Prove $sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$ is true [duplicate]Prove by induction:...

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Prove $sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$ is true [duplicate]


Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?Prove $f : Arightarrow B, g: Brightarrow C$ , and $gcirc f: A overset{1-1}{rightarrow}C$, then $f:Aoverset{1-1}{rightarrow}B$Solving complex eqautionsHow to plot a curve in complex plane that include e constantComplex Plane ( $arg(z)$)How to find the equations whose roots are equal to the following numbers?Complex Numbers: Im having a problem solving this.Find all real numbers m for which equation $z^3+(3+i)z^2-3z-(m+i)=0$ has ..Finding $1+frac{sin x}{sin x}+ frac{sin^2x}{sin 2x}+…+frac{sin^2x}{sin 2x}$ using this method.Finding Real & Distinct solutions in complex numbers for equation $x^2+4x-1+k(x^2+2x+1)=0$.Finding point of intersection of tangents through complex numbers













1












$begingroup$



This question already has an answer here:




  • Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$

    3 answers





If $x neq kpi$ and $n in mathbb Z$ then prove $$sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$$




So I stumbled into this problem in my textbook in the complex numbers chapter. It is an optional problem, so the teacher won't explain it to us. I have no idea on how to approach it in the first place. Could anyone please give me a clue on how to begin solving it?






complex-numbers proof-writing







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marked as duplicate by Martin R, Nosrati, José Carlos Santos, Lee David Chung Lin, Eevee Trainer Mar 15 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Also: math.stackexchange.com/q/17966/42969.
    $endgroup$
    – Martin R
    Mar 14 at 21:34
















1












$begingroup$



This question already has an answer here:




  • Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$

    3 answers





If $x neq kpi$ and $n in mathbb Z$ then prove $$sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$$




So I stumbled into this problem in my textbook in the complex numbers chapter. It is an optional problem, so the teacher won't explain it to us. I have no idea on how to approach it in the first place. Could anyone please give me a clue on how to begin solving it?






complex-numbers proof-writing







share|cite|improve this question











$endgroup$



marked as duplicate by Martin R, Nosrati, José Carlos Santos, Lee David Chung Lin, Eevee Trainer Mar 15 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Also: math.stackexchange.com/q/17966/42969.
    $endgroup$
    – Martin R
    Mar 14 at 21:34














1












1








1


1



$begingroup$



This question already has an answer here:




  • Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$

    3 answers





If $x neq kpi$ and $n in mathbb Z$ then prove $$sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$$




So I stumbled into this problem in my textbook in the complex numbers chapter. It is an optional problem, so the teacher won't explain it to us. I have no idea on how to approach it in the first place. Could anyone please give me a clue on how to begin solving it?






complex-numbers proof-writing







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$

    3 answers





If $x neq kpi$ and $n in mathbb Z$ then prove $$sin x + sin 3x + cdots + sin(2n-1)x = frac{sin^2 nx}{sin x}$$




So I stumbled into this problem in my textbook in the complex numbers chapter. It is an optional problem, so the teacher won't explain it to us. I have no idea on how to approach it in the first place. Could anyone please give me a clue on how to begin solving it?





This question already has an answer here:




  • Prove by induction: $sin{x}+sin{3x}+dots+sin{(2n-1)x}=frac{sin^2{nx}}{sin{x}}$

    3 answers







complex-numbers proof-writing




complex-numbers proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 21:33









Maria Mazur

48.3k1260121




48.3k1260121










asked Mar 14 at 21:18









MegaMathMegaMath

114




114




marked as duplicate by Martin R, Nosrati, José Carlos Santos, Lee David Chung Lin, Eevee Trainer Mar 15 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Martin R, Nosrati, José Carlos Santos, Lee David Chung Lin, Eevee Trainer Mar 15 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Also: math.stackexchange.com/q/17966/42969.
    $endgroup$
    – Martin R
    Mar 14 at 21:34


















  • $begingroup$
    Also: math.stackexchange.com/q/17966/42969.
    $endgroup$
    – Martin R
    Mar 14 at 21:34
















$begingroup$
Also: math.stackexchange.com/q/17966/42969.
$endgroup$
– Martin R
Mar 14 at 21:34




$begingroup$
Also: math.stackexchange.com/q/17966/42969.
$endgroup$
– Martin R
Mar 14 at 21:34










2 Answers
2






active

oldest

votes


















2












$begingroup$

For a complex numbers proof, note the LHS is the imaginary part
of
$$S=e^{ix}+e^{3ix}+cdots+e^{(2n-1)ix}.$$
That's a geometric progression:
$$S=e^{ix}frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}frac{e^{inx}
-e^{-inx}}{e^{i x}-e^{-ix}}
=e^{inx}frac{sin nx}{sin x}=(cos nx+isin nx)
frac{sin nx}{sin x}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @MariaMazur Not any more!
    $endgroup$
    – Lord Shark the Unknown
    Mar 14 at 21:38










  • $begingroup$
    I bet you have writen somwhere both solution?
    $endgroup$
    – Maria Mazur
    Mar 14 at 21:39



















3












$begingroup$

If we make an induction proof then it is enought to prove: $$sin(2n+1)x = frac{sin^2 (n+1)x}{sin x} -frac{sin^2 nx}{sin x} $$ or



$$sin(2n+1)x cdot sin x= Big(sin (n+1)x -sin nxBig)Big(sin (n+1)x +sin nxBig) $$



Now, since:
$$sin (n+1)x -sin nx = 2cos {(2n+1)xover 2}sin{xover 2}$$
and
$$sin (n+1)x +sin nx = 2sin {(2n+1)xover 2}cos{xover 2}$$



and using $$ sin2alpha = 2sin alphacos alpha $$ we are done.






share|cite|improve this answer











$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For a complex numbers proof, note the LHS is the imaginary part
    of
    $$S=e^{ix}+e^{3ix}+cdots+e^{(2n-1)ix}.$$
    That's a geometric progression:
    $$S=e^{ix}frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}frac{e^{inx}
    -e^{-inx}}{e^{i x}-e^{-ix}}
    =e^{inx}frac{sin nx}{sin x}=(cos nx+isin nx)
    frac{sin nx}{sin x}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @MariaMazur Not any more!
      $endgroup$
      – Lord Shark the Unknown
      Mar 14 at 21:38










    • $begingroup$
      I bet you have writen somwhere both solution?
      $endgroup$
      – Maria Mazur
      Mar 14 at 21:39
















    2












    $begingroup$

    For a complex numbers proof, note the LHS is the imaginary part
    of
    $$S=e^{ix}+e^{3ix}+cdots+e^{(2n-1)ix}.$$
    That's a geometric progression:
    $$S=e^{ix}frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}frac{e^{inx}
    -e^{-inx}}{e^{i x}-e^{-ix}}
    =e^{inx}frac{sin nx}{sin x}=(cos nx+isin nx)
    frac{sin nx}{sin x}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @MariaMazur Not any more!
      $endgroup$
      – Lord Shark the Unknown
      Mar 14 at 21:38










    • $begingroup$
      I bet you have writen somwhere both solution?
      $endgroup$
      – Maria Mazur
      Mar 14 at 21:39














    2












    2








    2





    $begingroup$

    For a complex numbers proof, note the LHS is the imaginary part
    of
    $$S=e^{ix}+e^{3ix}+cdots+e^{(2n-1)ix}.$$
    That's a geometric progression:
    $$S=e^{ix}frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}frac{e^{inx}
    -e^{-inx}}{e^{i x}-e^{-ix}}
    =e^{inx}frac{sin nx}{sin x}=(cos nx+isin nx)
    frac{sin nx}{sin x}.$$






    share|cite|improve this answer











    $endgroup$



    For a complex numbers proof, note the LHS is the imaginary part
    of
    $$S=e^{ix}+e^{3ix}+cdots+e^{(2n-1)ix}.$$
    That's a geometric progression:
    $$S=e^{ix}frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}frac{e^{inx}
    -e^{-inx}}{e^{i x}-e^{-ix}}
    =e^{inx}frac{sin nx}{sin x}=(cos nx+isin nx)
    frac{sin nx}{sin x}.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 14 at 21:38

























    answered Mar 14 at 21:32









    Lord Shark the UnknownLord Shark the Unknown

    107k1162135




    107k1162135












    • $begingroup$
      @MariaMazur Not any more!
      $endgroup$
      – Lord Shark the Unknown
      Mar 14 at 21:38










    • $begingroup$
      I bet you have writen somwhere both solution?
      $endgroup$
      – Maria Mazur
      Mar 14 at 21:39


















    • $begingroup$
      @MariaMazur Not any more!
      $endgroup$
      – Lord Shark the Unknown
      Mar 14 at 21:38










    • $begingroup$
      I bet you have writen somwhere both solution?
      $endgroup$
      – Maria Mazur
      Mar 14 at 21:39
















    $begingroup$
    @MariaMazur Not any more!
    $endgroup$
    – Lord Shark the Unknown
    Mar 14 at 21:38




    $begingroup$
    @MariaMazur Not any more!
    $endgroup$
    – Lord Shark the Unknown
    Mar 14 at 21:38












    $begingroup$
    I bet you have writen somwhere both solution?
    $endgroup$
    – Maria Mazur
    Mar 14 at 21:39




    $begingroup$
    I bet you have writen somwhere both solution?
    $endgroup$
    – Maria Mazur
    Mar 14 at 21:39











    3












    $begingroup$

    If we make an induction proof then it is enought to prove: $$sin(2n+1)x = frac{sin^2 (n+1)x}{sin x} -frac{sin^2 nx}{sin x} $$ or



    $$sin(2n+1)x cdot sin x= Big(sin (n+1)x -sin nxBig)Big(sin (n+1)x +sin nxBig) $$



    Now, since:
    $$sin (n+1)x -sin nx = 2cos {(2n+1)xover 2}sin{xover 2}$$
    and
    $$sin (n+1)x +sin nx = 2sin {(2n+1)xover 2}cos{xover 2}$$



    and using $$ sin2alpha = 2sin alphacos alpha $$ we are done.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      If we make an induction proof then it is enought to prove: $$sin(2n+1)x = frac{sin^2 (n+1)x}{sin x} -frac{sin^2 nx}{sin x} $$ or



      $$sin(2n+1)x cdot sin x= Big(sin (n+1)x -sin nxBig)Big(sin (n+1)x +sin nxBig) $$



      Now, since:
      $$sin (n+1)x -sin nx = 2cos {(2n+1)xover 2}sin{xover 2}$$
      and
      $$sin (n+1)x +sin nx = 2sin {(2n+1)xover 2}cos{xover 2}$$



      and using $$ sin2alpha = 2sin alphacos alpha $$ we are done.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        If we make an induction proof then it is enought to prove: $$sin(2n+1)x = frac{sin^2 (n+1)x}{sin x} -frac{sin^2 nx}{sin x} $$ or



        $$sin(2n+1)x cdot sin x= Big(sin (n+1)x -sin nxBig)Big(sin (n+1)x +sin nxBig) $$



        Now, since:
        $$sin (n+1)x -sin nx = 2cos {(2n+1)xover 2}sin{xover 2}$$
        and
        $$sin (n+1)x +sin nx = 2sin {(2n+1)xover 2}cos{xover 2}$$



        and using $$ sin2alpha = 2sin alphacos alpha $$ we are done.






        share|cite|improve this answer











        $endgroup$



        If we make an induction proof then it is enought to prove: $$sin(2n+1)x = frac{sin^2 (n+1)x}{sin x} -frac{sin^2 nx}{sin x} $$ or



        $$sin(2n+1)x cdot sin x= Big(sin (n+1)x -sin nxBig)Big(sin (n+1)x +sin nxBig) $$



        Now, since:
        $$sin (n+1)x -sin nx = 2cos {(2n+1)xover 2}sin{xover 2}$$
        and
        $$sin (n+1)x +sin nx = 2sin {(2n+1)xover 2}cos{xover 2}$$



        and using $$ sin2alpha = 2sin alphacos alpha $$ we are done.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 14 at 21:30

























        answered Mar 14 at 21:25









        Maria MazurMaria Mazur

        48.3k1260121




        48.3k1260121















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