Amplitude-Phase Form Proof

Again, I pulled formulas out of thin air when I showed amplitude-phase form. Your professor may want to see justification, so I will show an example here.

Say you have solved a differential equation, and you got an answer in the form y = c₁e^βxcos(γx) + c₂e^βxsin(γx). Now you're supposed to convert to amplitude phase form. As I stated in my previous post, this answer will be in the following form:  y(x) = Ae^βxcos(γx - θ). But why?

First, we will start by defining the constant A, which is equal to √(c₁² + c₂²). Now on the right-hand side of your equation, multiply and divide by A. This is the same as multiplying by 1, and does not change the actual problem. You'll get:

y = c₁√(c₁² + c₂²)/√(c₁² + c₂²)e^βxcos(γx) + c₂√(c₁² + c₂²)/√(c₁² + c₂²)e^βxsin(γx)

Now visualize a right triangle in which the legs have length c₁ and c₂. By the Pythagorean theorem, the hypotenuse will be √(c₁² + c₂²).

So you can replace several pieces of the above equation in terms of θ.

y = √(c₁² + c₂²)cos(θ)e^βxcos(γx) + √(c₁² + c₂²)sin(θ)e^βxsin(γx).

And what's left is equal to A:

y = Acos(θ)e^βxcos(γx) + Asin(θ)e^βxsin(γx).

Pull out common factors:

y = Ae^βx(cos(θ)cos(γx) + sin(θ)sin(γx)).

Now use a trigonometic identity to clean it up: cos(A)cos(B) + sin(A)sin(B) = cos(B-A)

y = Ae^βx(cos(γx - θ)

And that's the proof!