What is the geometric interpretation of the sine and cosine angle addition formula in terms of phasors?

Question

Although I knew about the formula of $\sin(a+b)$, its various derivations (including that by $e^{i\theta}$), phasors, etc, but after two years I realised (while doing wave optics) that the formula for $\sin(a+b)$ can be written as follows:

\begin{align*} R\sin(\omega t+\delta)&=R\sin\omega t \cos \delta+R\cos\omega t \sin\delta \\&= (R\cos\delta)\sin\omega t + (R\sin\delta)\cos\omega t \\&= R_x\sin \omega t +R_y\cos\omega t \end{align*}

where $R_x=\cos \delta$ and $R_y=\sin \delta$. That seems to imply that $R\sin(\omega t+\delta)$ somehow means the x component in y direction + y component in x direction. (In case of cos, it is x component in x direction + y component in y direction). What is the geometric interpretation of this in terms of phasors?

I tried to draw some diagrams where an angle is being added to a phasor but got confused about what should the $\sin \omega t$ "direction" represent in the diagram.

Answer 1

The formula you wrote simply says that $\sin(\omega t + \delta)$ is a linear combination of $\sin \omega t$ and $\cos \omega t$. Those two functions are real linearly independent and so they span a plane, where each point represents and is represented by a function of the type $A\sin(\omega t +\delta)$, $A\in\mathbb{R}$ and $0\le\delta< 2\pi$. In particular $R\sin(\omega t+\delta)$ is the point whose coordinates, w.r.t. the base $\{\mathbf{i}, \mathbf{j}\}=\{\cos\omega t, \sin\omega t\}$, are $$(x, y)=(R\sin\delta, R\cos\delta)$$ With those coordinates you can immediately draw the diagram.

Now let's turn our attention to phasors. Let's choose the convention whereby the phasor $1$ is associated to $\cos\omega t$, that is, $\cos\omega t$ can thus be reconstructed by the formula $$\cos\omega t=\Re\left[\mathbf{1}e^{j\omega t}\right]\tag{1}$$and the phasor $-j$ to $\sin\omega t$, that is, $\sin\omega t$ can thus be reconstructed by the formula $$\sin\omega t=\Re\left[-\boldsymbol{j}e^{j\omega t}\right]\tag{2}$$ Then it follows that the phasor $\bar R$ corresponding to $R\sin(\omega t+\delta)$ must be such that the following reconstruction formula holds: $R\sin(\omega t+\delta)=\Re\left[\bar Re^{j\omega t}\right]$. Then by $(1)$ and $(2)$

$$R\sin(\omega t+\delta)=R(\cos\delta)\Re\left[-je^{j\omega t}\right]+R(\sin\delta)\Re\left[e^{j\omega t}\right]=\Re\left[R(\sin\delta-j\cos\delta)e^{j\omega t}\right]$$ So the phasor corresponding to $R\sin(\omega t+\delta)$ is $R(\sin\delta-j\cos\delta)=Re^{j(\delta-\frac{\pi}{2})}$, that is $$-jRe^{j\delta}$$ The phasor diagram is drawn with the real component $R\sin\delta$ and the imaginary component $-R\cos\delta$.

Pay attention to the fact that this diagram is shaped differently w.r.t. that in the $\{\mathbf{i},\mathbf{j}\}$ plane. Phasors in the complex plane become real functions only when multiplied by $e^{j\omega t}$ and then projected on to the real axis, while points in the $\{\mathbf{i},\mathbf{j}\}$ plane represent, as is, real functions. So they're two completely different planes. In particular $\sin\omega t$ in the complex plane is represented by the unit phasor on the negative $y-$axis, while in the $\{\mathbf{i},\mathbf{j}\}$ plane by a unit vector in the positive $y-$axis. Vectors in the $\{\mathbf{i},\mathbf{j}\}$ plane are sometimes called instantaneous phasors (to distinguish them by conventional phasors of the complext plane) when such a real plane is embedded in $\mathbb{C}$, by sending $\mathbf{i}$ in $1$ and $\mathbf{j}$ in $j$.

Note also that for $R\cos(\omega t+\delta)$ you forgot a minus sign in your consideration. It has coordinates $(R\cos\delta,-R\sin\delta)$ in the chosen base, and phasor representation $Re^{j\delta}$ with the chosen convention.

Hope this help even though so much time has passed.