Trigonometric equation over unit circle [closed]












0












$begingroup$


Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?






calculus trigonometry







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$endgroup$



closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
    $endgroup$
    – Jam
    Dec 10 '18 at 14:11










  • $begingroup$
    Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
    $endgroup$
    – Ben W
    Dec 10 '18 at 14:14










  • $begingroup$
    @BenW yeah the second equation holds for all real s
    $endgroup$
    – pablo_mathscobar
    Dec 10 '18 at 14:15










  • $begingroup$
    The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
    $endgroup$
    – Alex Smart
    Dec 10 '18 at 14:34


















0












$begingroup$


Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?






calculus trigonometry







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
    $endgroup$
    – Jam
    Dec 10 '18 at 14:11










  • $begingroup$
    Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
    $endgroup$
    – Ben W
    Dec 10 '18 at 14:14










  • $begingroup$
    @BenW yeah the second equation holds for all real s
    $endgroup$
    – pablo_mathscobar
    Dec 10 '18 at 14:15










  • $begingroup$
    The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
    $endgroup$
    – Alex Smart
    Dec 10 '18 at 14:34
















0












0








0





$begingroup$


Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?






calculus trigonometry







share|cite|improve this question











$endgroup$




Let us consider the equations
$$
{p_0}^2 + {q_0}^2 =1 qquadtext{and}qquad {-a} p_0sin s + b q_0cos s = 0
$$
How do i solve for $p_0, q_0$?






calculus trigonometry




calculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 14:16









Harry49

6,17331132




6,17331132










asked Dec 10 '18 at 14:06









pablo_mathscobarpablo_mathscobar

996




996




closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Davide Giraudo, Jyrki Lahtonen, metamorphy, Andrei Dec 18 '18 at 16:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Davide Giraudo, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
    $endgroup$
    – Jam
    Dec 10 '18 at 14:11










  • $begingroup$
    Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
    $endgroup$
    – Ben W
    Dec 10 '18 at 14:14










  • $begingroup$
    @BenW yeah the second equation holds for all real s
    $endgroup$
    – pablo_mathscobar
    Dec 10 '18 at 14:15










  • $begingroup$
    The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
    $endgroup$
    – Alex Smart
    Dec 10 '18 at 14:34




















  • $begingroup$
    I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
    $endgroup$
    – Jam
    Dec 10 '18 at 14:11










  • $begingroup$
    Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
    $endgroup$
    – Ben W
    Dec 10 '18 at 14:14










  • $begingroup$
    @BenW yeah the second equation holds for all real s
    $endgroup$
    – pablo_mathscobar
    Dec 10 '18 at 14:15










  • $begingroup$
    The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
    $endgroup$
    – Alex Smart
    Dec 10 '18 at 14:34


















$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11




$begingroup$
I've edited your equation assuming you meant $asin(cdot)$ and not $operatorname{asin}(cdot)$. Please revert it if that is incorrect,
$endgroup$
– Jam
Dec 10 '18 at 14:11












$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14




$begingroup$
Presumably $a$ and $b$ are fixed constants. But does the second equation hold for all real $s$?
$endgroup$
– Ben W
Dec 10 '18 at 14:14












$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15




$begingroup$
@BenW yeah the second equation holds for all real s
$endgroup$
– pablo_mathscobar
Dec 10 '18 at 14:15












$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34






$begingroup$
The only way the second equation holds for all real s is: $a p_0 = 0$ and $b q_0 = 0$. If $a ne 0$ and $ b=0$, then $p_0 = 0$ and $q_0 = pm 1$, and if $b ne 0$ and $a=0$, then $q_0 = 0$ and $p_0 = pm 1$. If neither a nor b is zero, then both $p_0$ and $q_0$ must be $0$ which is a contradiction. If both a and b are zero, then the second equation is an identity and all points on the unit circle satisfy both equations.
$endgroup$
– Alex Smart
Dec 10 '18 at 14:34












3 Answers
3






active

oldest

votes


















3












$begingroup$

Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.



    You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.



    Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Let us rewrite the system in the simple form



      $$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$



      Then



      $$v^2p^2+v^2q^2=v^2$$ is



      $$(u^2+v^2)p^2=v^2.$$



      The rest is yours.






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.






            share|cite|improve this answer









            $endgroup$



            Let $s=0$ so that we have $bq_0=0$, and by letting $s=pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0in{-1,1}$ by the first equation. Similarly, if $p_0=0$ then $q_0in{-1,1}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 14:21









            Ben WBen W

            2,234615




            2,234615























                3












                $begingroup$

                The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.



                You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.



                Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.



                  You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.



                  Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.



                    You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.



                    Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.






                    share|cite|improve this answer









                    $endgroup$



                    The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.



                    You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (cos theta, sin theta) $ for $theta in [0, 2 pi)$.



                    Substituting in the second equation, you have $ tan theta = frac{a}{b} tan s$, or $theta = arctan(frac{a}{b} tan s)$ or $pi +arctan(frac{a}{b} tan s)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 '18 at 14:24









                    Alex SmartAlex Smart

                    614




                    614























                        1












                        $begingroup$

                        Let us rewrite the system in the simple form



                        $$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$



                        Then



                        $$v^2p^2+v^2q^2=v^2$$ is



                        $$(u^2+v^2)p^2=v^2.$$



                        The rest is yours.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Let us rewrite the system in the simple form



                          $$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$



                          Then



                          $$v^2p^2+v^2q^2=v^2$$ is



                          $$(u^2+v^2)p^2=v^2.$$



                          The rest is yours.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Let us rewrite the system in the simple form



                            $$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$



                            Then



                            $$v^2p^2+v^2q^2=v^2$$ is



                            $$(u^2+v^2)p^2=v^2.$$



                            The rest is yours.






                            share|cite|improve this answer









                            $endgroup$



                            Let us rewrite the system in the simple form



                            $$begin{cases}p^2+q^2=1,\up=vq.end{cases}$$



                            Then



                            $$v^2p^2+v^2q^2=v^2$$ is



                            $$(u^2+v^2)p^2=v^2.$$



                            The rest is yours.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 10 '18 at 15:08









                            Yves DaoustYves Daoust

                            125k671223




                            125k671223















                                -

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