How to do $arcsin(1/2)$ by hand?

0

$begingroup$

Can I get help on how I can calculate this without a calculator? I know it's the inverse of $sin(1/2)$, but I'm still a little confused on how I get $pi / 2$ from this.

Thanks!

calculus

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edited Jan 23 at 6:47

Ali

1,9812520

asked Jan 23 at 5:33

mingming

3906

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But $sin pi/2$ is $1,$ not $1/2.$
  $endgroup$
  – coffeemath
  Jan 23 at 5:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a triangle and use the definition of the sine function.
  $endgroup$
  – John Douma
  Jan 23 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
  $endgroup$
  – eyeballfrog
  Jan 23 at 6:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog I think a functional inverse was intended, not a reciprocal.
  $endgroup$
  – J.G.
  Jan 23 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right @J.G.
  $endgroup$
  – ming
  Jan 23 at 20:22
  

  
  
  
  

add a comment | 

0

$begingroup$

Can I get help on how I can calculate this without a calculator? I know it's the inverse of $sin(1/2)$, but I'm still a little confused on how I get $pi / 2$ from this.

Thanks!

calculus

share|cite|improve this question

edited Jan 23 at 6:47

Ali

1,9812520

asked Jan 23 at 5:33

mingming

3906

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But $sin pi/2$ is $1,$ not $1/2.$
  $endgroup$
  – coffeemath
  Jan 23 at 5:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a triangle and use the definition of the sine function.
  $endgroup$
  – John Douma
  Jan 23 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
  $endgroup$
  – eyeballfrog
  Jan 23 at 6:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog I think a functional inverse was intended, not a reciprocal.
  $endgroup$
  – J.G.
  Jan 23 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right @J.G.
  $endgroup$
  – ming
  Jan 23 at 20:22
  

  
  
  
  

add a comment | 

0

0

0

0

$begingroup$

Can I get help on how I can calculate this without a calculator? I know it's the inverse of $sin(1/2)$, but I'm still a little confused on how I get $pi / 2$ from this.

Thanks!

calculus

share|cite|improve this question

edited Jan 23 at 6:47

Ali

1,9812520

asked Jan 23 at 5:33

mingming

3906

$endgroup$

Can I get help on how I can calculate this without a calculator? I know it's the inverse of $sin(1/2)$, but I'm still a little confused on how I get $pi / 2$ from this.

Thanks!

calculus

calculus

share|cite|improve this question

edited Jan 23 at 6:47

Ali

1,9812520

asked Jan 23 at 5:33

mingming

3906

share|cite|improve this question

edited Jan 23 at 6:47

Ali

1,9812520

asked Jan 23 at 5:33

mingming

3906

share|cite|improve this question

share|cite|improve this question

edited Jan 23 at 6:47

Ali

1,9812520

edited Jan 23 at 6:47

Ali

1,9812520

edited Jan 23 at 6:47

Ali

1,9812520

1,9812520

asked Jan 23 at 5:33

mingming

3906

asked Jan 23 at 5:33

mingming

3906

asked Jan 23 at 5:33

mingming

3906

3906

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But $sin pi/2$ is $1,$ not $1/2.$
  $endgroup$
  – coffeemath
  Jan 23 at 5:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a triangle and use the definition of the sine function.
  $endgroup$
  – John Douma
  Jan 23 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
  $endgroup$
  – eyeballfrog
  Jan 23 at 6:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog I think a functional inverse was intended, not a reciprocal.
  $endgroup$
  – J.G.
  Jan 23 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right @J.G.
  $endgroup$
  – ming
  Jan 23 at 20:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  But $sin pi/2$ is $1,$ not $1/2.$
  $endgroup$
  – coffeemath
  Jan 23 at 5:38
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a triangle and use the definition of the sine function.
  $endgroup$
  – John Douma
  Jan 23 at 5:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
  $endgroup$
  – eyeballfrog
  Jan 23 at 6:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @eyeballfrog I think a functional inverse was intended, not a reciprocal.
  $endgroup$
  – J.G.
  Jan 23 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's right @J.G.
  $endgroup$
  – ming
  Jan 23 at 20:22
  

  
  
  
  

2

2

$begingroup$
But $sin pi/2$ is $1,$ not $1/2.$
$endgroup$
– coffeemath
Jan 23 at 5:38

$begingroup$
But $sin pi/2$ is $1,$ not $1/2.$
$endgroup$
– coffeemath
Jan 23 at 5:38

2

2

$begingroup$
Draw a triangle and use the definition of the sine function.
$endgroup$
– John Douma
Jan 23 at 5:55

$begingroup$
Draw a triangle and use the definition of the sine function.
$endgroup$
– John Douma
Jan 23 at 5:55

$begingroup$
When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
$endgroup$
– eyeballfrog
Jan 23 at 6:52

$begingroup$
When you say the inverse of $sin(1/2)$, are you implying that $arcsin(1/2) = 1/sin(1/2)$?
$endgroup$
– eyeballfrog
Jan 23 at 6:52

1

1

$begingroup$
@eyeballfrog I think a functional inverse was intended, not a reciprocal.
$endgroup$
– J.G.
Jan 23 at 7:26

$begingroup$
@eyeballfrog I think a functional inverse was intended, not a reciprocal.
$endgroup$
– J.G.
Jan 23 at 7:26

$begingroup$
That's right @J.G.
$endgroup$
– ming
Jan 23 at 20:22

$begingroup$
That's right @J.G.
$endgroup$
– ming
Jan 23 at 20:22

add a comment | 

2 Answers
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1

$begingroup$

Consider an equilateral triangle if side length $2$. An altitude cuts one of the $pi/3$ angles into two copies of $pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $pi/6=arcsin 1/2$.

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

$endgroup$

add a comment | 

0

$begingroup$

$sin30^{circ}=frac{1}{2}$ and for all $-1leq xleq 1$ we have $-90^{circ}leqarcsin xleq90^{circ}.$

Thus, $arcsinfrac{1}{2}=30^{circ}.$

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

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2 Answers
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2 Answers
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1

$begingroup$

Consider an equilateral triangle if side length $2$. An altitude cuts one of the $pi/3$ angles into two copies of $pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $pi/6=arcsin 1/2$.

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

$endgroup$

add a comment | 

1

$begingroup$

Consider an equilateral triangle if side length $2$. An altitude cuts one of the $pi/3$ angles into two copies of $pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $pi/6=arcsin 1/2$.

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

$endgroup$

add a comment | 

1

1

1

$begingroup$

Consider an equilateral triangle if side length $2$. An altitude cuts one of the $pi/3$ angles into two copies of $pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $pi/6=arcsin 1/2$.

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

$endgroup$

Consider an equilateral triangle if side length $2$. An altitude cuts one of the $pi/3$ angles into two copies of $pi/6$, in hypotenuse-$2$ right-angled triangles with opposite $1$. Thus $pi/6=arcsin 1/2$.

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

answered Jan 23 at 7:29

J.G.J.G.

29.7k22947

29.7k22947

add a comment | 

add a comment | 

0

$begingroup$

$sin30^{circ}=frac{1}{2}$ and for all $-1leq xleq 1$ we have $-90^{circ}leqarcsin xleq90^{circ}.$

Thus, $arcsinfrac{1}{2}=30^{circ}.$

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

0

$begingroup$

$sin30^{circ}=frac{1}{2}$ and for all $-1leq xleq 1$ we have $-90^{circ}leqarcsin xleq90^{circ}.$

Thus, $arcsinfrac{1}{2}=30^{circ}.$

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

0

0

0

$begingroup$

$sin30^{circ}=frac{1}{2}$ and for all $-1leq xleq 1$ we have $-90^{circ}leqarcsin xleq90^{circ}.$

Thus, $arcsinfrac{1}{2}=30^{circ}.$

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

$sin30^{circ}=frac{1}{2}$ and for all $-1leq xleq 1$ we have $-90^{circ}leqarcsin xleq90^{circ}.$

Thus, $arcsinfrac{1}{2}=30^{circ}.$

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

share|cite|improve this answer

share|cite|improve this answer

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

answered Jan 23 at 7:24

Michael RozenbergMichael Rozenberg

108k1895200

108k1895200

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

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