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Representing sum of cosine and sine as a single cosine expression

  1. Jul 12, 2010 #1
    a.cos(wt) + b.sin(wt) = M.cos(wt + ϕ)

    Can you give me M and ϕ in terms of a and b?
     
  2. jcsd
  3. Jul 12, 2010 #2

    arildno

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    cos(u+v)=cos(u)cos(v)-sin(u)sin(v)

    That 's all you need.
     
  4. Jul 12, 2010 #3
    The final representation was something like
    M = sqrt(a^2 + b^2)
    and
    ϕ = arctan(-b/a)
    but I'm no sure.

    Can anyone confirm it for me?
     
  5. Jul 12, 2010 #4

    arildno

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    No.

    Try to do it for yourself, and we can correct whatever mistakes you make.
     
  6. Jul 12, 2010 #5
    Code (Text):
    M.cos(wt + ϕ) = a.cos(wt) + b.sin(wt)

    cos(wt + ϕ) = (a/M).cos(wt) + (b/M).sin(wt)........(I)

    cos(wt + ϕ) = cos(wt).cos(ϕ) - sin(wt).sin(ϕ)........(II)

    From (I) and (II),
    cos(ϕ) = (a/M)
    sin(ϕ) = -(b/M)

    cos^2(ϕ) + sin^2(ϕ) = (a^2 + b^2)/(M^2) = 1

    We assume that M is always positive and we keep any negativity in the phase angle ϕ,
    M = sqrt(a^2 + b^2)

    sin(ϕ)/cos(ϕ) = tan(ϕ) = -(b/M)/(a/M) = -b/a

    tan(ϕ) = -b/a   ==>   ϕ = arctan(-b/a)
    Is there anything wrong in my derivation?
     
  7. Jul 12, 2010 #6

    arildno

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    No. :smile:
     
  8. Jul 12, 2010 #7
    I liked your way of "you must do it yourself if you want to success"... :)
     
  9. Jul 12, 2010 #8

    arildno

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    It's true, isn't it? :smile:
     
  10. Jul 12, 2010 #9
    Yeah, either I don't like the ones who ask a big problem and wait for others to solve it for him.
    In my question, this is a simple trigonometric identity, I expected a mathematician to write it for me since most of math guys have memorized and actively use these kind of identities.
     
  11. Jul 13, 2010 #10
    Last edited by a moderator: Apr 25, 2017
  12. Jul 13, 2010 #11

    Landau

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    You could also do the following:

    t=0 gives a=M cos(ϕ),
    t=pi/(2w) gives b=-M sin(ϕ).

    Hence

    a^2+b^2=M^2,
    tan(ϕ)=-b/a.
     
  13. Jul 13, 2010 #12
    Wow, that's super simple, thanks!
     
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