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1

$begingroup$

Hi everyone!

Can someone explain me this exercice? My solution was $thetain [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$, which is incorrect, but I can't understand why...

The other parts were correct.

I thought $x leq y Leftrightarrow rcdotcos(theta)sin(phi) le rcdot sin(theta)sin(phi) Leftrightarrow 1 le operatornametg(theta)$ so $theta in [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$

calculus

share|cite|improve this question

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

$endgroup$

add a comment | 

1

$begingroup$

Hi everyone!

Can someone explain me this exercice? My solution was $thetain [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$, which is incorrect, but I can't understand why...

The other parts were correct.

I thought $x leq y Leftrightarrow rcdotcos(theta)sin(phi) le rcdot sin(theta)sin(phi) Leftrightarrow 1 le operatornametg(theta)$ so $theta in [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$

calculus

share|cite|improve this question

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

$endgroup$

add a comment | 

$begingroup$

Hi everyone!

Can someone explain me this exercice? My solution was $thetain [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$, which is incorrect, but I can't understand why...

The other parts were correct.

I thought $x leq y Leftrightarrow rcdotcos(theta)sin(phi) le rcdot sin(theta)sin(phi) Leftrightarrow 1 le operatornametg(theta)$ so $theta in [fracpi4,fracpi2]cup[frac5pi4,frac3pi2)$

calculus

share|cite|improve this question

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

$endgroup$

share|cite|improve this question

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

share|cite|improve this question

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

edited 2 days ago

Ingix

4,867159

edited 2 days ago

Ingix

4,867159

asked 2 days ago

aluno20000aluno20000

446

asked 2 days ago

aluno20000aluno20000

446

1 Answer
1

active

oldest

votes

1

$begingroup$

You forgot that dividing (and multiplying) an inequality by some number requires checking the sign of the divisor (or factor), in order to find out if the inequality changes the direction.

That's because from $a le b$ it surely does not follow $-a le -b$, the opppsite is true: $a le b Leftrightarrow -a ge -b$.

That doesn't affect the common factor $sin(phi)$, as that is non-negative for $0 le phi < pi$. But $cos(theta)$ changes signs at $theta=fracpi2,frac3pi2$. That means when you divide by it to get the inequality involving $tan(theta)$ you need to consider 2 cases:

1) $cos(theta) ge 0$, or $theta in [0,pi over 2] cup [3pi over 2,2pi)$. Your correct solution set for $tan(theta) ge 1$ needs then be restricted to the given interval of $theta$, which gives $I_1(theta)=[pi over 4,pi over 2]$.

2) $cos(theta) < 0$, or $theta in (pi over 2, 3pi over 2)$. In that interval, then the inequality becomes $tan(theta) le 1$. The solution set of this, restricted to $(pi over 2, 3pi over 2)$ is $I_2(theta)=(pi over 2,5pi over 4]$.

Finally, the possible set for $theta$ is $I_1(theta) cup I_2(theta) = [pi over 4,5pi over 4]$, as given in the book.

So the main thing to remember is that multiplying/dividing an inequality requires extra steps to get the correct direction of the inequality!

share|cite|improve this answer

answered 2 days ago

IngixIngix

4,867159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!
  $endgroup$
  – aluno20000
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

You forgot that dividing (and multiplying) an inequality by some number requires checking the sign of the divisor (or factor), in order to find out if the inequality changes the direction.

That's because from $a le b$ it surely does not follow $-a le -b$, the opppsite is true: $a le b Leftrightarrow -a ge -b$.

That doesn't affect the common factor $sin(phi)$, as that is non-negative for $0 le phi < pi$. But $cos(theta)$ changes signs at $theta=fracpi2,frac3pi2$. That means when you divide by it to get the inequality involving $tan(theta)$ you need to consider 2 cases:

1) $cos(theta) ge 0$, or $theta in [0,pi over 2] cup [3pi over 2,2pi)$. Your correct solution set for $tan(theta) ge 1$ needs then be restricted to the given interval of $theta$, which gives $I_1(theta)=[pi over 4,pi over 2]$.

2) $cos(theta) < 0$, or $theta in (pi over 2, 3pi over 2)$. In that interval, then the inequality becomes $tan(theta) le 1$. The solution set of this, restricted to $(pi over 2, 3pi over 2)$ is $I_2(theta)=(pi over 2,5pi over 4]$.

Finally, the possible set for $theta$ is $I_1(theta) cup I_2(theta) = [pi over 4,5pi over 4]$, as given in the book.

So the main thing to remember is that multiplying/dividing an inequality requires extra steps to get the correct direction of the inequality!

share|cite|improve this answer

answered 2 days ago

IngixIngix

4,867159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!
  $endgroup$
  – aluno20000
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

You forgot that dividing (and multiplying) an inequality by some number requires checking the sign of the divisor (or factor), in order to find out if the inequality changes the direction.

That's because from $a le b$ it surely does not follow $-a le -b$, the opppsite is true: $a le b Leftrightarrow -a ge -b$.

That doesn't affect the common factor $sin(phi)$, as that is non-negative for $0 le phi < pi$. But $cos(theta)$ changes signs at $theta=fracpi2,frac3pi2$. That means when you divide by it to get the inequality involving $tan(theta)$ you need to consider 2 cases:

1) $cos(theta) ge 0$, or $theta in [0,pi over 2] cup [3pi over 2,2pi)$. Your correct solution set for $tan(theta) ge 1$ needs then be restricted to the given interval of $theta$, which gives $I_1(theta)=[pi over 4,pi over 2]$.

2) $cos(theta) < 0$, or $theta in (pi over 2, 3pi over 2)$. In that interval, then the inequality becomes $tan(theta) le 1$. The solution set of this, restricted to $(pi over 2, 3pi over 2)$ is $I_2(theta)=(pi over 2,5pi over 4]$.

Finally, the possible set for $theta$ is $I_1(theta) cup I_2(theta) = [pi over 4,5pi over 4]$, as given in the book.

So the main thing to remember is that multiplying/dividing an inequality requires extra steps to get the correct direction of the inequality!

share|cite|improve this answer

answered 2 days ago

IngixIngix

4,867159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer!
  $endgroup$
  – aluno20000
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

You forgot that dividing (and multiplying) an inequality by some number requires checking the sign of the divisor (or factor), in order to find out if the inequality changes the direction.

That's because from $a le b$ it surely does not follow $-a le -b$, the opppsite is true: $a le b Leftrightarrow -a ge -b$.

That doesn't affect the common factor $sin(phi)$, as that is non-negative for $0 le phi < pi$. But $cos(theta)$ changes signs at $theta=fracpi2,frac3pi2$. That means when you divide by it to get the inequality involving $tan(theta)$ you need to consider 2 cases:

1) $cos(theta) ge 0$, or $theta in [0,pi over 2] cup [3pi over 2,2pi)$. Your correct solution set for $tan(theta) ge 1$ needs then be restricted to the given interval of $theta$, which gives $I_1(theta)=[pi over 4,pi over 2]$.

2) $cos(theta) < 0$, or $theta in (pi over 2, 3pi over 2)$. In that interval, then the inequality becomes $tan(theta) le 1$. The solution set of this, restricted to $(pi over 2, 3pi over 2)$ is $I_2(theta)=(pi over 2,5pi over 4]$.

Finally, the possible set for $theta$ is $I_1(theta) cup I_2(theta) = [pi over 4,5pi over 4]$, as given in the book.

So the main thing to remember is that multiplying/dividing an inequality requires extra steps to get the correct direction of the inequality!

share|cite|improve this answer

answered 2 days ago

IngixIngix

4,867159

$endgroup$

share|cite|improve this answer

answered 2 days ago

IngixIngix

4,867159

answered 2 days ago

IngixIngix

4,867159

answered 2 days ago

IngixIngix

4,867159