Solving sin⁡x=sin⁡y by using the prosthaphaeresis formulas sin⁡(x)−sin⁡(y)=0 2cos⁡(x+y2),sin⁡(x−y2)=0 There are two possibilities cos⁡(x+y2)=0 or...

Wendy Gutierrez

Wendy Gutierrez

Answered

2022-01-27

Solving sinx=siny by using the prosthaphaeresis formulas
sin(x)sin(y)=0
2cos(x+y2),sin(xy2)=0
There are two possibilities cos(x+y2)=0 or sin(xy2)=0
In the first case:
x+y2=π2+kπ
In the second case:
xy2=kπ
The above two equations give the result:
x=π2+2kπ
y=π2

Answer & Explanation

waijazar1

waijazar1

Expert

2022-01-28Added 13 answers

In the first case:
x+y2=π2+kπ
This is fine and after multiplying by 2, clearly equivalent to supplementary angles having the same sine:
x+y=π+2kπx=πy+2kπ
In the second case:
xy2=kπ
This is fine and after multiplying by 2, clearly equivalent to the same angles (obviously) having the same sine:
xy=2kπx=y+2kπ
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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