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

[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
$\Rightarrow$cos2$\pi$+isin$\pi$=1
$\Rightarrow$cos2$\pi$+isin$\pi$=1

 

[I like Serena]

Applying Euler's formula:-
e^i2$\pi$=1
$\Rightarrow$cos2$\pi$+isin$\pi$=1
$\Rightarrow$cos2$\pi$+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)