HELP PRECALC I DO NOT UNDERSTAND AT AL

  φ ≈ 1.19029 radians   (≈ 68.2°)

Step-by-step explanation:

There are simple formulas for A and φ in this conversion, but it can be instructive to see how they are derived.

We want to compare ...

  y(t) = Asin(ωt +φ)

to

  y(t) = Psin(ωt) +Qcos(ωt)

Using trig identities to expand the first equation, we have ...

  y(t) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)

Matching coefficients with the second equation, we have ...

  P = Acos(φ)

  Q = Asin(φ)

The ratio of these eliminates A and gives a relation for φ:

  Q/P = sin(φ)/cos(φ)

  Q/P = tan(φ)

  φ = arctan(Q/P) . . . . taking quadrant into account

__

We can also use our equations for P and Q to find A:

  P² +Q² = (Acos(φ))² +(Asin(φ))² = A²(cos(φ)² +sin(φ)²) = A²

  A = √(P² +Q²)

Here, we want φ.

  φ = arctan(Q/P) = arctan(5/2)

  φ ≈ 1.19029 . . . radians

d? maybe

step-by-step explanation: