How to get the following equation in the form $A \cos(ωt+φ)$

Question

$$v = [1+Γ] \cos(ωt)\cos(ωx/c) - [1-Γ] \sin(ωt)\sin(ωx/c) = A(x)\cos\left(ωt+\varphi\right)$$

I need to express the above function as a product of two functions, f(x) and f(t) where one is dependent on distance, the other dependent on time.

The final form will therefore be: $A\sin(ωt + φ)$ or $A\cos(ωt + φ)$ either is fine.

Where $A$ is the function dependent on distance, $x$.

I need to do this so I can isolate the ‘A’ function that will in essence describe the amplitude function of the above wave-equation with transmission line distance.

I’m sure this is an easy problem but I have been really struggleing with doing this. It must be possible.

I would greatly apprciante your help!

Answer 1

We can use the linear combination formula for sine and cosine, that is

$$a\cos x+b\sin x=c\cos(x+\varphi)$$

with

• $c = \operatorname{sign(a)} \sqrt{a^2 + b^2}$

• $\varphi = \operatorname{arctan} \left(-\frac{b}{a}\right)$

that is

$$v = [1+Γ] \cos(ωt)\cos(ωx/c) - [1-Γ] \sin(ωt)\sin(ωx/c) = A(x)\cos\left(ωt+\varphi\right)$$

with

• $A(x) = \operatorname{sign\left([1+Γ] \cos(ωx/c)\right)} \sqrt{([1+Γ] \cos(ωx/c))^2+ ([1-Γ] \sin(ωx/c))^2}=$

$$= \operatorname{sign\left([1+Γ] \cos(ωx/c)\right)}\sqrt{1+Γ^2+2Γ\cos(2ωx/c)}$$

• $\varphi = \operatorname{arctan} \left(-\frac{[1-Γ] \sin(ωx/c)}{[1+Γ] \cos(ωx/c)}\right)$