To express the given function in the form A cos (omega t + phi)Acos(ωt+ϕ), notice that according to the trigonometric sum rule
A cos (omega t + phi) = A [cos (omega t )cos (phi)-sin(omega t)sin(phi)] = -A sin(phi) sin(omega t) + A cos(phi)cos(omega t)Acos(ωt+ϕ)=A[cos(ωt)cos(ϕ)−sin(ωt)sin(ϕ)]=−Asin(ϕ)sin(ωt)+Acos(ϕ)cos(ωt)
Comparing with the expression given, we have omega = piω=π nd
A sin(phi) = -1.2, qquad A cos(phi) = 3.5
Thus
A^2 = = (A sin(phi))^2 + (A cos(phi))^2= (-1.2)^2+(3.5)^2 = 13.69 = 3.7^2
So, A = 3.7
Now,
tan(phi) = {A sin(phi)}/{A cos(phi)} = -{1.2}/{3.5} = -0.3429
Since sin(phi) is negative while cos(phi) is positive, the angle phi must be in the fourth quadrant. So
phi =2pi-tan^{-1}(12/35)=5.953