Wendy Gutierrez

2022-01-27

Solving $\mathrm{sin}x=\mathrm{sin}y$ by using the prosthaphaeresis formulas
$\mathrm{sin}\left(x\right)-\mathrm{sin}\left(y\right)=0$
$2\mathrm{cos}\left(\frac{x+y}{2}\right),\mathrm{sin}\left(\frac{x-y}{2}\right)=0$
There are two possibilities
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:
$x=\frac{\pi }{2}+2k\pi$
$y=\frac{\pi }{2}$

waijazar1

Expert

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:
$x+y=\pi +2k\pi ⇔x=\pi -y+2k\pi$
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:
$x-y=2k\pi ⇔x=y+2k\pi$
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

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