Question #0a796

1 Answer
Feb 17, 2018

3.7 cos(pi t-5.953)3.7cos(πt5.953) i.e. the constants are A = 3.7A=3.7, omega = piω=π and phi =2pi-tan^{-1}(12/35)=5.953 ϕ=2πtan1(1235)=5.953

Explanation:

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