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

[Saitama]

Hi!
I had read the complex numbers chapter from my textbook and i think i now have a basic idea of them. So may i know how ehild has got this relationship:-
$$sinx=\frac{e^{ix}-e^{-ix}}{2i}$$

 

[I like Serena]

Hi again!

Did you find Euler's formula e^ix = cos x + i sin x in your chapter?

 

[Saitama]

Hi again!

Did you find Euler's formula e^ix = cos x + i sin x in your chapter?

No there was no "Euler's formula" in my book.

 

[I like Serena]

No there was no "Euler's formula" in my book.

That's a problem then, since this identity is the most important one in the entire field of the complex numbers.

Richard Feynman called Euler's formula "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."

The formula ehild mentioned follows from this formula.

 

[Saitama]

That's a problem then, since this identity is the most important one in the entire field of the complex numbers.

Richard Feynman called Euler's formula "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."

The formula ehild mentioned follows from this formula.

I read it on Wikipedia, i think its not that hard as i thought of.

 

[I like Serena]

Good! Indeed I never found it hard to understand! :)

Do you see how ehild's formula follows from it?

 

[Saitama]

Good! Indeed I never found it hard to understand! :)

Do you see how ehild's formula follows from it?

I was checking out the ehild's solution, i got confused at one step when ehild multiplied -8 with 2 in the denominator. In the next step there was no minus sign.
Also how did ehild get -8?

 

[I like Serena]

I was checking out the ehild's solution, i got confused at one step when ehild multiplied -8 with 2 in the denominator. In the next step there was no minus sign.
Also how did ehild get -8?

The other thing you need to know about complex numbers is the definition of i:
i² = -1

So (2i)³ = 2³ i² i = 8 (-1) i = -8i.

 

[Saitama]

The other thing you need to know about complex numbers is the definition of i:
i² = -1

So (2i)³ = 2³ i² i = 8 (-1) i = -8i.

Oh sorry! I am really a fool to ask this question.
Now i need to practice more questions on Complex Numbers.

 

[I like Serena]

Well here's a nice (and fundamental) one:

Can you find the solutions of z³ = 1?

 

[Saitama]

Well here's a nice (and fundamental) one:

Can you find the solutions of z³ = 1?

Real solutions or complex solutions?
Real solution z=1
Complex Solution=?

 

[I like Serena]

Complex solutions of course.

To find them substitute z = e^iϕ and remember that ϕ is an angle with a period of 2π.

 

[Saitama]

Complex solutions of course.

To find them substitute z = e^iϕ and remember that ϕ is an angle with a period of 2π.

Do i have to find the angle ϕ?

 

[I like Serena]

Yes.

 

[Saitama]

Here are my steps:-
$$(e^{iϕ})^3=1$$
$$e^{3iϕ}=e^0$$
$$3iϕ=0$$
$$ϕ=0$$

Right..?

 

[I like Serena]

Almost. :)

As I said, you have to take into account that ϕ is an angle with a period of 2π.
It means that there are more solutions than just this one.

So e^iϕ = e^{i(ϕ + 2kπ)} for any integer k.

Can you repeat what you just did with e^{i(ϕ + 2kπ)}?

 

[Saitama]

Almost. :)

As I said, you have to take into account that ϕ is an angle with a period of 2π.
It means that there are more solutions than just this one.

So e^iϕ = e^{i(ϕ + 2kπ)} for any integer k.

Can you repeat what you just did with e^{i(ϕ + 2kπ)}?

Ok, i repeat the steps:-
$$i(ϕ+2kπ)=0$$
$$ϕ=-2kπ$$

Btw, i still didn't get why there are more solutions?

 

[I like Serena]

Btw, i still didn't get why there are more solutions?

Aren't you forgetting a power or rather factor of 3 here? ;)

 

[Saitama]

Aren't you forgetting a factor 3 here? ;)

What?

 

[I like Serena]

What?

Try:
(e^{i(ϕ + 2kπ)})³ = 1

Or rather:
i(3ϕ + 2kπ) = 0

 

[Saitama]

Try:
(e^{i(ϕ + 2kπ)})³ = 1

But i am still getting -2kπ.

 

[I like Serena]

Yeah, I just edited my previous post...

 

[Saitama]

This time i am getting (-2kπ)/3...
But how do you get i(3ϕ+2kπ)=0?

 

[I like Serena]

Look at it this way:

You already had:
e^i(3ϕ) = 1

So:
cos(3ϕ) + i sin(3ϕ) = 1

This means that
cos(3ϕ) = 1 and sin(3ϕ) = 0

Can you solve that?

 

[I like Serena]

And here's another way to look at it:

A complex number is represented as in the following picture.

An exponentiation with a power of 3, means a multiplication of the angle with a factor 3.
If you have an angle of 120 degrees, a factor of 3 will make it fully go round.

 

[Saitama]

Look at it this way:

You already had:
e^i(3ϕ) = 1

So:
cos(3ϕ) + i sin(3ϕ) = 1

This means that
cos(3ϕ) = 1 and sin(3ϕ) = 0

Can you solve that?

ϕ=0.
I want to know why you take z=e^iϕ?

 

[I like Serena]

ϕ=0.

Err... no.

The solution of cos(3ϕ) = 1
is 3ϕ=0 mod 2π,
which comes out as ϕ = 0 mod (2/3)π.

My bad, I thought you already knew this.
But now that I think about it, I recall that you did not know the "mod" notation yet.
So are you aware of the periodicity of the cosine function?
That it has a period 2π?

I want to know why you take z=e^iϕ?

Actually, I didn't do that quite right.
It should be:
z = r e^iϕ

Any complex number z has 2 representations:
z = x + i y
z = r e^iϕ

Are you aware of the first representation?
The second form follows from the first combined with Euler's formula.

 

[Saitama]

Err... no.

The solution of cos(3ϕ) = 1
is 3ϕ=0 mod 2π,
which comes out as ϕ = 0 mod (2/3)π.

My bad, I thought you already knew this.
But now that I think about it, I recall that you did not know the "mod" notation yet.
So are you aware of the periodicity of the cosine function?
That is has a period 2π?

Yes, i know about the periodicity of cosine function which is 2π.

Actually, I didn't do that quite right.
It should be:
z = r e^iϕ

Any complex number z has 2 representations:
z = x + i y
z = r e^iϕ

Are you aware of the first representation?
The second form follows from the first combined with Euler's formula.

I know about the first representation.
But i don't know about the second one.

 

[I like Serena]

Yes, i know about the periodicity of cosine function which is 2π.

So do you get that cos(3ϕ) = 1 has 3 solutions?

I know about the first representation.
But i don't know about the second one.

Are you familiar with so called polar coordinates?
That is:
x = r cos(ϕ)
y = r sin(ϕ)

 

[Saitama]

So do you get that cos(3ϕ) = 1 has 3 solutions?

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

Are you familiar with so called polar coordinates?
That is:
x = r cos(ϕ)
y = r sin(ϕ)

No, i don't know about polar coordinates.

 

[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?