Closed form solution for $t_1$ and $t_2$ in the system $\sin(t_1 - t_2)=c_1\sin(t_1)$, $\sin(t_2)=c_2\sin(t_1)$

Question

I have the following system of equations

$$\sin(t_1 - t_2) = c_1 \sin(t_1)$$ $$\sin(t_2) = c_2 \sin(t_1)$$

where $c_1, c_2$ are known constants.

And I would like to find all the solutions in closed form over $(t_1, t_2) \in (0, 2\pi) \times (0, 2\pi) $

Note that it is required that $t_1 \ne 0$ and $t_2 \ne 0 $ and $t_1 \ne t_2 $.

Context:

The context of this problem is finding conjugate diameters of an ellipse passing through three given points, and that has a known center.

In particular, suppose the center is the vector $C$ and the three points through which the ellipse passes are $P_1, P_2, P_3$ then a parametric equation of this ellipse is

$ P(t) = C + F_1 \cos(t) + F_2 \sin(t) $

where $F_1$ and $F_2$ are called the conjugate semi-diameters, and are not unique. So now define

$V_1 = P_1 - C , V_2 = P_2 - C , P_3 - C $ then

Then the shifted ellipse has the equation

$ V(t) = F_1 \cos(t) + F_2 \sin(t) $

We can choose $V(0)$ to be $V_1$ , then

$ V(t) = V_1 \cos(t) + F_2 \sin(t) $

At times $t_1$ and $t_2$ we have

$ V(t_1) = V_2 = V_1 \cos(t_1) + F_2 \sin(t_1) $

Now, since we are in $2D$, then $V_3 = c_1 V_1 + c_2 V_2$, and $F_2 = a V_1 + b V_2 $ where $c_1$ and $c_2$ are known, while $a$ and $b$ are unknown.

With this, and from the linear independence of $V_1$ and $V_2$ , it follows that

$ 0 = \cos(t_1) + a \sin(t_1) \hspace{15pt}(1)$

$ 1 = b \sin(t_1) \hspace{15pt}(2)$

And finally, at $t = t_2$ we have

$ V(t_2) = V_3 = c_1 V_1 + c_2 V_2 = V_1 \cos(t_2)+ (a V_1 + b V_2) \sin(t_2) $

From which we get another two equations

$ c_1 = \cos(t_2) + a \sin(t_2) \hspace{15pt}(3) $

$ c_2 = b \sin(t_2) \hspace{15pt}(4)$

Eliminating $a$ and $b$ from equations $(1)-(4)$ yields the system of two equations above in $t_1 $ and $t_2 $.

Answer 1

$$\begin{align} \sin (t_1)\cos(t_2)-\cos(t_1)\sin(t_2)=c_1 \sin(t_1) \end{align}$$

Plug in the second equation: $\sin(t_2)=c_2 \sin(t_1),~~\cos(t_2)=\pm\sqrt{1-c_2^2\sin^2(t_1)}$

$$\sin (t_1)\left(\pm\sqrt{1-c_2^2\sin^2(t_1)}\right)-\cos(t_1)c_2 \sin(t_1)=c_1 \sin(t_1)$$

Note $t_1\neq\pi,$ otherwise, $t_1=\pi\Rightarrow t_1=t_2$, which gives contradictions. $$\begin{align} \left(\pm\sqrt{1-c_2^2\sin^2(t_1)}\right)-c_2\cos(t_1)&=c_1 \\ \\ \left(\pm\sqrt{1-c_2^2\sin^2(t_1)}\right)&=c_1+c_2 \cos(t_1)\\ \\ 1-c_2^2\sin^2(t_1)&=c_1^2+c_2^2 \cos^2(t_1)+2c_1c_2\cos(t_1)\\ \\ 1-c_1^2-c_2^2&=2c_1c_2\cos(t_1)\\ \\ \cos(t_1)&=\frac{1-c_1^2-c_2^2}{2c_1c_2}\\ \end{align}$$

Similarly, we can solve $\cos(t_2)$

$$\cos(t_2)=\frac{1+c_1^2-c_2^2}{2c_1}$$

Answer 2

We will transform the first equation first and then the second equation to get a linear system.

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Expand $\sin(t_1-t_2)$ so the first equation becomes

$$ \sin t_1\cos t_2-\cos t_1\color{Green}{\sin t_2}= c_1\sin t_1~.$$

Divide by $\sin t_1$ (noting $\color{Green}{\sin t_2}=c_2\sin t_1$ from the first equation) to get

$$ \cos t_2-c_2\cos t_1=c_1~. $$

Define $x_1=\cos t_1$, $x_2=\cos t_2$ for simplicity:

$$ x_2-c_2x_1=c_1~. \tag{1} $$

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Square the second equation $\sin t_2=c_2\sin t_1$ to get

$$1-x_2^2=c_2^2(1-x_1^2)$$

and rearrange it to become

$$1-c_2^2=x_2^2-c_2^2x_1^2 ~.$$

Divide this by $c_1=x_2-c_2x_1$ to get

$$ \frac{1-c_2^2}{c_1}=x_2+c_2x_1 \tag{2}$$

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Now $(1)$ and $(2)$ make a system

$$ \begin{cases} x_2-c_2x_1 &=c_1 \\ x_2+c_2x_1 & =(1-c_2^2)/c_1 \end{cases} $$

By adding or substracting the equations and rescaling we get the solutions

$$ \begin{cases} \cos t_1 & \displaystyle =\frac{1}{2}\left(c_1+\frac{1-c_2^2}{c_1}\right), \\[5pt] \cos t_2 & \displaystyle =\frac{1}{2c_2}\left(-c_1+\frac{1-c_2^2}{c_1}\right) \end{cases} $$

This determines $t_1$, $t_2$ up to reflex angle at least.