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

[Saitama]

I think it's rather good handwriting …

large and clear ​

(except I think the "4" should be more angular ) chunky: []

curvey: ()

curly: {}

pointy: <>

Thanks for appreciating my handwriting.

But are my attempts correct?

 

[tiny-tim]

hmm …

i'm reluctant to go through all that and check it …

you seem to have started by making it more complicated …

i'd have started by using the basic https://www.physicsforums.com/library.php?do=view_item&itemid=18" to get formulas like

2sinxsin3x = cos(3x - x) - cos(3x + x) = cos2x - cos4x

isn't that simpler?

try that ​

 

[Saitama]

I am correcting my attempts after $$\frac{2cos^32x-3cos^22x+1}{4}$$

$$\frac{2cos^32x-3cos^22x+1}{4}$$=$$\frac{2(\frac{cos6x+3cos2x}{4})-3(\frac{cos4x+1}{2})+1}{4}$$=$$\frac{(\frac{cos6x+3cos2x}{2})-3(\frac{cos4x+1}{2})+1}{4}$$=$$\frac{cos6x+3cos2x-3cos4x-1}{8}$$

 

[Saitama]

hmm …

i'm reluctant to go through all that and check it …

you seem to have started by making it more complicated …

i'd have started by using the basic https://www.physicsforums.com/library.php?do=view_item&itemid=18" to get formulas like

2sinxsin3x = cos(3x - x) - cos(3x + x) = cos2x - cos4x

isn't that simpler?

try that ​

Wow! :
I tried that way, it was much simpler. Thank you!
But what should i do next?

I get:-
$$\frac{cos6x+3cos2x-3cos4x-1}{8}$$

 

[tiny-tim]

I tried that way, it was much simpler.

he he

now use a similar formula for sinxcos(something), to use up one more sinx

the use up the final sinx with a coscos formula again

(oh, and yes i get the same final result as you do! …

and to make sure I checked it by putting x = 0 and π/2)

btw, have you done complex numbers yet?

if so, try using https://www.physicsforums.com/library.php?do=view_item&itemid=162" to rewrite the answer as a binomial expansion

 

[Saitama]

he he

now use a similar formula for sinxcos(something), to use up one more sinx

the use up the final sinx with a coscos formula again

(oh, and yes i get the same final result as you do! …

and to make sure I checked it by putting x = 0 and π/2)

btw, have you done complex numbers yet?

if so use https://www.physicsforums.com/library.php?do=view_item&itemid=162" to rewrite the answer as a binomial expansion

Here's no more sine now.

And i haven't done complex numbers yet.

 

[tiny-tim]

Here's no more sine now.

yes there is …

you started with sin³xsin3x = sinx(sinx(sinxsin3x))), and the first step was to expand the sinxsin3x,

so now you use up the middle sinx, and when you've done that you use up the left-hand sinx

And i haven't done complex numbers yet.

doesn't matter, you don't need them

 

[Saitama]

yes there is …

you started with sin³xsin3x = sinx(sinx(sinxsin3x))), and the first step was to expand the sinxsin3x,

so now you use up the middle sinx, and when you've done that you use up the left-hand sinx


doesn't matter, you don't need them

No, there's no sin x.
I started with (sin²x)(sinxsin3x).
As you said 2sinxsin3x = cos(3x - x) - cos(3x + x) = cos2x - cos4x. I used this same process and at the place of "sinxsin3x" i wrote $\frac{cos2x-cos4x}{2}$. Then i was left with sin²x. I expanded this left over sin²x, solved the whole equation and got:-
$$\frac{cos6x+3cos2x-3cos4x-1}{8}$$

 

[tiny-tim]

oh I see!

I thought you hadn't done those extra steps, and you were just repeating the answer you got by your original method, and you were asking how to get there, because you asked "What should I do next?"

you do nothing next, that is the answer, isn't it?

 

[Saitama]

oh I see!

I thought you hadn't done those extra steps, and you were just repeating the answer you got by your original method, and you were asking how to get there, because you asked "What should I do next?"

you do nothing next, that is the answer, isn't it?

No, i still have to find the value of n

 

[tiny-tim]

Nooo …

you have to find the value of n …

which is of course 6 ​

 

[Saitama]

Nooo …

you have to find the value of n …

which is of course 6 ​

Yes, i have to find the value of n, but how do you get 6?

 

[tiny-tim]

i think you've lost the plot

the original question (with that wrong "n" removed) was …

Suppose $$sin^3xsin3x=\sum^n_{m=0}C_mcosmx$$ is an identity in x, where C₀, C₁, ...C_n are constants, and C_n $\neq$0, then what is the value of n?

your answer is (cos6x -3cos4x +3cos2x - 1)/8,

which is ∑ C_mcosmx with C₆ = 1/8, C₄ = -3/8, C₂ = 3/8, C₀ = -1/8

 

[Saitama]

your answer is (cos6x -3cos4x +3cos2x - 1)/8,

which is ∑ C_mcosmx with C₆ = 1/8, C₄ = -3/8, C₂ = 3/8, C₀ = 1/8

I still don't get it...

 

[tiny-tim]

"∑ C_mcosmx" is another way of writing

C₀cos0x + C₁cosx + C₂cos2x + C₃cos3x + ...

(with each C being an ordinary number)

do you see that?​

 

[Saitama]

"∑ C_mcosmx" is another way of writing

C₀cos0x + C₁cosx + C₂cos2x + C₃cos3x + ...

(with each C being an ordinary number)

do you see that?​

That i already know..
But how did you get C₆=1/8 and other values

 

[tiny-tim]

I just read it off your answer of (1/8)cos6x -(3/8)cos4x +(3/8)cos2x - (1/8)

 

[Saitama]

I just read it off your answer of (1/8)cos6x -(3/8)cos4x +(3/8)cos2x - (1/8)

Okay, i understand how you get C₆=(1/8).
But how do you get n=6??

 

[tiny-tim]

Because the question defines n as being the highest value of m.

The "highest" one is C₆ (any C above C₆ is zero), so n = 6.

 

[Saitama]

Because the question defines n as being the highest value of m.

The "highest" one is C₆ (any C above C₆ is zero), so n = 6.

Okay, but if i wanted to find out the value of C₁?

 

[tiny-tim]

C₁ is the coefficient of cosx, so it's 0.

 

[Saitama]

Because the question defines n as being the highest value of m.

The "highest" one is C₆ (any C above C₆ is zero), so n = 6.

Ok thanks..
You helped me a lot!
Thank you tiny-tim!

 

[tiny-tim]

As an "appendix", here's the alternative result using complex numbers and https://www.physicsforums.com/library.php?do=view_item&itemid=162" …

you won't understand it yet, but anyone else following this might be interested …

sin³xsin3x

= sin³x Im(e^3ix)

= sin³x Im((cosx + isinx)³)

= Im((cosxsinx + isin²x)³)

= (1/8) Im((sin2x + i(1 - cos2x))³)

= (1/8) Re((isin2x + 1 - cos2x)³)

= (1/8) Re((1 - e^2ix)³)

= (1/8) Re(1 - 3e^2ix + 3e^4ix - e^6ix)

= (1/8) (1 - 3cos2x + 3cos4x - cos6x) ​

This works for sin^kxsinkx for any value of k …

sin^kxsinkx = (-1/2^k) ∑ ^kC_m cos2mx ​

(so what's sin^kxcoskx, and is there a quicker way of doing it?)

 

[Saitama]

As an "appendix", here's the alternative result using complex numbers and https://www.physicsforums.com/library.php?do=view_item&itemid=162" …

you won't understand it yet, but anyone else following this might be interested …

sin³xsin3x

= sin³x Im(e^3ix)

= sin³x Im((cosx + isinx)³)

= Im((cosxsinx + isin²x)³)

= (1/8) Im((sin2x + i(1 - cos2x))³)

= (1/8) Re((isin2x + 1 - cos2x)³)

= (1/8) Re((1 - e^2ix)³)

= (1/8) Re(1 - 3e^2ix + 3e^4ix - e^6ix)

= (1/8) (1 - 3cos2x + 3cos4x - cos6x) ​

This works for sin^kxsinkx for any value of k …

sin^kxsinkx = (-1/2^k) ∑ ^kC_m cos2mx ​

(so what's sin^kxcoskx, and is there a quicker way of doing it?)

I think i would take a look at it again. My teacher may start up with complex numbers after 2 or 3 chapters. I would save this a text file in my computer.

 

[I like Serena]

As another aside, this series is called the Fourier cosine series expansion.
Fourier was a mathematician who found that any function can be expressed as a cosine series of this type, and that the coefficients are unique.

Many mathematical programmatic tools have this built in, such as Mathematica.
It's also available in WolframAlpha:
http://www.wolframalpha.com/input/?i=FourierCosSeries+sin^3x+sin3x
although I found that you won't be able to get the full expansion here, because WolframAlpha will time out before you'll get there. :(

However, I'm afraid it will be quite a while before you get to learn this, and maybe you will never learn this, depending on what you're going to study exactly.

The application of Fourier analysis is typically in spectral analysis, signal processing, image processing and electronical engineering, where it's used for frequency related analysis and operations.

 

[ehild]

Hi, tiny-tim, is not there a sign error in your derivation?

Just to show off my way:

$$sinx=\frac{e^{ix}-e^{-ix}}{2i}$$,
$$cosx=\frac{e^{ix}+e^{-ix}}{i}$$,

$$sin(x)^3 sin(3x)=(\frac {e^{ix}-e^{-ix}}{2i})^3 \frac{e^{3ix}-e^{-3ix}}{2i}=$$

$$=\frac {e^{3ix}-3 e^{ix}+3 e^{-ix}-e^{-3ix}}{-8i}\frac{e^{3ix}-e^{-3ix}}{2i}=$$
$$1/16(e^{6ix}-3 e^{4ix}+3 e^{2ix}-1-1+3 e^{-2ix}-3 e^{-4ix}+e^{-6ix})=$$
$$1/8(cos(6x)-3cos(4x)+3cos(2x)-1)$$

ehild

 

[I like Serena]

@tiny-tim: Nice improvement on the calculation btw. It's way less work!

Just to show off my way:

Thank you ehild, this is a nice solution, and shows a couple of things.

First, I think that Pranav-Arora is very well able to follow this one and do it himself.
It only requires a basic grasp on complex numbers, and beyond that it's just exponentiation rules, which Pranav-Arora's has already practiced quite a bit.

And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete.
I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier. :shy:

I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations...
Does any of you have any?

 

[ehild]

And it also shows that in my humble opinion calculating with sines, cosines and angles is kind of obsolete.
I found that calculating with imaginary e-powers (or vectors in geometric problems) is much easier. :shy:

I'm interested in examples where you would really want to use sines, cosines and angles to do your calculations...
Does any of you have any?

What about determining the height of a tower from the angle of view at a known distance?

ehild

 

[I like Serena]

What about determining the height of a tower from the angle of view at a known distance?

ehild

Yes, it's still useful for immediate applications of the definitions of sines, cosines and tangents based on an angle.
What I'm looking for is actually calculating with it.
Usually I convert an angle as soon as I can to a vector, then calculate with it, and possibly convert it back into an angle to present a result.
To me it feels a bit like converting inches to metric before calculating, and converting back to present a result.

And even here...
If you measure an angle, you use a tangent to convert distance to height.
But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio.
Multiply by the ratio in distance and you get the height without ever using an angle or the tangent function.

 

[tiny-tim]

hi ehild!

yes that's nice ​

and it's certainly shorter for k = 3 (but not so good for large k)

 

[icystrike]

Please try to use euler's forumla and u will realize that it is made of some even m (2,4,6).

So n=6

 

[I like Serena]

and it's certainly shorter for k = 3 (but not so good for large k)

Let's see.

For k is odd we have:
$$\begin{eqnarray} \sin^kx \sin kx &=& \left(\frac {e^{ix} - e^{-ix}} {2i} \right)^k \left(\frac {e^{ikx} - e^{-ikx}} {2i} \right) \\ &=& \frac 1 {(2i)^{k+1}} \left(\sum_{m=0}^k \binom k m (-1)^m e^{i(k-2m)x} \right) (e^{ikx} - e^{-ikx}) \\ &=& \frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x}) \\ &=& \frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x \\ &=& (-1)^{\frac {k+1} 2} \frac 1 {2^k} \left(\binom k 0 \cos 2kx - \binom k 1 \cos(2k-2)x + \binom k 2 \cos(2k-4)x - ... + \binom k {k-1} \cos 2x - 1 \right) \end{eqnarray}$$

As you can see, I've also corrected the sign error.

 

[tiny-tim]

mmm … let's compare length …

sin^kxsinkx = sin^kx Im((cosx + isinx)^k)

= Im((cosxsinx + isin²x)^k)

= (1/2^k) Im((±sin2x + i(1 - cos2x))^k)

= (1/2^k) Im(i^k(1 - cos2x ± isin2x)^k)

(I lost an i² here originally )

= (1/2^k) Im(i^k(1 - e^2ix)^k)

= (1/2^k) Im(i^k ∑ ^kC_m (-1^m) e^2imx)

= (1/2^k)(-1)^(k-1)/2 ∑ ^kC_m (-1^m) cos2mx for k odd

= (1/2^k)(-1)^k/2 ∑ ^kC_m (-1^m) sin2mx for k even​

ooh, yes, yours is 2 lines shorter

$$\frac 1 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m (e^{i(2k-2m)x} + e^{-i(2k-2m)x})$$
$$\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^m \cos(2k-2m)x$$

oh, that's how it works! nice!

(and for k even, there's a sign-change somewhere)

though we can then use ^kC_m = ^kC_k-m to go straight to …

$$\frac 2 {(2i)^{k+1}} \sum_{m=0}^k \binom k m (-1)^{k-m} \cos2mx$$

 

[ehild]

I just wanted to show this, but you both beat me while I watched Poirot on TV

ehild

 

[ehild]

If you measure an angle, you use a tangent to convert distance to height.
But what you actually measure, is a distance on a measuring device that is marked as an angle, but it might as well be marked as a distance at a specific short distance, or as a tangent ratio..

As far as I know the angle of view is measure of the length of an arc between the arms of a compass, when those arms point to the desired directions. And the tangent of that angle is used to get distance or height.

ehild