Can I Transform x[t] into the Form x[t]=A(t)*cos[ωo*t + θ(t)]?

[brad sue]

Homework Statement

Hi,
I need to prove that:
x[t]=cos(ω0*t)+ cos( ωo*t + Δω*t)

can be transformed into the form:
x[t]=A(t)*cos[ωo*t + θ(t)]

where A(t) and θ(t) are function of Δω.

I have the solution but I cannot find out the way to solve it
A(t)=2|cos(Δω*t)|

and
θ(t)= ArcTan[sin(Δω*t)/(1+cos(Δω*t))]

here I can not figure out how to fin A(t) and θ(t).

please can someone help me ?
thank you
B

The Attempt at a Solution

I have started by using the trigon identity cos(a+b) expansion.

Then, I factor cos[ωo*t] to have 1+cos(Δω*t) and I factor 1+cos(Δω*t) to have the expression under the Arctan.

OK I have:
$$[1+\cos (\Delta \omega t) ] [\cos (\omega_0 t) - \sin (\omega_0 t)\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)}]$$

Now let
$$\theta(t)=\arctan(\frac{\sin (\Delta \omega t)}{1+\cos (\Delta \omega t)})$$

After that I am stuck..
I don't know how to continue the transfromation to have another expansion od the type cos(a+b).

 

[robphy]

In addition to cos(a+b), can you use cos(a-b)?

 

[brad sue]

In addition to cos(a+b), can you use cos(a-b)?

Yes I think I can.