Wave formula

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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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.
 
  • #3
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i got C[cos(wt)cos(sigma)+sin(wt)sin(sigma)]...
am i missing a step?
@@a
 
  • #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?
 
  • #5
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C[cos(wt)A/C + sin(wt)B/C]= the orignal equation!
thanks!
 
  • #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.
 
  • #7
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cool! thanks again! :)
 

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