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How do I go from the form sin(6x) to the form A sin(x)?

  1. Nov 17, 2012 #1
    I am working on a homework problem: sin(6x) + cos(6x). I have to represent it in the form:
    k sin(x + [itex]\phi[/itex])

    I have the formulas to change A sin(x) + B cos(x) int k sin(x + [itex]\phi[/itex]), however, I don't know how to go from sin(6x) to A sin(x). Is it just 6 sin(x)?

    I don't think it is, because if so, k will be equal to √(62+62), which is not a simple whole number, so I suspect my approach is wrong.
     
  2. jcsd
  3. Nov 17, 2012 #2

    SteamKing

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    Try applying the angle addition formulas for sin and cos first until you obtain sin(x) and cos(x).
     
  4. Nov 18, 2012 #3

    Mentallic

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    If you let

    [tex]\sin(6x)+\cos(6x)=k\cdot\sin(6x+\phi)[/tex]

    with some constants k and [itex]\phi[/itex], then upon expanding the RHS we get

    [tex]\sin(6x)+\cos(6x)=k\left(\sin(6x)\cos(\phi)+\cos(6x)\sin(\phi)\right)[/tex]

    [tex]\sin(6x)+\cos(6x)=k\cdot\cos(\phi)\sin(6x)+k\cdot\sin(\phi)\cos(6x)[/tex]

    Now notice that for each side to be equal, the coefficient of [itex]\sin(6x)[/itex] on the LHS must equal the coefficient of [itex]\sin(6x)[/itex] on the RHS, and similarly for [itex]\cos(6x)[/itex]

    This will give you 2 equations in 2 unknowns, so you should be able to find the value of each.
     
  5. Nov 18, 2012 #4

    Ray Vickson

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    You can't go from sin(6x) + cos(6x) to ksin(x+##\phi##), at least not for all x. The formula involving 6x has period π/3, while the other one has period 2π, no matter what values of k and ##\phi## you pick. And no, sin(6x) is not 6sin(x), and for exactly the same reason. Note, however, that you *can* go to ksin(6x+##\phi##).

    RGV
     
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