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RCos(a+b) Clarification

  1. Jun 5, 2008 #1
    Am i right in saying that when your comparing co-efficients for RSin(a+b) you swap over for the coefficients for sin and cos, while in RCos(a+b) you dont swap the co-efficients?

    RSin(x+a)= RSinxCosa + RCosxSina
    4Sin + 1Cos = RSinxCosa + RCosxSina

    RSina = 1
    RCosa = 4

    BUT for RCos(x+a) = RSinxCosa + RCosxSina


    4Sin + 1Cos = RSinxCosa + RCosxSina

    RSina = 4
    RCosa = 1

    Do you see what i mean?

    Thanks
     
  2. jcsd
  3. Jun 5, 2008 #2
    1. sin(x + a) = sin(x)cos(a) + cos(x)sin(a) so Rsin(x+a) = Rsin(x)cos(a)+Rcos(x)sin(a)
    2. cos(x + a) = cos(x)cos(a) − sin(x)sin(a) so Rcos(x+a) = Rcos(x)cos(a)-Rsin(x)sin(a)

    [tex]sin(x+a) \neq cos(x+a)[/tex]
     
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