$\Re(e^{-j\theta}A)-\Re(e^{-j\theta}B)\gt0 \Leftrightarrow cos(\theta-\angle(A-B))\gt0$

Question

I have $\cos(\theta cdot (A-B)$.

Is it valid to say that $\cos(\theta) \cdot (A-B)= \cos(\theta -\angle{(A-B)})$ and if yes which rule is used το come up with such result?

edit:

We know that $\theta \in [0,2\pi)$ and that

\begin{equation} \Re(e^{-j\theta}A)-\Re(e^{-j\theta}B)\gt0 \Leftrightarrow cos(\theta-\angle(A-B))\gt0 \end{equation}

This is the proposition from which my question came up.

$\Re(e^{-j\theta}A)=Acos(\theta)$

$\Re(e^{-j\theta}B)=Bcos(\theta)$

So $LHS = (A-B)cos(\theta)\gt0$

and $RHS = cos(\theta-\angle(A-B))\gt0$

Can somebody explain to me this equivalence?

Answer 1

Try with $\theta=A=B=\pi/4$. This is a good method to see if a formula is valid: give values to the variables. If it fails, the formula is false. If it holds after saveral tries, you can start to think in a proof.

Answer 2

Hint:

In $LHS=\cos(\theta)\cdot (A-B)$, $(A-B)$ is treated as an arbitrary constant multiplied to $\cos (\theta)$

While in $RHS=\cos(\theta-\angle (A-B))$, $(A-B)$ is treated as an angle

For checking validity, use the values of both $A$ & $B$ in radians. It may hold true for some values but not for all the values of $A$ & $B$.