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How to know whether √(x-y)² is right or √(y-x)² is right? [closed]

-3

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sin⁻¹x + sin⁻¹y = sin⁻¹{x√(1-y²) + y√(1-x²)} 2sin⁻¹x =sin⁻¹{2x√(1-x²)} 3sin⁻¹x = 2sin⁻¹x + sin⁻¹x ⇒sin⁻¹{2x√(1-x²)}+ sin⁻¹x y=2x√(1-x²) x=x ⇒sin⁻¹[x√(1-{2x√(1-x²)}²+2x√(1-x²)√(1-x²)]
⇒ sin⁻¹[x√(1-2(2x²)+(2x²)²)+ 2x - 2x³] ⇒sin⁻¹[x√({2x²-1}²)+ 2x - 2x³]

⇒sin⁻¹[x{2x²-1}+ 2x - 2x³] ⇒sin⁻¹[x] This is clearly wrong. But I dont understand why is this wrong.

algebra-precalculus

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asked Mar 17 at 8:04

user654700user654700

585

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closed as off-topic by José Carlos Santos, Song, Alex Provost, Thomas Shelby, Eevee Trainer Mar 17 at 9:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Song, Alex Provost, Thomas Shelby, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Hi and welcome to MSE! Could you please use Mathjax in your posts? It's easier to read. Here you have a nice tutorial: math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Dr. Mathva
  Mar 17 at 8:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember that $sqrt{u^2} = color{red}{|}ucolor{red}{|}$ for all real $u$ (absolute value sign required).
  $endgroup$
  – Minus One-Twelfth
  Mar 17 at 8:22
  
  
  

  
  
  
  

add a comment | 

-3

$begingroup$

sin⁻¹x + sin⁻¹y = sin⁻¹{x√(1-y²) + y√(1-x²)} 2sin⁻¹x =sin⁻¹{2x√(1-x²)} 3sin⁻¹x = 2sin⁻¹x + sin⁻¹x ⇒sin⁻¹{2x√(1-x²)}+ sin⁻¹x y=2x√(1-x²) x=x ⇒sin⁻¹[x√(1-{2x√(1-x²)}²+2x√(1-x²)√(1-x²)]
⇒ sin⁻¹[x√(1-2(2x²)+(2x²)²)+ 2x - 2x³] ⇒sin⁻¹[x√({2x²-1}²)+ 2x - 2x³]

⇒sin⁻¹[x{2x²-1}+ 2x - 2x³] ⇒sin⁻¹[x] This is clearly wrong. But I dont understand why is this wrong.

algebra-precalculus

share|cite|improve this question

asked Mar 17 at 8:04

user654700user654700

585

$endgroup$

closed as off-topic by José Carlos Santos, Song, Alex Provost, Thomas Shelby, Eevee Trainer Mar 17 at 9:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Song, Alex Provost, Thomas Shelby, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Hi and welcome to MSE! Could you please use Mathjax in your posts? It's easier to read. Here you have a nice tutorial: math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Dr. Mathva
  Mar 17 at 8:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember that $sqrt{u^2} = color{red}{|}ucolor{red}{|}$ for all real $u$ (absolute value sign required).
  $endgroup$
  – Minus One-Twelfth
  Mar 17 at 8:22
  
  
  

  
  
  
  

add a comment | 

$begingroup$

sin⁻¹x + sin⁻¹y = sin⁻¹{x√(1-y²) + y√(1-x²)} 2sin⁻¹x =sin⁻¹{2x√(1-x²)} 3sin⁻¹x = 2sin⁻¹x + sin⁻¹x ⇒sin⁻¹{2x√(1-x²)}+ sin⁻¹x y=2x√(1-x²) x=x ⇒sin⁻¹[x√(1-{2x√(1-x²)}²+2x√(1-x²)√(1-x²)]
⇒ sin⁻¹[x√(1-2(2x²)+(2x²)²)+ 2x - 2x³] ⇒sin⁻¹[x√({2x²-1}²)+ 2x - 2x³]

⇒sin⁻¹[x{2x²-1}+ 2x - 2x³] ⇒sin⁻¹[x] This is clearly wrong. But I dont understand why is this wrong.

algebra-precalculus

share|cite|improve this question

asked Mar 17 at 8:04

user654700user654700

585

$endgroup$

share|cite|improve this question

asked Mar 17 at 8:04

user654700user654700

585

share|cite|improve this question

asked Mar 17 at 8:04

user654700user654700

585

asked Mar 17 at 8:04

user654700user654700

585

asked Mar 17 at 8:04

user654700user654700

585

2 Answers
2

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1

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We have $$sqrt{(x-y)^2}=sqrt{(-1)^2(y-x)^2}=sqrt{(y-x)^2}$$

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answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

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add a comment | 

1

$begingroup$

$$sqrt{(x-y)^2}=sqrt{(y-x)^2}=mid x-y mid $$

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answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

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add a comment | 

1

$begingroup$

We have $$sqrt{(x-y)^2}=sqrt{(-1)^2(y-x)^2}=sqrt{(y-x)^2}$$

share|cite|improve this answer

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

1

$begingroup$

We have $$sqrt{(x-y)^2}=sqrt{(-1)^2(y-x)^2}=sqrt{(y-x)^2}$$

share|cite|improve this answer

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

$begingroup$

We have $$sqrt{(x-y)^2}=sqrt{(-1)^2(y-x)^2}=sqrt{(y-x)^2}$$

share|cite|improve this answer

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

share|cite|improve this answer

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 17 at 8:12

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

1

$begingroup$

$$sqrt{(x-y)^2}=sqrt{(y-x)^2}=mid x-y mid $$

share|cite|improve this answer

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

$endgroup$

add a comment | 

1

$begingroup$

$$sqrt{(x-y)^2}=sqrt{(y-x)^2}=mid x-y mid $$

share|cite|improve this answer

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

$endgroup$

add a comment | 

$begingroup$

$$sqrt{(x-y)^2}=sqrt{(y-x)^2}=mid x-y mid $$

share|cite|improve this answer

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

$endgroup$

share|cite|improve this answer

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423

answered Mar 17 at 8:13

Paras KhoslaParas Khosla

2,701423