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How to show that Acosx+Bsinx=Ccos(x+a) ?

  1. Apr 8, 2009 #1
    How to show that Acosx+Bsinx=Ccos(x+a) ??

    I'm really bad with trig identities, how do we show the following:
    Acosx+Bsinx=Ccos(x+a)
     
  2. jcsd
  3. Apr 8, 2009 #2
    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    Try to work out what cos(x)+sin(x) is. I'll give you a hint. sin(x)=cos(x-Pi/2), so you have cos(x)+sin(x)=cos(x)+cos(x-Pi/2). Now use the formula 2*cos(a)cos(b)=cos(a+b)+cos(a-b).
     
  4. Apr 9, 2009 #3
    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    I still get crap, I don't know how to get to the answer they want. What do I do with the A and B in front of both of the cos terms?
     
  5. Apr 9, 2009 #4
    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    What are you trying to do? Are you trying to find (A,B,C) such that [tex]A \cos{x}+B \sin{x}=C \cos{(x+a)}[/tex]?
     
  6. Apr 9, 2009 #5
    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    No, this has to be true for any arbitrary constants A,B,C,a (a is the phase angle). We use it all the time. I just don't know the trig identities and I don't know how to show that Acosx+Bsinx=Ccos(x+a) is true.
     
  7. Apr 9, 2009 #6

    Integral

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    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    This cannot be true for abitrary A, B, and C.
    Prove by example.
    A=1, B=1, C= 0

    There has to be a relationship between A,B and C.
     
  8. Apr 10, 2009 #7
    Re: How to show that Acosx+Bsinx=Ccos(x+a) ??

    Expand the RHS using the cos(A+B) identity.

    cos(A+B)=cos(A)cos(B)-sin(A)sin(B) or

    Ccos(x+a)=Ccos(x)cos(a)-Csin(x)sin(a)

    If Acosx+Bsinx = Ccos(x)cos(a)-Csin(x)sin(a) then

    Ccos(a)=A and
    B=-Csin(a)

    Then you can solve those simultaneous equations.
     
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