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Acos(x)+bsin(x) = Rsin(x+t)

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data

    acos(x)+bsin(x)=Rsin(x+t)

    2. Relevant equations
    3. The attempt at a solution

    Is there any way to show how R is "placed" in acos(x)+bsin(x)=Rsin(x+t) algebraically?
    I mean I could, probably, do acos(x)+bsin(x)=sin(t)cos(x)+ cos(t)sinx(x), but still somehow need R in it. Does R give the equation more balance? ;)

    Well, we also have x=Rcost and y=Rsint in addition to double angle identities, but I still can't seem to find satisfying algebraic justification for R's existence in f(x)= Rsin(x+t).

    Please, help me figure it out.

    Thanks.
     
  2. jcsd
  3. Jun 12, 2012 #2

    tiny-tim

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    hi solve! :smile:

    just expand sin(x+t), and equate coefficients :wink:
     
  4. Jun 12, 2012 #3
    Re: acos(x)+bsin(x)=Rsin(x+t)

    Hi,tiny-tim.

    See, sin(x+t)=sin(x)cos(t)+ cos(x)sin(t)= acos(x)+ bsin(x)

    Ok, a=cos(t) and b=sin(t), but how or why does R end up there?
     
    Last edited: Jun 12, 2012
  5. Jun 12, 2012 #4

    tiny-tim

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    you left something out! :biggrin:
     
  6. Jun 12, 2012 #5

    Mentallic

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    Re: acos(x)+bsin(x)=Rsin(x+t)

    You have it backwards! If

    [tex]\sin(x)\cos(t)+\cos(x)\sin(t)\equiv a\cdot\cos(x)+b\cdot\sin(x)[/tex]

    then [itex]a=\sin(t)[/itex] and [itex]b=\cos(t)[/itex].
     
  7. Jun 12, 2012 #6
    Re: acos(x)+bsin(x)=Rsin(x+t)

    Does that happen to be R? If yes, I'd like to know why and how R got there.
     
  8. Jun 12, 2012 #7
    Re: acos(x)+bsin(x)=Rsin(x+t)

    Thanks.
     
  9. Jun 12, 2012 #8

    tiny-tim

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    Rsin(x+t) = Rsin(x)cos(t) + Rcos(x)sin(t) = acos(x)+ bsin(x) :wink:
     
  10. Jun 12, 2012 #9
    Re: acos(x)+bsin(x)=Rsin(x+t)

    Ok.

    acos(x)+bsin(x)=Rsin(x+t)

    Why is there R in the above equation? Yes, you can expand it, but that doesn't explain what kind of reasoning keeps R in f(x)=Rsin(x+t)
     
  11. Jun 12, 2012 #10

    HallsofIvy

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    Re: acos(x)+bsin(x)=Rsin(x+t)

    Of course, sin(x+ t)= sin(x)cos(t)+ cos(x)sin(t) so that a= sin(t) and b= cos(t). But note that sin(t) and cos(t) are between -1 and 1 and that [itex]sin^2(t)+ cos^2(t)= 1[/itex]. In order to be able to write A sin(x)+ B cos(x), for any A and B, as "R sin(x+ t)" we must "normalize" the coefficients: letting [itex]R= \sqrt{A^2+ B^2}[/itex] so that Asin(x)+ Bcos(x)= R((A/R)sin(x)+ (B/R)cos(x)) and then let cos(t)= A/R and sin(t)= B/R.
     
  12. Jun 12, 2012 #11

    tiny-tim

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    Rsin(x+t)

    expand sin(x+t) …

    Rsin(x+t)

    :wink:
     
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