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Wave formula

  1. Sep 5, 2005 #1
    why can y(t) = A cos[omega(t)]+bsin[omega(t)]
    equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and
    tan(sigma)=B/A?
     
  2. jcsd
  3. Sep 6, 2005 #2

    James R

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    Well

    [tex]\cos(a-b) = \cos a \cos b + \sin a \sin b[/tex]

    Try expanding out [itex]C\cos(\omega t - \sigma)[/itex] using that formula, and see what you get.
     
  4. Sep 6, 2005 #3
    i got C[cos(wt)cos(sigma)+sin(wt)sin(sigma)]...
    am i missing a step?
    @@a
     
  5. Sep 7, 2005 #4

    James R

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    Now define A=Ccos(sigma) and B=Csin(sigma).

    What do you get if you eliminate sigma from these two equations?
     
  6. Sep 8, 2005 #5
    C[cos(wt)A/C + sin(wt)B/C]= the orignal equation!
    thanks!
     
  7. Sep 10, 2005 #6

    James R

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    Note also that [itex]A^2 + B^2 = C^2[/itex] and tan(sigma) = B/A, with the given definitions.
     
  8. Sep 12, 2005 #7
    cool! thanks again! :)
     
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