Calculating exact values of “weird” functions like arcsin(sin 100)












0












$begingroup$


This is pretty much the last thing I need to know for now.



Tasks (calculate):




  1. $arccos{(cos{12})}$


  2. $arctan{(tan{sqrt{5}})}$


  3. $arcsin{(sin{100})}$



Answers:




  1. $4pi-12$

  2. $sqrt{5}-pi$

  3. $100-32pi$


All the answers are a little bit weird too "complicated" for me to deduce proper way on how to approach these tasks.



If only I had $pi$ values in there, for example: $arccos{(cos{(12pi)})}$, then I would know how to apply "the shift" rule in there.



Thanks for taking your time in educating total noob. :D






real-analysis trigonometry inverse-function







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$endgroup$












  • $begingroup$
    How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
    $endgroup$
    – Chris Culter
    Jan 23 at 20:16
















0












$begingroup$


This is pretty much the last thing I need to know for now.



Tasks (calculate):




  1. $arccos{(cos{12})}$


  2. $arctan{(tan{sqrt{5}})}$


  3. $arcsin{(sin{100})}$



Answers:




  1. $4pi-12$

  2. $sqrt{5}-pi$

  3. $100-32pi$


All the answers are a little bit weird too "complicated" for me to deduce proper way on how to approach these tasks.



If only I had $pi$ values in there, for example: $arccos{(cos{(12pi)})}$, then I would know how to apply "the shift" rule in there.



Thanks for taking your time in educating total noob. :D






real-analysis trigonometry inverse-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
    $endgroup$
    – Chris Culter
    Jan 23 at 20:16














0












0








0





$begingroup$


This is pretty much the last thing I need to know for now.



Tasks (calculate):




  1. $arccos{(cos{12})}$


  2. $arctan{(tan{sqrt{5}})}$


  3. $arcsin{(sin{100})}$



Answers:




  1. $4pi-12$

  2. $sqrt{5}-pi$

  3. $100-32pi$


All the answers are a little bit weird too "complicated" for me to deduce proper way on how to approach these tasks.



If only I had $pi$ values in there, for example: $arccos{(cos{(12pi)})}$, then I would know how to apply "the shift" rule in there.



Thanks for taking your time in educating total noob. :D






real-analysis trigonometry inverse-function







share|cite|improve this question









$endgroup$




This is pretty much the last thing I need to know for now.



Tasks (calculate):




  1. $arccos{(cos{12})}$


  2. $arctan{(tan{sqrt{5}})}$


  3. $arcsin{(sin{100})}$



Answers:




  1. $4pi-12$

  2. $sqrt{5}-pi$

  3. $100-32pi$


All the answers are a little bit weird too "complicated" for me to deduce proper way on how to approach these tasks.



If only I had $pi$ values in there, for example: $arccos{(cos{(12pi)})}$, then I would know how to apply "the shift" rule in there.



Thanks for taking your time in educating total noob. :D






real-analysis trigonometry inverse-function




real-analysis trigonometry inverse-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 at 20:13









wenoweno

29211




29211












  • $begingroup$
    How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
    $endgroup$
    – Chris Culter
    Jan 23 at 20:16


















  • $begingroup$
    How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
    $endgroup$
    – Chris Culter
    Jan 23 at 20:16
















$begingroup$
How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
$endgroup$
– Chris Culter
Jan 23 at 20:16




$begingroup$
How would you apply "the shift" if the argument were a multiple of $pi$? What prevents you from doing the same thing when it isn't?
$endgroup$
– Chris Culter
Jan 23 at 20:16










2 Answers
2






active

oldest

votes


















1












$begingroup$

The first thing you would want to do is see what signs $cos(12)$, $tanleft(sqrt{5}right)$, and $sin(100)$ have. From there, using the range of inverse functions, you find the answer.



For the first example, $dfrac{7pi}{2} < 12 < 4pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $arccos(x)$ is $y in(0, pi)$, the angle in quadrant $1$ is desired: $4pi-12$.



For the second example, $dfrac{pi}{2} < sqrt{5} < pi$, so the angle lies in quadrant $2$. However, recall that the range of $arctan(x)$ is $y in left(-frac{pi}{2}, frac{pi}{2}right)$, so the quadrant $4$ angle is desired. Also recall that $tan(x)$ is periodic by $pi$ radians, so if $sqrt{5}$ is in quadrant $2$, then $sqrt{5}+pi$ is in quadrant $4$. Note that this angle is also represented by $sqrt{5}-pi$, as they are $2pi$ radians apart.



For the third example, I would start by finding the greatest integer $n$ such that $100-npi > 0$. From here, $n < dfrac{100}{pi} approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30pi$ ($2pi$ for each revolution) from the angle: $100-30pi approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $dfrac{3pi}{2} < 100-30pi < 2pi$. Fortunately, the range of $arcsin(x)$ is $y in left[-frac{pi}{2}, frac{pi}{2}right]$, so $100-30pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2pi$: $100-32pi$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
    $endgroup$
    – weno
    Jan 23 at 21:15










  • $begingroup$
    That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
    $endgroup$
    – KM101
    Jan 23 at 21:18



















1












$begingroup$

hint



If $Xin [0,pi]$, then



$$arccos(cos(X))=X=$$
$$arccos(cos(X+2kpi))=$$
$$arccos(cos(-X+2kpi))$$



observe that
$$4pi-12in[0,pi]$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    The first thing you would want to do is see what signs $cos(12)$, $tanleft(sqrt{5}right)$, and $sin(100)$ have. From there, using the range of inverse functions, you find the answer.



    For the first example, $dfrac{7pi}{2} < 12 < 4pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $arccos(x)$ is $y in(0, pi)$, the angle in quadrant $1$ is desired: $4pi-12$.



    For the second example, $dfrac{pi}{2} < sqrt{5} < pi$, so the angle lies in quadrant $2$. However, recall that the range of $arctan(x)$ is $y in left(-frac{pi}{2}, frac{pi}{2}right)$, so the quadrant $4$ angle is desired. Also recall that $tan(x)$ is periodic by $pi$ radians, so if $sqrt{5}$ is in quadrant $2$, then $sqrt{5}+pi$ is in quadrant $4$. Note that this angle is also represented by $sqrt{5}-pi$, as they are $2pi$ radians apart.



    For the third example, I would start by finding the greatest integer $n$ such that $100-npi > 0$. From here, $n < dfrac{100}{pi} approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30pi$ ($2pi$ for each revolution) from the angle: $100-30pi approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $dfrac{3pi}{2} < 100-30pi < 2pi$. Fortunately, the range of $arcsin(x)$ is $y in left[-frac{pi}{2}, frac{pi}{2}right]$, so $100-30pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2pi$: $100-32pi$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
      $endgroup$
      – weno
      Jan 23 at 21:15










    • $begingroup$
      That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
      $endgroup$
      – KM101
      Jan 23 at 21:18
















    1












    $begingroup$

    The first thing you would want to do is see what signs $cos(12)$, $tanleft(sqrt{5}right)$, and $sin(100)$ have. From there, using the range of inverse functions, you find the answer.



    For the first example, $dfrac{7pi}{2} < 12 < 4pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $arccos(x)$ is $y in(0, pi)$, the angle in quadrant $1$ is desired: $4pi-12$.



    For the second example, $dfrac{pi}{2} < sqrt{5} < pi$, so the angle lies in quadrant $2$. However, recall that the range of $arctan(x)$ is $y in left(-frac{pi}{2}, frac{pi}{2}right)$, so the quadrant $4$ angle is desired. Also recall that $tan(x)$ is periodic by $pi$ radians, so if $sqrt{5}$ is in quadrant $2$, then $sqrt{5}+pi$ is in quadrant $4$. Note that this angle is also represented by $sqrt{5}-pi$, as they are $2pi$ radians apart.



    For the third example, I would start by finding the greatest integer $n$ such that $100-npi > 0$. From here, $n < dfrac{100}{pi} approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30pi$ ($2pi$ for each revolution) from the angle: $100-30pi approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $dfrac{3pi}{2} < 100-30pi < 2pi$. Fortunately, the range of $arcsin(x)$ is $y in left[-frac{pi}{2}, frac{pi}{2}right]$, so $100-30pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2pi$: $100-32pi$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
      $endgroup$
      – weno
      Jan 23 at 21:15










    • $begingroup$
      That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
      $endgroup$
      – KM101
      Jan 23 at 21:18














    1












    1








    1





    $begingroup$

    The first thing you would want to do is see what signs $cos(12)$, $tanleft(sqrt{5}right)$, and $sin(100)$ have. From there, using the range of inverse functions, you find the answer.



    For the first example, $dfrac{7pi}{2} < 12 < 4pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $arccos(x)$ is $y in(0, pi)$, the angle in quadrant $1$ is desired: $4pi-12$.



    For the second example, $dfrac{pi}{2} < sqrt{5} < pi$, so the angle lies in quadrant $2$. However, recall that the range of $arctan(x)$ is $y in left(-frac{pi}{2}, frac{pi}{2}right)$, so the quadrant $4$ angle is desired. Also recall that $tan(x)$ is periodic by $pi$ radians, so if $sqrt{5}$ is in quadrant $2$, then $sqrt{5}+pi$ is in quadrant $4$. Note that this angle is also represented by $sqrt{5}-pi$, as they are $2pi$ radians apart.



    For the third example, I would start by finding the greatest integer $n$ such that $100-npi > 0$. From here, $n < dfrac{100}{pi} approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30pi$ ($2pi$ for each revolution) from the angle: $100-30pi approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $dfrac{3pi}{2} < 100-30pi < 2pi$. Fortunately, the range of $arcsin(x)$ is $y in left[-frac{pi}{2}, frac{pi}{2}right]$, so $100-30pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2pi$: $100-32pi$.






    share|cite|improve this answer









    $endgroup$



    The first thing you would want to do is see what signs $cos(12)$, $tanleft(sqrt{5}right)$, and $sin(100)$ have. From there, using the range of inverse functions, you find the answer.



    For the first example, $dfrac{7pi}{2} < 12 < 4pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $arccos(x)$ is $y in(0, pi)$, the angle in quadrant $1$ is desired: $4pi-12$.



    For the second example, $dfrac{pi}{2} < sqrt{5} < pi$, so the angle lies in quadrant $2$. However, recall that the range of $arctan(x)$ is $y in left(-frac{pi}{2}, frac{pi}{2}right)$, so the quadrant $4$ angle is desired. Also recall that $tan(x)$ is periodic by $pi$ radians, so if $sqrt{5}$ is in quadrant $2$, then $sqrt{5}+pi$ is in quadrant $4$. Note that this angle is also represented by $sqrt{5}-pi$, as they are $2pi$ radians apart.



    For the third example, I would start by finding the greatest integer $n$ such that $100-npi > 0$. From here, $n < dfrac{100}{pi} approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30pi$ ($2pi$ for each revolution) from the angle: $100-30pi approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $dfrac{3pi}{2} < 100-30pi < 2pi$. Fortunately, the range of $arcsin(x)$ is $y in left[-frac{pi}{2}, frac{pi}{2}right]$, so $100-30pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2pi$: $100-32pi$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 23 at 20:47









    KM101KM101

    6,0701525




    6,0701525












    • $begingroup$
      About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
      $endgroup$
      – weno
      Jan 23 at 21:15










    • $begingroup$
      That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
      $endgroup$
      – KM101
      Jan 23 at 21:18


















    • $begingroup$
      About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
      $endgroup$
      – weno
      Jan 23 at 21:15










    • $begingroup$
      That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
      $endgroup$
      – KM101
      Jan 23 at 21:18
















    $begingroup$
    About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
    $endgroup$
    – weno
    Jan 23 at 21:15




    $begingroup$
    About the first example, can I denote it as $12 - 4pi$ instead of $4pi - 12$?
    $endgroup$
    – weno
    Jan 23 at 21:15












    $begingroup$
    That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
    $endgroup$
    – KM101
    Jan 23 at 21:18




    $begingroup$
    That would be the quadrant $4$ angle, which isn’t included in the range of $arccos(x)$.
    $endgroup$
    – KM101
    Jan 23 at 21:18











    1












    $begingroup$

    hint



    If $Xin [0,pi]$, then



    $$arccos(cos(X))=X=$$
    $$arccos(cos(X+2kpi))=$$
    $$arccos(cos(-X+2kpi))$$



    observe that
    $$4pi-12in[0,pi]$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      hint



      If $Xin [0,pi]$, then



      $$arccos(cos(X))=X=$$
      $$arccos(cos(X+2kpi))=$$
      $$arccos(cos(-X+2kpi))$$



      observe that
      $$4pi-12in[0,pi]$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        hint



        If $Xin [0,pi]$, then



        $$arccos(cos(X))=X=$$
        $$arccos(cos(X+2kpi))=$$
        $$arccos(cos(-X+2kpi))$$



        observe that
        $$4pi-12in[0,pi]$$






        share|cite|improve this answer









        $endgroup$



        hint



        If $Xin [0,pi]$, then



        $$arccos(cos(X))=X=$$
        $$arccos(cos(X+2kpi))=$$
        $$arccos(cos(-X+2kpi))$$



        observe that
        $$4pi-12in[0,pi]$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 23 at 20:26









        hamam_Abdallahhamam_Abdallah

        38.1k21634




        38.1k21634






























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