1. The problem statement, all variables and given/known data Show that Ccos(wt+phi) = Acos(wt)+Bsin(wt) 2. Relevant equations Trig identity that states cos(wt+phi) = cos(wt)cos(phi)-sin(wt)sin(phi) 3. The attempt at a solution Ccos(wt+phi)=(Ccos(phi))cos(wt)+(-Csin(phi))sin(wt) let A = Ccos(phi) Let B = -Csin(phi) Ccos(wt+phi) = Acos(wt)+Bsin(wt) and done. Is this as simple as I have shown? Or am I making a critical mistake in letting A = Ccos(phi) and B = -Csin(phi)? Is there a more rigorous way of doing this that would be expected? Since phi is a constant, C is a constant, I would think that this is a suitable way to prove that these two sides are equal, but I can't help but feel a bit weak here about this.