I have a function of $x$ with 5 parameters $$f(x)=\frac{A + B\sin(x)}{C - D\sin(x-\phi)}$$ which I would like to represent in the form $$f(x)=E\sin(x-\theta).$$
What interests me in particular are lower and upper phase bounds of $f(x)$ (i.e. parameter $\theta$) given that:
- $0<\phi<\pi/2$ and
- $B>D$.
I am fully aware that I probably could not get explicit expressions for $E$ and $\theta$.
In which direction should I proceed, which tool should I use? I am totally new to this kind of analysis.
Assuming that $b$ is significantly larger than $d$, you could generate a table $\left[x_i,F(x_i) \right]$ over the range of interest and try to curve fit the points using as a model $$F(x)=e\sin(x-\theta)+\color{red}{f}$$ but the result will just be an approximation.
For illustration purposes, I used $a=5$, $b=6$, $c=7$, $d=1$, $\phi=\frac \pi 3$ and generated $25$ equally spaced values in the range $0\leq x \leq 2\pi$.
A standard nonlinear regression lead to $$F(x)=0.92056 \sin (x-0.10377)+0.75191$$ to which corresponds $R^2=0.9978$ and the following results $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\e & 0.92056 & 0.01452 & \{0.89037,0.95075\} \\ \theta & 0.10377 & 0.01519 & \{0.07218,0.13536\} \\ f & 0.75191 & 0.01008 & \{0.73095,0.77287\} \\ \end{array}$$
Edit
I do not know where we should go with the following.
Assuming, as you wrote in comments, that, in the model $$F(x)=\frac{a + b\sin(x)}{c - d\sin(x-\phi)}$$ $c$ is much larger than $d$. So, by Taylor, $$F(x)\approx G(x)=\frac{(a+b \sin (x)) (c+d \sin (x-\phi ))}{c^2}$$ Now computing $$\Psi=\int_0^{2\pi}\left(G(x)-e\sin(x-\theta)-f\right)^2\,dx$$ which is doable, compute $$\frac{d\Psi}{de}\qquad \frac{d\Psi}{d\theta}\qquad \frac{d\Psi}{df}$$ and try to make them equal to zero.
The formulae are too long to type and I shall go to the results (obtained by successive eliminations) $$f=\frac{2 a c+b d \cos (\phi )}{2 c^2}$$ $$e=\frac{a d \cos (\theta -\phi )+b c \cos (\theta )}{c^2}$$ and we end with an equation for $\theta$ $$a d (a d \sin (2 (\theta -\phi ))+2 b c \sin (2 \theta -\phi ))+b^2 c^2 \sin (2 \theta )=0$$ that we need to solve using a numerical method such as Newton's.
Using the same numbers as before, we get $$e=\frac{5}{49} \sin \left(\theta +\frac{\pi }{6}\right)+\frac{6 }{7}\cos (\theta )\qquad \qquad f=\frac{73}{98}$$ and the equation to be solved is $$1764 \sin (2 \theta )-25 \sin \left(2 \theta+\frac{\pi}{3} \right)-420 \cos \left(2 \theta +\frac{\pi }{6}\right)=0$$ the first solution of which being $\theta\approx 0.0970008$, making $e=0.912453$ and $f=0.744898$ which is quite close to what was obtained by nonlinear regression.
Edit
For sure, we could continue the initial Taylor expansion and use $$G(x)=\frac{(a+b \sin (x)) \left(c^2+c d \sin (x-\phi )+d^2 \sin ^2(x-\phi )\right)}{c^3}$$ and use the same procedure as above.
This would lead to $$f=\frac{2 a c^2+a d^2+b c d \cos (\phi )}{2 c^3}$$ $$e=\frac{d (4 a c \cos (\theta -\phi )+b d \cos (\theta -2 \phi ))+2 b \left(2 c^2+d^2\right) \cos (\theta )}{4 c^3}$$ leaving us with the equation $$d \left(d \left(4 \sin (2 (\theta -\phi )) \left(4 a^2 c^2+b^2 \left(2 c^2+d^2\right)\right)+b d (8 a c \sin (2 \theta -3 \phi )+b d \sin (2 (\theta -2 \phi )))\right)+16 a b c \left(2 c^2+d^2\right) \sin (2 \theta -\phi )\right)+4 b^2 \left(2 c^2+d^2\right)^2 \sin (2 \theta )=0$$ to be solved for $\theta$.
Using the same values as before, $\theta\approx 0.100410$, making $e=0.919354$ and $f=0.752187$ which is still closer to what was obtained by nonlinear regression.