RCos(a+b) Clarification

[thomas49th]

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 don't 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

 

[konthelion]

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)

$$sin(x+a) \neq cos(x+a)$$