Solving a system of equations with trig

Question

How do you solve a system of the following form: $$a = 2\sin(x) - \sin(y) + \sin(x+y)\tag1$$ $$b = 2\sin(y) - \sin(x) + \sin(x+y)\tag2$$ where $a,b$ are constants, and $x,y$ the variables I'd like to solve for. Subtracting $(1)-(2)$ gives an expression for $\sin(y)$. However, rewriting $$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$$ requires an expression for $\cos(y)$. Using $\cos(y)=\sqrt{1-\sin^2(y)}$ results in a complicated equation, which I cannot solve.

Is there an easier way to solve this system?

Answer 1

You are on the right track. Subtracting $(2)$ from $(1)$ gives $\sin x=c+\sin y$ where $c=\frac{a-b}3$ and putting this into $(1)$ gives $$a=2c+\sin y+\sin x\cos y+\sin y\cos x$$ so $$\cos x=-1-\cos y+\frac{b+c-c\cos y}{\sin y}$$ Thus $$\small \sin^2x+\cos^2x\\=\\\small c^2+2c\sin y+\sin^2y+1+\cos^2y+\left(\frac{b+c-c\cos y}{\sin y}\right)^2+2\cos y-2\frac{b+c-c\cos y}{\sin y}-2\cos y\frac{b+c-c\cos y}{\sin y}$$ so $$\left(\frac{b+c-c\cos y}{\sin y}-1\right)^2-2\cos y\left(\frac{b+c-c\cos y}{\sin y}-1\right)+c^2+2c\sin y=0$$ giving $$\frac{b+c-c\cos y}{\sin y}-1=\frac{2\cos y\pm\sqrt{4\cos^2y-4(c^2+2c\sin y)}}2$$ and the half-angle identities give $$\frac b{\sin y}+c\tan\frac y2=2\cos^2\frac y2\pm\sqrt{1-(c+\sin y)^2}$$ I highly doubt there is an analytical solution to this, but once $y$ is found perhaps numerically, $x=\sin^{-1}(c+\sin y)$.

Answer 2

Hint: Multiplying the first equation by $-1$ and adding to the second equation we get $$b-a=3\sin(y)-3\sin(x)$$, solving this equation for $y$ we obtain $$y=-\arcsin(\frac{1}{3}a-\frac{1}{3}b-\sin(x))$$ this equation can be used to eliminate the variable $y$

Answer 3

With a couple of rounds of squaring, one can re-write the two equations using, say, sines alone. The polynomials are degree $6$ in both $\sin x$ and $\sin y$. From there, the Method of Resultants (eg, invoking Mathematica's Resultant function) can eliminate one, leaving a polynomial in the other; when I try this, the final polynomial has four degree-six factors (presumably with many extraneous solutions).

Alternatively, we can convert to complex exponentials via $$\cos \theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right)\qquad \sin \theta = \frac12\left(e^{i\theta} - e^{-i\theta}\right)$$ Defining $u := e^{ix}$ and $v := e^{iy}$, and noting that a comment from OP expresses $a=3c$ and $b=3d$, the equations can be written $$\begin{align} u^2 v^2 + 2 u^2 v - u v^2 + u - 2 v - 1 - 6 i c u v &= 0 \tag{1} \\ u^2 v^2 - u^2 v + 2 u v^2 - 2 u + v - 1 - 6 i d u v &= 0 \tag{2} \end{align}$$ Using resultants to eliminate $v$ gives $$\begin{align} 0 &= \phantom{2}u^6 \\ &+ \phantom{2}u^5 \left(1 - 6 ci\right) \\ &- \phantom{2}u^4 \left( 1 + \phantom{1}8 c^2 - 4 c d - 4 d^2 + 12 ci \right) \\ &- 2 u^3 \left( 1 + 10 c^2 + 4 c d + 4 d^2 \right) \\ &- \phantom{2}u^2 \left( 1 + \phantom{1}8 c^2 - 4 c d - 4 d^2 - 12 ci \right) \\ &+ \phantom{2} u\phantom{^2} \left(1 + 6 ci \right) \\ &+ 1 \end{align} \tag{3}$$ So, we still obtain a sixth-degree polynomial to solve (by numerical methods, unless there's some super-convenient relation between $c$ and $d$ that causes $(3)$ to factor). Of course, by $(1)$, each value of $u$ gives rise to up to two values of $v$; and all candidate $u$s and $v$s will have to tested for viability.

Answer 4

The easiest way to solve the system exists.

We can greatly reduce the degree of the equation (from 6 to 2), simply by calculating the $cos(x+y)$, and then apply the fundamental formula of trigonometry.

The resulting equation is:

$9B sin(y)^{2}+6C sin(y)+F=0$,

with

$B=4(a-b)^{2}+9$,

$C=2a^{3}-6a^{2}b+6ab^{2}-9ª-2b^{3}-18b$,

$F=H+9s^{2}-81$,

$H=(a-b)^{4}$,

$s=a+2b$.

The solutions of the equation are:

$sin(y)=+\frac{-C+\sqrt{C^{2}-BF}}{3B}$,

$sin(y)=-\frac{+C+\sqrt{C^{2}-BF}}{3B}$.

Choosing the first solution we have:

$sin(x)=+\frac{-C+B(a-b)+\sqrt{C^2-BF}}{3B}$,

$sin(x+y)=-\frac{-C-B(a-b)+\sqrt{C^2-BF}}{3B}$.