For $0 < x < 1$, express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$












2












$begingroup$


I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.



I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.



For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$



That's where I'm stuck.






algebra-precalculus trigonometry







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  • 2




    $begingroup$
    socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:12






  • 3




    $begingroup$
    Hint: $arcsin x+ arccos x=frac{pi}{2}$.
    $endgroup$
    – Anurag A
    Nov 23 '18 at 19:12


















2












$begingroup$


I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.



I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.



For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$



That's where I'm stuck.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:12






  • 3




    $begingroup$
    Hint: $arcsin x+ arccos x=frac{pi}{2}$.
    $endgroup$
    – Anurag A
    Nov 23 '18 at 19:12
















2












2








2





$begingroup$


I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.



I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.



For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$



That's where I'm stuck.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.



I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.



For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$



That's where I'm stuck.






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 19:35









Robert Howard

1,9161822




1,9161822










asked Nov 23 '18 at 19:09









LuminousNutriaLuminousNutria

1709




1709








  • 2




    $begingroup$
    socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:12






  • 3




    $begingroup$
    Hint: $arcsin x+ arccos x=frac{pi}{2}$.
    $endgroup$
    – Anurag A
    Nov 23 '18 at 19:12
















  • 2




    $begingroup$
    socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
    $endgroup$
    – lab bhattacharjee
    Nov 23 '18 at 19:12






  • 3




    $begingroup$
    Hint: $arcsin x+ arccos x=frac{pi}{2}$.
    $endgroup$
    – Anurag A
    Nov 23 '18 at 19:12










2




2




$begingroup$
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:12




$begingroup$
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:12




3




3




$begingroup$
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
$endgroup$
– Anurag A
Nov 23 '18 at 19:12






$begingroup$
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
$endgroup$
– Anurag A
Nov 23 '18 at 19:12












3 Answers
3






active

oldest

votes


















2












$begingroup$

Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm sorry, I don't understand how that helps.
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:28










  • $begingroup$
    Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:29










  • $begingroup$
    So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:32












  • $begingroup$
    I clarified my answer a bit more. Hope it help now!
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:36










  • $begingroup$
    So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:41





















1












$begingroup$

Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.



Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint: Use



    $$arcsin x = y_1 implies sin y_1 = x$$



    $$arccos x = y_2 implies cos y_2 = x$$



    and



    $$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$



    to get



    $$implies y_2 = frac{pi}{2}-y_1$$



    So what does $y_1+y_2$ become?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
      $endgroup$
      – Anurag A
      Nov 23 '18 at 19:24










    • $begingroup$
      But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
      $endgroup$
      – KM101
      Nov 23 '18 at 19:27












    • $begingroup$
      what do you mean by $2pi=0$? !!!
      $endgroup$
      – Anurag A
      Nov 23 '18 at 19:30










    • $begingroup$
      (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
      $endgroup$
      – KM101
      Nov 23 '18 at 19:32










    • $begingroup$
      No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
      $endgroup$
      – Anurag A
      Nov 23 '18 at 19:34











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I'm sorry, I don't understand how that helps.
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:28










    • $begingroup$
      Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:29










    • $begingroup$
      So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:32












    • $begingroup$
      I clarified my answer a bit more. Hope it help now!
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:36










    • $begingroup$
      So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:41


















    2












    $begingroup$

    Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I'm sorry, I don't understand how that helps.
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:28










    • $begingroup$
      Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:29










    • $begingroup$
      So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:32












    • $begingroup$
      I clarified my answer a bit more. Hope it help now!
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:36










    • $begingroup$
      So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:41
















    2












    2








    2





    $begingroup$

    Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$






    share|cite|improve this answer











    $endgroup$



    Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 23 '18 at 19:35

























    answered Nov 23 '18 at 19:13









    Mostafa AyazMostafa Ayaz

    14.7k3937




    14.7k3937












    • $begingroup$
      I'm sorry, I don't understand how that helps.
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:28










    • $begingroup$
      Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:29










    • $begingroup$
      So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:32












    • $begingroup$
      I clarified my answer a bit more. Hope it help now!
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:36










    • $begingroup$
      So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:41




















    • $begingroup$
      I'm sorry, I don't understand how that helps.
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:28










    • $begingroup$
      Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:29










    • $begingroup$
      So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:32












    • $begingroup$
      I clarified my answer a bit more. Hope it help now!
      $endgroup$
      – Mostafa Ayaz
      Nov 23 '18 at 19:36










    • $begingroup$
      So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
      $endgroup$
      – LuminousNutria
      Nov 23 '18 at 19:41


















    $begingroup$
    I'm sorry, I don't understand how that helps.
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:28




    $begingroup$
    I'm sorry, I don't understand how that helps.
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:28












    $begingroup$
    Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:29




    $begingroup$
    Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:29












    $begingroup$
    So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:32






    $begingroup$
    So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:32














    $begingroup$
    I clarified my answer a bit more. Hope it help now!
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:36




    $begingroup$
    I clarified my answer a bit more. Hope it help now!
    $endgroup$
    – Mostafa Ayaz
    Nov 23 '18 at 19:36












    $begingroup$
    So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:41






    $begingroup$
    So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
    $endgroup$
    – LuminousNutria
    Nov 23 '18 at 19:41













    1












    $begingroup$

    Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.



    Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.



      Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.



        Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.






        share|cite|improve this answer











        $endgroup$



        Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.



        Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 23 '18 at 19:57









        Timothy Cho

        789519




        789519










        answered Nov 23 '18 at 19:19









        AlexAlex

        111




        111























            0












            $begingroup$

            Hint: Use



            $$arcsin x = y_1 implies sin y_1 = x$$



            $$arccos x = y_2 implies cos y_2 = x$$



            and



            $$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$



            to get



            $$implies y_2 = frac{pi}{2}-y_1$$



            So what does $y_1+y_2$ become?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:24










            • $begingroup$
              But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:27












            • $begingroup$
              what do you mean by $2pi=0$? !!!
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:30










            • $begingroup$
              (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:32










            • $begingroup$
              No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:34
















            0












            $begingroup$

            Hint: Use



            $$arcsin x = y_1 implies sin y_1 = x$$



            $$arccos x = y_2 implies cos y_2 = x$$



            and



            $$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$



            to get



            $$implies y_2 = frac{pi}{2}-y_1$$



            So what does $y_1+y_2$ become?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:24










            • $begingroup$
              But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:27












            • $begingroup$
              what do you mean by $2pi=0$? !!!
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:30










            • $begingroup$
              (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:32










            • $begingroup$
              No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:34














            0












            0








            0





            $begingroup$

            Hint: Use



            $$arcsin x = y_1 implies sin y_1 = x$$



            $$arccos x = y_2 implies cos y_2 = x$$



            and



            $$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$



            to get



            $$implies y_2 = frac{pi}{2}-y_1$$



            So what does $y_1+y_2$ become?






            share|cite|improve this answer











            $endgroup$



            Hint: Use



            $$arcsin x = y_1 implies sin y_1 = x$$



            $$arccos x = y_2 implies cos y_2 = x$$



            and



            $$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$



            to get



            $$implies y_2 = frac{pi}{2}-y_1$$



            So what does $y_1+y_2$ become?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 '18 at 19:24

























            answered Nov 23 '18 at 19:20









            KM101KM101

            5,8711423




            5,8711423












            • $begingroup$
              You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:24










            • $begingroup$
              But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:27












            • $begingroup$
              what do you mean by $2pi=0$? !!!
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:30










            • $begingroup$
              (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:32










            • $begingroup$
              No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:34


















            • $begingroup$
              You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:24










            • $begingroup$
              But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:27












            • $begingroup$
              what do you mean by $2pi=0$? !!!
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:30










            • $begingroup$
              (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
              $endgroup$
              – KM101
              Nov 23 '18 at 19:32










            • $begingroup$
              No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
              $endgroup$
              – Anurag A
              Nov 23 '18 at 19:34
















            $begingroup$
            You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:24




            $begingroup$
            You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:24












            $begingroup$
            But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
            $endgroup$
            – KM101
            Nov 23 '18 at 19:27






            $begingroup$
            But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
            $endgroup$
            – KM101
            Nov 23 '18 at 19:27














            $begingroup$
            what do you mean by $2pi=0$? !!!
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:30




            $begingroup$
            what do you mean by $2pi=0$? !!!
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:30












            $begingroup$
            (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
            $endgroup$
            – KM101
            Nov 23 '18 at 19:32




            $begingroup$
            (Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
            $endgroup$
            – KM101
            Nov 23 '18 at 19:32












            $begingroup$
            No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:34




            $begingroup$
            No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
            $endgroup$
            – Anurag A
            Nov 23 '18 at 19:34


















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