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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

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

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

$endgroup$

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
  

  
  
  
  

add a comment | 

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

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

$endgroup$

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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

$endgroup$

share|cite|improve this question

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

share|cite|improve this question

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

edited Mar 14 at 21:33

Maria Mazur

48.3k1260121

asked Mar 14 at 21:18

MegaMathMegaMath

114

asked Mar 14 at 21:18

MegaMathMegaMath

114

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

edited Mar 14 at 21:38

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

add a comment | 

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

edited Mar 14 at 21:30

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

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

edited Mar 14 at 21:38

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

add a comment | 

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

edited Mar 14 at 21:38

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Mar 14 at 21:38

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

$endgroup$

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

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

answered Mar 14 at 21:32

Lord Shark the UnknownLord Shark the Unknown

107k1162135

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

edited Mar 14 at 21:30

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

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

edited Mar 14 at 21:30

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

$begingroup$

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

edited Mar 14 at 21:30

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

share|cite|improve this answer

edited Mar 14 at 21:30

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121

answered Mar 14 at 21:25

Maria MazurMaria Mazur

48.3k1260121