If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and...












0












$begingroup$


If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?



I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?






calculus algebra-precalculus trigonometry







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$endgroup$












  • $begingroup$
    I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
    $endgroup$
    – Matti P.
    Dec 14 '18 at 12:17










  • $begingroup$
    The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
    $endgroup$
    – David K
    Dec 14 '18 at 12:19


















0












$begingroup$


If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?



I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?






calculus algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
    $endgroup$
    – Matti P.
    Dec 14 '18 at 12:17










  • $begingroup$
    The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
    $endgroup$
    – David K
    Dec 14 '18 at 12:19
















0












0








0





$begingroup$


If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?



I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?






calculus algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?



I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?






calculus algebra-precalculus trigonometry




calculus algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 12:19









amWhy

192k28225439




192k28225439










asked Dec 14 '18 at 12:14









hopefullyhopefully

166112




166112












  • $begingroup$
    I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
    $endgroup$
    – Matti P.
    Dec 14 '18 at 12:17










  • $begingroup$
    The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
    $endgroup$
    – David K
    Dec 14 '18 at 12:19




















  • $begingroup$
    I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
    $endgroup$
    – Matti P.
    Dec 14 '18 at 12:17










  • $begingroup$
    The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
    $endgroup$
    – David K
    Dec 14 '18 at 12:19


















$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17




$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17












$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19






$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19












5 Answers
5






active

oldest

votes


















4












$begingroup$

Use Prosthaphaeresis Formulas,



$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



What if $cosdfrac{alpha-beta}2$$ne0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
    $endgroup$
    – hopefully
    Dec 14 '18 at 12:28






  • 1




    $begingroup$
    cosine and sine functions never equal 0 at the same time
    $endgroup$
    – hopefully
    Dec 14 '18 at 12:29






  • 1




    $begingroup$
    @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 12:29












  • $begingroup$
    @hopefully, We must have $cosdfrac{alpha-beta}2=0$
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 12:30






  • 1




    $begingroup$
    equals one, so we have a contradiction ...... correct?
    $endgroup$
    – hopefully
    Dec 14 '18 at 12:31





















4












$begingroup$

$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.

The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.

This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
    $$
    sinalphacosbeta-cosalphasinbeta=0
    $$
    that is, $sin(alpha-beta)=0$. This entails
    $$
    alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
    $$
    ($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.



    For the second set of solutions, we have
    begin{align}
    cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
    sinalpha&=sin(beta+pi+2kpi)=-sinbeta
    end{align}
    for every $beta$. So the conclusion is that your conditions are equivalent to
    $$
    alpha=beta+pi+2kpi qquadtext{($k$ integer)}
    $$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      By the definition we have



      $$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$



      $$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$



      then we have





      • $beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible


      • $beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$


      and therefore finally we have




      • $alpha =hpi$

      • $beta=pi-hpi+2kpi=(1-h)pi+2kpi$


      or




      • $alpha+beta=pi + 2kpi$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        $$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$



        $impliesalpha=npi+beta$ where $n$ is any integer



        Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$






        share|cite|improve this answer









        $endgroup$













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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Use Prosthaphaeresis Formulas,



          $$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          $$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          What if $cosdfrac{alpha-beta}2$$ne0$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:28






          • 1




            $begingroup$
            cosine and sine functions never equal 0 at the same time
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:29






          • 1




            $begingroup$
            @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:29












          • $begingroup$
            @hopefully, We must have $cosdfrac{alpha-beta}2=0$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:30






          • 1




            $begingroup$
            equals one, so we have a contradiction ...... correct?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:31


















          4












          $begingroup$

          Use Prosthaphaeresis Formulas,



          $$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          $$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          What if $cosdfrac{alpha-beta}2$$ne0$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:28






          • 1




            $begingroup$
            cosine and sine functions never equal 0 at the same time
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:29






          • 1




            $begingroup$
            @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:29












          • $begingroup$
            @hopefully, We must have $cosdfrac{alpha-beta}2=0$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:30






          • 1




            $begingroup$
            equals one, so we have a contradiction ...... correct?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:31
















          4












          4








          4





          $begingroup$

          Use Prosthaphaeresis Formulas,



          $$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          $$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          What if $cosdfrac{alpha-beta}2$$ne0$






          share|cite|improve this answer









          $endgroup$



          Use Prosthaphaeresis Formulas,



          $$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          $$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$



          What if $cosdfrac{alpha-beta}2$$ne0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 12:19









          lab bhattacharjeelab bhattacharjee

          224k15156274




          224k15156274












          • $begingroup$
            Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:28






          • 1




            $begingroup$
            cosine and sine functions never equal 0 at the same time
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:29






          • 1




            $begingroup$
            @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:29












          • $begingroup$
            @hopefully, We must have $cosdfrac{alpha-beta}2=0$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:30






          • 1




            $begingroup$
            equals one, so we have a contradiction ...... correct?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:31




















          • $begingroup$
            Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:28






          • 1




            $begingroup$
            cosine and sine functions never equal 0 at the same time
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:29






          • 1




            $begingroup$
            @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:29












          • $begingroup$
            @hopefully, We must have $cosdfrac{alpha-beta}2=0$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 12:30






          • 1




            $begingroup$
            equals one, so we have a contradiction ...... correct?
            $endgroup$
            – hopefully
            Dec 14 '18 at 12:31


















          $begingroup$
          Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:28




          $begingroup$
          Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:28




          1




          1




          $begingroup$
          cosine and sine functions never equal 0 at the same time
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:29




          $begingroup$
          cosine and sine functions never equal 0 at the same time
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:29




          1




          1




          $begingroup$
          @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
          $endgroup$
          – lab bhattacharjee
          Dec 14 '18 at 12:29






          $begingroup$
          @hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
          $endgroup$
          – lab bhattacharjee
          Dec 14 '18 at 12:29














          $begingroup$
          @hopefully, We must have $cosdfrac{alpha-beta}2=0$
          $endgroup$
          – lab bhattacharjee
          Dec 14 '18 at 12:30




          $begingroup$
          @hopefully, We must have $cosdfrac{alpha-beta}2=0$
          $endgroup$
          – lab bhattacharjee
          Dec 14 '18 at 12:30




          1




          1




          $begingroup$
          equals one, so we have a contradiction ...... correct?
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:31






          $begingroup$
          equals one, so we have a contradiction ...... correct?
          $endgroup$
          – hopefully
          Dec 14 '18 at 12:31













          4












          $begingroup$

          $(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.

          The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.

          This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$






          share|cite|improve this answer











          $endgroup$


















            4












            $begingroup$

            $(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.

            The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.

            This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$






            share|cite|improve this answer











            $endgroup$
















              4












              4








              4





              $begingroup$

              $(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.

              The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.

              This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$






              share|cite|improve this answer











              $endgroup$



              $(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.

              The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.

              This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 14 '18 at 13:27

























              answered Dec 14 '18 at 12:20









              user376343user376343

              3,2982825




              3,2982825























                  1












                  $begingroup$

                  Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
                  $$
                  sinalphacosbeta-cosalphasinbeta=0
                  $$
                  that is, $sin(alpha-beta)=0$. This entails
                  $$
                  alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
                  $$
                  ($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.



                  For the second set of solutions, we have
                  begin{align}
                  cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
                  sinalpha&=sin(beta+pi+2kpi)=-sinbeta
                  end{align}
                  for every $beta$. So the conclusion is that your conditions are equivalent to
                  $$
                  alpha=beta+pi+2kpi qquadtext{($k$ integer)}
                  $$






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
                    $$
                    sinalphacosbeta-cosalphasinbeta=0
                    $$
                    that is, $sin(alpha-beta)=0$. This entails
                    $$
                    alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
                    $$
                    ($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.



                    For the second set of solutions, we have
                    begin{align}
                    cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
                    sinalpha&=sin(beta+pi+2kpi)=-sinbeta
                    end{align}
                    for every $beta$. So the conclusion is that your conditions are equivalent to
                    $$
                    alpha=beta+pi+2kpi qquadtext{($k$ integer)}
                    $$






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
                      $$
                      sinalphacosbeta-cosalphasinbeta=0
                      $$
                      that is, $sin(alpha-beta)=0$. This entails
                      $$
                      alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
                      $$
                      ($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.



                      For the second set of solutions, we have
                      begin{align}
                      cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
                      sinalpha&=sin(beta+pi+2kpi)=-sinbeta
                      end{align}
                      for every $beta$. So the conclusion is that your conditions are equivalent to
                      $$
                      alpha=beta+pi+2kpi qquadtext{($k$ integer)}
                      $$






                      share|cite|improve this answer









                      $endgroup$



                      Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
                      $$
                      sinalphacosbeta-cosalphasinbeta=0
                      $$
                      that is, $sin(alpha-beta)=0$. This entails
                      $$
                      alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
                      $$
                      ($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.



                      For the second set of solutions, we have
                      begin{align}
                      cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
                      sinalpha&=sin(beta+pi+2kpi)=-sinbeta
                      end{align}
                      for every $beta$. So the conclusion is that your conditions are equivalent to
                      $$
                      alpha=beta+pi+2kpi qquadtext{($k$ integer)}
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 14 '18 at 13:30









                      egregegreg

                      180k1485202




                      180k1485202























                          1












                          $begingroup$

                          By the definition we have



                          $$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$



                          $$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$



                          then we have





                          • $beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible


                          • $beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$


                          and therefore finally we have




                          • $alpha =hpi$

                          • $beta=pi-hpi+2kpi=(1-h)pi+2kpi$


                          or




                          • $alpha+beta=pi + 2kpi$






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            By the definition we have



                            $$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$



                            $$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$



                            then we have





                            • $beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible


                            • $beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$


                            and therefore finally we have




                            • $alpha =hpi$

                            • $beta=pi-hpi+2kpi=(1-h)pi+2kpi$


                            or




                            • $alpha+beta=pi + 2kpi$






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              By the definition we have



                              $$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$



                              $$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$



                              then we have





                              • $beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible


                              • $beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$


                              and therefore finally we have




                              • $alpha =hpi$

                              • $beta=pi-hpi+2kpi=(1-h)pi+2kpi$


                              or




                              • $alpha+beta=pi + 2kpi$






                              share|cite|improve this answer











                              $endgroup$



                              By the definition we have



                              $$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$



                              $$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$



                              then we have





                              • $beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible


                              • $beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$


                              and therefore finally we have




                              • $alpha =hpi$

                              • $beta=pi-hpi+2kpi=(1-h)pi+2kpi$


                              or




                              • $alpha+beta=pi + 2kpi$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 14 '18 at 13:32

























                              answered Dec 14 '18 at 13:18









                              gimusigimusi

                              1




                              1























                                  1












                                  $begingroup$

                                  $$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$



                                  $impliesalpha=npi+beta$ where $n$ is any integer



                                  Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    $$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$



                                    $impliesalpha=npi+beta$ where $n$ is any integer



                                    Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      $$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$



                                      $impliesalpha=npi+beta$ where $n$ is any integer



                                      Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$



                                      $impliesalpha=npi+beta$ where $n$ is any integer



                                      Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 14 '18 at 17:46









                                      lab bhattacharjeelab bhattacharjee

                                      224k15156274




                                      224k15156274






























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