1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution $$3cos(20t+10°)-5cos(20t-30°)\\ =3\angle 10°-5\angle -30°\\ =-1.376+3.0209j\\ =3.32\angle -65.51°$$ In the last step, the textbook actually got ##3.32\angle 114.49°##. I checked both answers and it seems that the textbook's answer is correct. All I did to get -65.51° is using the arctan function on my calculator. It looks like their angle is 180° greater than that. How did they choose the correct angle? I'm rusty on my trigonometry.