Does the equation
$$A\cos(\theta) + B\sin(\theta) = \sqrt{A^2+B^2}\cos(\theta + \gamma) \label{1} \tag{1}$$
with $\gamma = \arg(A-jB)$
require that $A$ and $B$ be real, or can they be complex?
Consider the case $B= jA$ which results in:
$$A\cos(\theta) + jA\sin(\theta) = Ae^{j\theta}$$
Using $\ref{1}$ this results in:
$$=\sqrt{A^2-A^2}\cos(\theta + arg(2A)) = 0$$
Which appears to confirm the answer is no. So then, is there a unified relationship for $A\cos(\theta) + B\sin(\theta)$ that is closest to the form of \ref{1} and allows $A$ and $B$ to be real, imaginary or complex: $A, B \in \mathbb{C}$, $\theta \in \mathbb{R}$ (and \ref{1} is just a simplification of this for A, B real)?
I got this far toward a geometric solution with two cases with A and B both real and with A real and B imaginary as shown below in case this helps toward the analytic result, along with subsequent more significant progress which I provided as an answer. However I would be very interested in a more concise formulation toward a solution or comments on how the answer I provided may be further simplified (toward the form in \ref{1}).
$$A\cos(\theta) + B\sin(\theta)$$
$$= \frac{A}{2}e^{j\theta} + \frac{A}{2}e^{-j\theta} + \frac{B}{2j}e^{j\theta} - \frac{B}{2j}e^{-j\theta}$$ $$= \frac{A}{2}e^{j\theta} + \frac{A}{2}e^{-j\theta} - \frac{jB}{2}e^{j\theta} + \frac{jB}{2}e^{-j\theta}$$
Case with A, B real to confirm known relationship resulting in $A\cos(\theta) + B\sin(\theta) = \sqrt{A^2+B^2}\cos(\theta + \gamma)$:
Case with real A and imaginary B resulting in $\frac{A+jB}{2}\cos(\theta) - jBe^{j\theta}$:
from Hot Weekly Questions - Mathematics Stack Exchange
Dan Boschen
Post a Comment