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RCos(a+b) Clarification

  • Thread starter thomas49th
  • Start date
  • #1
655
0
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
 

Answers and Replies

  • #2
238
0
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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