Some weird results in complex number computing












3












$begingroup$


The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$



In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$



So $Re(frac{z-1}{z+1})=0$



If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$



So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.



Which step I made mistake or they are equivalent?






complex-numbers







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$endgroup$












  • $begingroup$
    There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 9:44












  • $begingroup$
    Demoivre's theory, the arguments of two complex number are added.
    $endgroup$
    – yuanming luo
    Jan 8 at 9:47










  • $begingroup$
    How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
    $endgroup$
    – 5xum
    Jan 8 at 9:55










  • $begingroup$
    @5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
    $endgroup$
    – Arthur
    Jan 8 at 9:56












  • $begingroup$
    @Arthur " That's what I asked earlier"... I don't see your question anywhere...
    $endgroup$
    – 5xum
    Jan 8 at 9:58
















3












$begingroup$


The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$



In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$



So $Re(frac{z-1}{z+1})=0$



If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$



So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.



Which step I made mistake or they are equivalent?






complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 9:44












  • $begingroup$
    Demoivre's theory, the arguments of two complex number are added.
    $endgroup$
    – yuanming luo
    Jan 8 at 9:47










  • $begingroup$
    How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
    $endgroup$
    – 5xum
    Jan 8 at 9:55










  • $begingroup$
    @5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
    $endgroup$
    – Arthur
    Jan 8 at 9:56












  • $begingroup$
    @Arthur " That's what I asked earlier"... I don't see your question anywhere...
    $endgroup$
    – 5xum
    Jan 8 at 9:58














3












3








3





$begingroup$


The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$



In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$



So $Re(frac{z-1}{z+1})=0$



If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$



So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.



Which step I made mistake or they are equivalent?






complex-numbers







share|cite|improve this question











$endgroup$




The question I met is to show that if $z=cos (theta)+isin(theta)$ with $i=sqrt{-1}$,then $ Re(frac{z-1}{z+1})=0$



In the normal way, we found that: $$frac{z-1}{z+1}=frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}\
=frac{bigl(cos(theta)-1+isin(theta)bigl)bigl(cos(theta)+1-isin(theta)bigl)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{cos^2(theta)+cos(theta)-cos(theta)-1+sin^2(theta)+2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}\
=frac{2isin(theta)}{bigl(cos(theta)+1bigl)^2+sin^2(theta)}$$



So $Re(frac{z-1}{z+1})=0$



If we do it in another way:
$$
frac{cos(theta)-1+isin(theta)}{cos(theta)+1+isin(theta)}=frac{-2sin^2(frac{theta}{2})+2icos(frac{theta}{2})sin(frac{theta}{2})}{2cos^2(frac{theta}{2})+2isin(frac{theta}{2})cos(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{sin(frac{theta}{2})-icos(frac{theta}{2})}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}\
=-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)
$$



So the real part of it will be $-tan(frac{theta}{2})cos(-theta-frac{pi}{2})$ which is not $0$.



Which step I made mistake or they are equivalent?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 9:47







yuanming luo

















asked Jan 8 at 9:38









yuanming luoyuanming luo

10211




10211












  • $begingroup$
    There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 9:44












  • $begingroup$
    Demoivre's theory, the arguments of two complex number are added.
    $endgroup$
    – yuanming luo
    Jan 8 at 9:47










  • $begingroup$
    How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
    $endgroup$
    – 5xum
    Jan 8 at 9:55










  • $begingroup$
    @5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
    $endgroup$
    – Arthur
    Jan 8 at 9:56












  • $begingroup$
    @Arthur " That's what I asked earlier"... I don't see your question anywhere...
    $endgroup$
    – 5xum
    Jan 8 at 9:58


















  • $begingroup$
    There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
    $endgroup$
    – Yves Daoust
    Jan 8 at 9:44












  • $begingroup$
    Demoivre's theory, the arguments of two complex number are added.
    $endgroup$
    – yuanming luo
    Jan 8 at 9:47










  • $begingroup$
    How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
    $endgroup$
    – 5xum
    Jan 8 at 9:55










  • $begingroup$
    @5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
    $endgroup$
    – Arthur
    Jan 8 at 9:56












  • $begingroup$
    @Arthur " That's what I asked earlier"... I don't see your question anywhere...
    $endgroup$
    – 5xum
    Jan 8 at 9:58
















$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44






$begingroup$
There must be a sign error, the $theta$ should simplify instead, and the fraction turn to $e^{ipi/2}$.
$endgroup$
– Yves Daoust
Jan 8 at 9:44














$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47




$begingroup$
Demoivre's theory, the arguments of two complex number are added.
$endgroup$
– yuanming luo
Jan 8 at 9:47












$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55




$begingroup$
How did you get from $$-tan(frac{theta}{2})frac{cosbigl(-(frac{theta}{2}+frac{pi}{2})big)+isinbigl(-(frac{theta}{2}+frac{pi}{2})big)}{cos(frac{theta}{2})+isin(frac{theta}{2})}$$ to $$-tan(frac{theta}{2})bigl(cos(-theta-frac{pi}{2})+isin(-theta-frac{pi}{2})bigl)?$$
$endgroup$
– 5xum
Jan 8 at 9:55












$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56






$begingroup$
@5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another.
$endgroup$
– Arthur
Jan 8 at 9:56














$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58




$begingroup$
@Arthur " That's what I asked earlier"... I don't see your question anywhere...
$endgroup$
– 5xum
Jan 8 at 9:58










2 Answers
2






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oldest

votes


















2












$begingroup$

You should be more careful when applying transformations of sines into cosines.



It's much easier than that:
$$
sinfrac{theta}{2}-icosfrac{theta}{2}=
ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
$$
You could write this in trigonometric form
$$
=left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
=cosleft(frac{theta}{2}+frac{pi}{2}right)+
isinleft(frac{theta}{2}+frac{pi}{2}right)
$$
and now you can go and chase for your mistake.



Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
$$
bar{w}=frac{bar{z}-1}{bar{z}+1}
$$
but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
$$
bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
$$
From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.



If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
$$
w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
=frac{u-bar{u}}{u+bar{u}}=
frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We have
    $$
    sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
    = -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
    = -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
    $$
    so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      You should be more careful when applying transformations of sines into cosines.



      It's much easier than that:
      $$
      sinfrac{theta}{2}-icosfrac{theta}{2}=
      ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
      $$
      You could write this in trigonometric form
      $$
      =left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
      left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
      =cosleft(frac{theta}{2}+frac{pi}{2}right)+
      isinleft(frac{theta}{2}+frac{pi}{2}right)
      $$
      and now you can go and chase for your mistake.



      Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
      $$
      bar{w}=frac{bar{z}-1}{bar{z}+1}
      $$
      but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
      $$
      bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
      $$
      From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.



      If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
      $$
      w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
      =frac{u-bar{u}}{u+bar{u}}=
      frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
      $$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You should be more careful when applying transformations of sines into cosines.



        It's much easier than that:
        $$
        sinfrac{theta}{2}-icosfrac{theta}{2}=
        ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
        $$
        You could write this in trigonometric form
        $$
        =left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
        left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
        =cosleft(frac{theta}{2}+frac{pi}{2}right)+
        isinleft(frac{theta}{2}+frac{pi}{2}right)
        $$
        and now you can go and chase for your mistake.



        Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
        $$
        bar{w}=frac{bar{z}-1}{bar{z}+1}
        $$
        but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
        $$
        bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
        $$
        From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.



        If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
        $$
        w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
        =frac{u-bar{u}}{u+bar{u}}=
        frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
        $$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You should be more careful when applying transformations of sines into cosines.



          It's much easier than that:
          $$
          sinfrac{theta}{2}-icosfrac{theta}{2}=
          ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
          $$
          You could write this in trigonometric form
          $$
          =left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
          left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
          =cosleft(frac{theta}{2}+frac{pi}{2}right)+
          isinleft(frac{theta}{2}+frac{pi}{2}right)
          $$
          and now you can go and chase for your mistake.



          Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
          $$
          bar{w}=frac{bar{z}-1}{bar{z}+1}
          $$
          but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
          $$
          bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
          $$
          From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.



          If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
          $$
          w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
          =frac{u-bar{u}}{u+bar{u}}=
          frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
          $$






          share|cite|improve this answer









          $endgroup$



          You should be more careful when applying transformations of sines into cosines.



          It's much easier than that:
          $$
          sinfrac{theta}{2}-icosfrac{theta}{2}=
          ileft(cosfrac{theta}{2}+isinfrac{theta}{2}right)
          $$
          You could write this in trigonometric form
          $$
          =left(cosfrac{pi}{2}+isinfrac{pi}{2}right)
          left(cosfrac{theta}{2}+isinfrac{theta}{2}right)
          =cosleft(frac{theta}{2}+frac{pi}{2}right)+
          isinleft(frac{theta}{2}+frac{pi}{2}right)
          $$
          and now you can go and chase for your mistake.



          Another way to do the same: the conjugate of your number $w=frac{z-1}{z+1}$ is
          $$
          bar{w}=frac{bar{z}-1}{bar{z}+1}
          $$
          but, since $|z|=1$, we have $bar{z}=z^{-1}$; therefore
          $$
          bar{w}=frac{z^{-1}-1}{z^{-1}+1}=frac{1-z}{1+z}=-w
          $$
          From $bar{w}=-w$ it follows that $operatorname{Re}(w)=0$.



          If you want to find the imaginary part in term of $theta$, you can consider $z=u^2$, where $u=cos(theta/2)+isin(theta/2)$; then
          $$
          w=frac{z-1}{z+1}=frac{u^2-1}{u^2+1}=frac{u-u^{-1}}{u+u^{-1}}
          =frac{u-bar{u}}{u+bar{u}}=
          frac{2isin(theta/2)}{2cos(theta/2)}=itanfrac{theta}{2}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 10:05









          egregegreg

          186k1486208




          186k1486208























              1












              $begingroup$

              We have
              $$
              sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
              = -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
              = -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
              $$
              so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                We have
                $$
                sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
                = -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
                = -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
                $$
                so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  We have
                  $$
                  sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
                  = -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
                  = -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
                  $$
                  so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.






                  share|cite|improve this answer









                  $endgroup$



                  We have
                  $$
                  sinleft(fractheta2right) - icosleft(fractheta2right) = -sinleft(-fractheta2right) - icosleft(fractheta2right)\
                  = -cosleft(-fractheta2-fracpi2right) - isinleft(fractheta2 + fracpi2right)\
                  = -left(cosleft(fractheta2 + fracpi2right) + isinleft(fractheta2 + fracpi2right)right)
                  $$
                  so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 9:56









                  ArthurArthur

                  123k7122211




                  123k7122211






























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