Solving $\sin x = \sin y$ by using the prosthaphaeresis formulas

Question

I'd like to solve the equation $\sin x = \sin y$ by using the prosthaphaeresis formulas:

$$ \sin(x)-\sin(y)=0 $$

$$ 2 \cos \left( \frac{x+y}{2} \right) \, \sin \left( \frac{x-y}{2} \right) = 0 $$

There are two possibilities $\cos \left( \frac{x+y}{2} \right)=0$ or $\sin \left( \frac{x-y}{2} \right)=0$.

In the first case:

$$ \frac{x+y}{2} = \frac{\pi}{2} + k \, \pi$$

In the second case:

$$ \frac{x-y}{2} = k \, \pi $$

The above two equations give the result:

$$ x = \frac{\pi}{2} + 2 \, k \, \pi $$

$$ y = \frac{\pi}{2} $$

It seems that the solution is wrong; can you show me the right solution and the way to get it (by using the prosthaphaeresis formulas if possible) please?

Thank you for your willingness.

Answer 1

In the first case:

$$ \frac{x+y}{2} = \frac{\pi}{2} + k \, \pi$$

This is fine and after multiplying by $2$, clearly equivalent to supplementary angles having the same sine: $$x+y=\pi+2k\pi \iff x=\pi-y+2k\pi $$

In the second case:

$$ \frac{x-y}{2} = k \, \pi $$

This is fine and after multiplying by $2$, clearly equivalent to the same angles (obviously) having the same sine: $$x-y=2k\pi \iff x=y+2k\pi $$

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What you did afterwards is considering the system of equations consisting of your two (partial) solution sets, but that doesn't make sense: the solution set is the combination (union) of these solutions. There was no system to solve, so you were already there and went too far ;-).

Answer 2

The two cases you have are separated by an or, not a and. If one of the two equations is true, then you already have the solution.

Your mistake is looking for a solution that solves both equations, when really, any solution that solves at least one of the equations is already enough.