo
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    \sin(3x)=\sin(2x)\cos(x)+\cos(2x)\sin(x)

    \sin(2x)=2\sin(x) \cos(x)

    ---> \sin(3x)=2\sin(x) \cos^2(x)+\cos(2x)\sin(x)

    Replace \cos^2(x)=1-sin^2(x) and \cos(2x)=1-2\sin^2(x)
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  7. #7
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    Using all that I "simplified" it to:
    3sin(x)-4sin^3(x)

    Now I am stuck again...
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  8. #8
    Moo
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    This is correct

    Now you want to solve for y'=0


    So 3\sin(x)-4\sin^3(x)=0 \Longleftrightarrow \sin(x)(3-4\sin^2(x))=0

    Can you solve it ?
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  9. #9
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    I get:
    sin(x)=0
    sin^2(x)=3/4

    I solved the first one but how do I solve the second one.
    I know that I need to do square root so that I get
    sin(x)=sqrt(3/4)
    But thats not a number that I can express... Or can I? How?
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  10. #10
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    \sin^{2} x = \frac{3}{4}
    \sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}
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