Jtdylktuy

How are systems such as the following solved analytically?

$$begin{align}
cos x-cos y &=phantom{-}0.2187 \
sin x-sin y &=-0.469
end{align}$$

Wolfram alpha gives the analytical solution so there has to be away, but I cant figure it out.

From the comments:

I am stuck at this point. It comes from a complex equation problem where I have complex numbers $A = B + Ge^{iC}$ and $D = B + Ge^{iE}$, where I know $A$ and $D$ and want to find $B$. I simplified it to this problem but don't know where to go next.

Label your right-hand sides as $a$ and $b$.

Using the prosthaphaeresis formulae, we have
$$begin{align*}
-2sin{tfrac{1}{2}(x+y)} sin{tfrac{1}{2}(x-y)} &= a \
2cos{tfrac{1}{2}(x+y)} sin{tfrac{1}{2}(x-y)} &= b
end{align*}$$
Squaring and adding gives
$$ sin^2{tfrac{1}{2}(x-y)} = frac{a^2+b^2}{4}. tag{*} $$

We now have a small pile of cases:

1. $a=b=0$. We must have $sin{tfrac{1}{2}(x-y)} = 0$, so $x-y=0 pmod{2pi}$, and we can say nothing else.


2. $b neq a=0$. The only possibility is $sin{tfrac{1}{2}(x+y)} =0$, so $x+y=0 pmod{2pi}$. Then $cos{tfrac{1}{2}(x+y)} = pm 1$, so we would have to solve $sin{tfrac{1}{2}(x-y)} = pm b/2$, and we end up with 2 possible solutions in $[0,2pi)$.


3. $a neq b=0$. Left as an exercise: you can do it in exactly the same way as 2.


4. Dividing the prosthaphaeresis equations gives
   $$ tan{tfrac{1}{2}(x+y)} = frac{-a}{b}, tag{**} $$
   and you have to solve (*) and (**) simultaneously. My geometrical argument below suggests this case has 4 solutions in $[0,2pi)$.

@abel's answer gives me another thought: take the unit circle and a rectangle of side lengths $a$ and $b$ with sides parallel to the axes. If we put the opposite vertices on the unit circle, their coordinates are $(cos{x},sin{x})$ and $(cos{y},sin{y})$ for some angles $x$ and $y$, and $x$, $y$ satisfy the given equations.

There are generically four ways to place this rectangle to satisfy this (2 diagonals that can be the chord, and 2 chords in the circle of that length and angle for each chord), so there are normally four solutions to the equations.

If one of the side lengths is zero, we have only two solutions (since we then have a horizontal (or vertical) line, which means only one choice of "diagonal"). If both are zero, then it's just a point on the circle, which can be anywhere, and hence $x+y$ is not determined.

Hint: If you square the two equations and add them, the square terms sum to $1$ and the cross terms give you $cos(x-y)$ so you can get $x-y$. If you multiply the two you get $sin(x+y)+cos(x-y)$. Not the most elegant, but it will work.

i don't think there is a unique answer to this problem. here is a reason why. take a chord $AB$ of length $sqrt{0.2187^2 + (-0.469)^2} = 0.5175$ on the unit circle. let $A=(cos x, sin x), B = cos y, sin y)$ only the difference $x - y$ is unique and equals $2 sin^{-1}(0.5175) = 62.327^circ$

I don’t have a solution, but maybe this can help you or someone along?

Squaring both equations gives:

$$0.2187^2=(cos x-cos y)^2implies\ 0.04782969=cos^2x-2cos xcos y+cos^2y$$

$$(-0.469)^2=(sin x-sin y)^2implies\ 0.1025703=sin^2x-2sin xsin y+sin^2y$$

Add them together:

$$0.15039999=(sin^2x+cos^2x)+2sin xsin y+2cos xcos y+(sin^2y+cos^2y)$$

Terms in parentheses add to $1$, leaving:

$$0.15039999=2+2sin xsin y+2cos xcos y\ 0.15039999=2+cos x-y +cos x+y+cos x-y -cos x+y\ −1.84960001 =2cos x-y$$