Using the sum formula for sine, which states that sin(a+b) = sin(a)cos(b) + cos(a)sin(b):
Asin(Bt + φ) = A(sin(Bt)cos(φ) + cos(Bt)sin(φ)) = Asin(Bt)cos(φ) + Acos(Bt)sin(φ) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)
Since we are to write -6sin(8t) + 5cos(8t) in this form:
-6sin(8t) + 5cos(8t) = Asin(Bt + φ) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)
-6sin(8t) + 5cos(8t) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)
We can now see that to write the left side of the equation as the right side, the following must be true:
Acos(φ) = -6, Asin(φ) = 5, B = 8
We now need to solve for A and φ!
We can solve for φ by solving for tan(φ) and cancelling out A as follows:
Asin(φ)/Acos(φ) = 5/-6
sin(φ)/cos(φ) = -5/6
tan(φ) = -5/6
φ = tan-1(-5/6) = -39.81°
Edit: As Doug C. points out, since A is positive and based on what we know about Asin(φ) and Acos(φ), φ should be in Quadrant II.
To get there from Quadrant IV, we add by 180° to get 140.19°.
Therefore, φ = 140.19°.
Now we can solve for A by using the Pythagorean formula (sin2Θ + cos2Θ= 1).
[Asin(φ)]2 + [Acos(φ)]2 = 52 + (-6)2
A2sin2(φ) + A2cos2(φ) = 25 + 36
A2[sin2(φ) + cos2(φ)] = 61
A2 = 61
A = √(61)
By plugging in A, B, and φ, we can arrive at our final answer:
-6sin(8t) + 5cos(8t) = √(61)sin(8t + 140.19°)