What is the value of n in the identity sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx?

[I like Serena]

No, i don't know about polar coordinates.

Well, I guess we're chunking off a bit more than I originally thought.
But then, you seem so knowledgeable already!

Well, as long as you want to learn, that's fine by me.

Here's a picture that shows which r and phi I'm talking about (in the picture they use theta instead of phi though).

If you have a point (x, y) it has a distance to the origin, which we call "r".
As you can see in the picture, you can define a rectangular triangle with an angle.
If you apply the definition of the cosine and the sine, you should be able to see that
x = r cos angle
y = r sin angle

That's it!
This is what we call polar coordinates, which is a different way to identify points in a plane.

Yep, i found out three solutions using graph.
Is there any other way to find the number of solutions?

Erm... what are you thinking of?

The key is that any angle has a period of 2pi.

 

[Saitama]

Well, I guess we're chunking off a bit more than I originally thought.
But then, you seem so knowledgeable already!

Well, as long as you want to learn, that's fine by me.

Here's a picture that shows which r and phi I'm talking about (in the picture they use theta instead of phi though).

If you have a point (x, y) it has a distance to the origin, which we call "r".
As you can see in the picture, you can define a rectangular triangle with an angle.
If you apply the definition of the cosine and the sine, you should be able to see that
x = r cos angle
y = r sin angle

That's it!
This is what we call polar coordinates, which is a different way to identify points in a plane.

Thank you for your explanation I Like Serena!
So now let's get back to the question i.e. z=re^iϕ.

Erm... what are you thinking of?

The key is that any angle has a period of 2pi.

To find the solutions of cos3x=1, what i did was that i draw a graph of cos 3x, then draw a line parallel to x-axis at y=1 on the graph of cos 3x. The points where both the graphs intersect are the solutions for cos 3x=1.
This method is graphical but i want to know the other way to find the number of solutions for cos3x=1, i don't want to use graphical method.

 

[I like Serena]

Thank you for your explanation I Like Serena!
So now let's get back to the question i.e. z=re^iϕ.

What do you get if you substitute Euler's formula?

To find the solutions of cos3x=1, what i did was that i draw a graph of cos 3x, then draw a line parallel to x-axis at y=1 on the graph of cos 3x. The points where both the graphs intersect are the solutions for cos 3x=1.
This method is graphical but i want to know the other way to find the number of solutions for cos3x=1, i don't want to use graphical method.

Well, the algebraic method is to use the "mod 2pi" notation, or the "+2 k pi" notation, meaning that a cosine will be the same if you use an angle that is a multiple of 2pi bigger or smaller.

And another graphical method, is to look at the polar coordinates and consider what the angle should be to come out at (1, 0).

What do you know already of this?

 

[Saitama]

What do you get if you substitute Euler's formula?

Substituting Euler's formula:-
z=r(cosx+isinx)
What's this "r"?

Well, the algebraic method is to use the "mod 2pi" notation, or the "+2 k pi" notation, meaning that a cosine will be the same if you use an angle that is a multiple of 2pi bigger or smaller.

And another graphical method, is to look at the polar coordinates and consider what the angle should be to come out at (1, 0).

What do you know already of this?

I think i should not go to algebraic method.

 

[I like Serena]

Substituting Euler's formula:-
z=r(cosx+isinx)
What's this "r"?

The same r as in the polar coordinates.
It is the distance of point (X, Y) to the origin, where z = X + i Y.
Btw, I used capital letters to distinguish them from the "x" in your equation which is actually phi.

Taking it one step further, you have:
z = r cosx + i r sinx = X + i Y

Do you see now?

I think i should not go to algebraic method.

Nice icon!

So you have not learned yet how to solve cos x = c?

 

[Saitama]

The same r as in the polar coordinates.
It is the distance of point (X, Y) to the origin, where z = X + i Y.
Btw, I used capital letters to distinguish them from the "x" in your equation which is actually phi.

Taking it one step further, you have:
z = r cosx + i r sinx = X + i Y

Do you see now?

Got it now! I would print this disccussion to keep it for my reference.
So what should be the answer of z³=1?

Nice icon!

So you have not learned yet how to solve cos x = c?

Thanks!

No, i have not learned the algebraic method.

 

[I like Serena]

Got it now! I would print this disccussion to keep it for my reference.
So what should be the answer of z³=1?

In exponential notation:
z₁ = 1
z₂ = e^i(2/3)π
z₃ = e^i(4/3)π

In cartesian notation:
z₁ = 1
z₂ = -½ + i ½√3
z₃ = -½ - i ½√3


And to top it off, here's a cool pic:
http://www.wolframalpha.com/input/?i=polar%20plot%20Re%28%28e^%28i%20phi%29%29^3-1%29&lk=2

 

[Saitama]

In exponential notation:
z₁ = 1
z₂ = e^i(2/3)π
z₃ = e^i(4/3)π

In cartesian notation:
z₁ = 1
z₂ = -½ + i ½√3
z₃ = -½ - i ½√3

How did you get
z₂ = e^i(2/3)π
z₃ = e^i(4/3)π

 

[I like Serena]

How did you get
z₂ = e^i(2/3)π
z₃ = e^i(4/3)π

What do you get if you substitute them in the equation z³ = 1?

 

[Saitama]

What do you get if you substitute them in the equation z³ = 1?

Substituting z2, eⁱ²ⁿ=1
Substituting z3, eⁱ⁴ⁿ=1

Solving both the equations i get n=0.
But how do you get the solutions?

 

[I like Serena]

Substituting z2, eⁱ²ⁿ=1
Substituting z3, eⁱ⁴ⁿ=1

Solving both the equations i get n=0.
But how do you get the solutions?

Ah, I see, there is some confusion about π.
That might explain some of the confusion in the previous posts.

Well, π is π and not n.
See the difference?
It's a bit of a font problem I'm afraid.

 

[Saitama]

Ah, I see, there is some confusion about π.
That might explain some of the confusion in the previous posts.

Well, π is π and not n.
See the difference?
It's a bit of a font problem I'm afraid.

Okay i again substitute the values,
Substituting z2, e^i2\pi=1
Substituting z3, e^i4\pi=1

How do you get these solutions?

 

[I like Serena]

Okay i again substitute the values,
Substituting z2, e^i2\pi=1
Substituting z3, e^i4\pi=1

How do you get these solutions?

First you will need to understand why these are solutions.
Can you use Euler's formula to convert them to cosine/sine form?

 

[Saitama]

First you will need to understand why these are solutions.
Can you use Euler's formula to convert them to cosine/sine form?

Applying Euler's formula:-
e^i2\pi=1
\Rightarrowcos2\pi+isin\pi=1
\Rightarrowcos2\pi+isin\pi=1

 

[I like Serena]

Applying Euler's formula:-
e^i2\pi=1
\Rightarrowcos2\pi+isin\pi=1
\Rightarrowcos2\pi+isin\pi=1

Erm... you did not substitute correctly.
But then, do you understand why this is a solution?
What is cos(2pi)?

 

[Saitama]

Erm... you did not substitute correctly.
And then, do you understand why this is a solution?
What is cos(2pi)?

Oops sorry, it should be sin 2pi.
I think this is the solution because if we further solve the sine and cosine we get 1 which is equal to the right hand side of the equation.
cos(2pi) is 1.

 

[I like Serena]

Oops sorry, it should be sin 2pi.
I think this is the solution because if we further solve the sine and cosine we get 1 which is equal to the right hand side of the equation.
cos(2pi) is 1.

Exactly!

What about the other solution?
What is cos(4pi)? And sin(4pi)?

Do you perhaps understand now how to I found these solutions?

 

[Saitama]

Exactly!

What about the other solution?
What is cos(4pi)? And sin(4pi)?

Do you perhaps understand now how to I found these solutions?

cos(4pi) is 1 and sin(4pi) is 0, but i still don't understand how you find out the solution?

 

[I like Serena]

cos(4pi) is 1 and sin(4pi) is 0, but i still don't understand how you find out the solution?

Try to work backward then.
You already know that 2π is a solution as cos(2π)=1
That means that if 3ϕ= 2π that we have solution, so ϕ= (2/3)π.

 

[I like Serena]

Due to the periodicity of angles in general we have:

1 = e^{i \cdot 0} = e^{i \cdot 2\pi} = e^{i \cdot 4\pi} = e^{i \cdot 6\pi} = ...
So starting from z^3 = (e^{i\phi})^3 = e^{i \cdot 3\phi} = 1

We find 3\phi = 0 or 3\phi = 2\pi or 3\phi = 4\pi or 3\phi = 6\pi or ...

This means \phi = 0 or \phi = \frac 2 3\pi or \phi = \frac 4 3 \pi or \phi = \frac 6 3 \pi = 2\pi or ...

However, the last solution is just the same as the first solution, being exactly one period along, and the same holds for all the following solutions.
So we only have 3 unique solutions.

 

[Saitama]

Due to the periodicity of angles in general we have:

1 = e^{i \cdot 0} = e^{i \cdot 2\pi} = e^{i \cdot 4\pi} = e^{i \cdot 6\pi} = ...
So starting from z^3 = (e^{i\phi})^3 = e^{i \cdot 3\phi} = 1

We find 3\phi = 0 or 3\phi = 2\pi or 3\phi = 4\pi or 3\phi = 6\pi or ...

This means \phi = 0 or \phi = \frac 2 3\pi or \phi = \frac 4 3 \pi or \phi = \frac 6 3 \pi = 2\pi or ...

However, the last solution is just the same as the first solution, being exactly one period along, and the same holds for all the following solutions.
So we only have 3 unique solutions.

Sorry for the late reply.
I thought i would post after going through our discussion once again.

May i know how you get the following relation:-
1 = e^{i \cdot 0} = e^{i \cdot 2\pi} = e^{i \cdot 4\pi} = e^{i \cdot 6\pi} = ...

 

[I like Serena]

Can you ask your question(s) a bit more specific?

I'm really not sure what I should explain, and what you already get.

 

[Saitama]

Can you ask your question(s) a bit more specific?

I'm really not sure what I should explain, and what you already get.

Oops sorry! I edited my previous post.

 

[I like Serena]

Let's just pick one.

We have because of Euler's formula and the periodicity of sine and cosine:
e^{i \cdot 6\pi} = \cos 6\pi + i \sin 6\pi = 1 + i \cdot 0 = 1

That's all.

 

[Saitama]

So are we done?

 

[I like Serena]

So are we done?

I don't know.

Are we?

 

[Saitama]

I don't know.

Are we?

Yes, i think.
I would practice complex numbers more and more now.

(Never thought complex numbers would be so interesting)