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

[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)