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












0












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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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  • 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
















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











$endgroup$








  • 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














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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 6:47









Ali

1,9832520




1,9832520










asked Jan 23 at 5:33









mingming

3886




3886








  • 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




    $begingroup$
    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










2 Answers
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$begingroup$

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









$endgroup$





















    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









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      2 Answers
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      2 Answers
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      $begingroup$

      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









      $endgroup$


















        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









        $endgroup$
















          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









          $endgroup$



          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












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 7:29









          J.G.J.G.

          28.4k22845




          28.4k22845























              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









              $endgroup$


















                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









                $endgroup$
















                  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









                  $endgroup$



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



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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 7:24









                  Michael RozenbergMichael Rozenberg

                  106k1893198




                  106k1893198






























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