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 sin⁡(x−y2)=0 In the first case: x+y2=π2+kπ In the second case: x−y2=kπ The above two equations give the result: x=π2+2kπ y=π2

Name: 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 sin⁡(x−y2)=0 In the first case: x+y2=π2+kπ In the second case: x−y2=kπ The above two equations give the result: x=π2+2kπ y=π2
Uploaded: 2022-02-23T17:22:57+00:00
Description: 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 sin⁡(x−y2)=0 In the first case: x+y2=π2+kπ In the second case: x−y2=kπ The above two equations give the result: x=π2+2kπ y=π2

Question

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 sin⁡(x−y2)=0 In the first case: x+y2=π2+kπ In the second case: x−y2=kπ The above two equations give the result: x=π2+2kπ y=π2

waijazar1 · Accepted Answer

In the first case:x+y2=π2+kπThis is fine and after multiplying by 2, clearly equivalent to supplementary angles having the same sine:x+y=π+2kπ⇔x=π−y+2kπIn the second case:x−y2=kπThis is fine and after multiplying by 2, clearly equivalent to the same angles (obviously) having the same sine:x−y=2kπ⇔x=y+2kπWhat you did afterwards is considering the system of equations consisting of your two (partial) solution sets, but that doesn&#039;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