Writing linear combination of exponentials as cosine












0














Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.



So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$



Am I on the right track? I can't seem to figure out where to go next.










share|cite|improve this question






















  • Should the coefficients be conjugate? Otherwise $y$ might not be real.
    – xbh
    Dec 5 '18 at 2:58










  • The question didn't specify that the coefficients are conjugate...
    – 51n84d
    Dec 5 '18 at 3:19
















0














Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.



So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$



Am I on the right track? I can't seem to figure out where to go next.










share|cite|improve this question






















  • Should the coefficients be conjugate? Otherwise $y$ might not be real.
    – xbh
    Dec 5 '18 at 2:58










  • The question didn't specify that the coefficients are conjugate...
    – 51n84d
    Dec 5 '18 at 3:19














0












0








0







Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.



So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$



Am I on the right track? I can't seem to figure out where to go next.










share|cite|improve this question













Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.



So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$



Am I on the right track? I can't seem to figure out where to go next.







differential-equations complex-numbers






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 2:51









51n84d51n84d

383




383












  • Should the coefficients be conjugate? Otherwise $y$ might not be real.
    – xbh
    Dec 5 '18 at 2:58










  • The question didn't specify that the coefficients are conjugate...
    – 51n84d
    Dec 5 '18 at 3:19


















  • Should the coefficients be conjugate? Otherwise $y$ might not be real.
    – xbh
    Dec 5 '18 at 2:58










  • The question didn't specify that the coefficients are conjugate...
    – 51n84d
    Dec 5 '18 at 3:19
















Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58




Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58












The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19




The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19










1 Answer
1






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oldest

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0














Hint



Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.






share|cite|improve this answer





















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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Hint



    Since
    $$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
    must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
    $$A_1+A_2=A cos (delta )$$
    $$i(A_1-A_2)=A sin (delta )$$
    Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.






    share|cite|improve this answer


























      0














      Hint



      Since
      $$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
      must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
      $$A_1+A_2=A cos (delta )$$
      $$i(A_1-A_2)=A sin (delta )$$
      Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.






      share|cite|improve this answer
























        0












        0








        0






        Hint



        Since
        $$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
        must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
        $$A_1+A_2=A cos (delta )$$
        $$i(A_1-A_2)=A sin (delta )$$
        Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.






        share|cite|improve this answer












        Hint



        Since
        $$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
        must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
        $$A_1+A_2=A cos (delta )$$
        $$i(A_1-A_2)=A sin (delta )$$
        Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 3:54









        Claude LeiboviciClaude Leibovici

        119k1157132




        119k1157132






























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