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

[Saitama]

Homework Statement

Suppose sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx is an identity in x, where C₀, C₁, ...C_n are constants, and C_n \neq0, then what is the value of n?

Homework Equations

The Attempt at a Solution

I expanded the sigma notation and got:-
sin^3xsin3x={}^nC_0cos0+{}^nC_1cosx+{}^nC_2cos2x...
I wasn't able to think what should i do next?
Please help!

Thanks!

 

[tiny-tim]

Hi Pranav-Arora!

Suppose sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx is an identity in x …

That doesn't look right …

shouldn't that be sin^3xsin3x=\sum^n_{m=0}C_mcosmx ?

 

[Saitama]

Hi Pranav-Arora!


That doesn't look right …

shouldn't that be sin^3xsin3x=\sum^n_{m=0}C_mcosmx ?

Yep, you're right.
In my book too, it is of the same form. I thought adding a "n" before "C" wouldn't make any difference.

 

[tiny-tim]

I thought adding a "n" before "C" wouldn't make any difference.

No, ⁿC_m means the binomial coefficient n!/m!(n-m)!, with ∑ⁿC_mx^my^n-m = (x+y)ⁿ.

Anyway, use standard trigonometric identities to write sin³x and sin3x in terms of cosx cos2x cos3x etc. ​

 

[Saitama]

No, ⁿC_m means the binomial coefficient n!/m!(n-m)!, with ∑ⁿC_mx^my^n-m = (x+y)ⁿ.

Anyway, use standard trigonometric identities to write sin³x and sin3x in terms of cosx cos2x cos3x etc. ​

Which identity should i use?

 

[SammyS]

Try sin²x + cos²x = 1 to help break-down sin³x .

Write sin(3x) as sin(x + 2x) and use angle addition for the sine function.

Then see what the result is & proceed further.

 

[Saitama]

Try sin²x + cos²x = 1 to help break-down sin³x .

Write sin(3x) as sin(x + 2x) and use angle addition for the sine function.

Then see what the result is & proceed further.

I got:-
(\sqrt{1-cos^2x})^3(sinxcos2x+sin2xcosx)

Am i right? What should i do next?

 

[I like Serena]

Hi Pranav-Arora!

Let's not go into roots and stuff.
That way the expression becomes more complex and starts looking less like a sum of cosines.

I think SammyS intended you to split (\sin^3 x) into (\sin^2 x \sin x) and only apply the squared sum formula to the first part.

Furthermore, can you break up sin(2x) further?

 

[Saitama]

Hi Pranav-Arora!

Let's not go into roots and stuff.
That way the expression becomes more complex and starts looking less like a sum of cosines.

I think SammyS intended you to split the sin³x into sin²x sin x and only apply the squared sum formula to the first part.

Furthermore, can you break up sin(2x) further?

Hi I like Serena!
I did it as you said and got:-
(1-cos^2x)(sinx)(sinxcos2x+2sinxcos^2x)
Am i right now..?

 

[I like Serena]

Yep! :)
Now get rid of the round thingies...

 

[Saitama]

Yep! :)
Now get rid of the round thingies...

How??
I still have a "cos2x".

 

[I like Serena]

How??
I still have a "cos2x".

Yes, and you want to keep that, since it matches the cosine expression you're working towards.
I meant doing stuff like a(b + c) = ab + ac

And actually, now that I think about it, the squared sum formula does not really help you forward.
What you need is the cos 2x = 2cos²x - 1 and cos 2x = 1 - 2sin²x formulas, or rather use them the other way around.
That is, cos² x = (cos 2x + 1)/2.

 

[Saitama]

Yes, and you want to keep that, since it matches the cosine expression you're working towards.
I meant doing stuff like a(b + c) = ab + ac

And actually, now that I think about it, the squared sum formula does not really help you forward.
What you need is the cos 2x = 2cos²x - 1 and cos 2x = 1 - 2sin²x formulas, or rather use them the other way around.
That is, cos² x = (cos 2x + 1)/2.

Should i substitute cos²x = (cos 2x + 1)/2 in this:-
(1-cos^2x)(sinx)(sinxcos2x+2sinxcos^2x)
Or should i go from start again?

 

[I like Serena]

At this stage it doesn't matter much.

 

[Saitama]

I tried solving it and got:-
(\frac{1-cos2x}{2})^2(2cos2x+1)
Now i am stuck.

 

[I like Serena]

Multiply the round thingies away?

 

[Saitama]

Multiply the round thingies away?

Multiplied and it resulted to be:-
\frac{4cos2x-2cos^32x+3cos^22x+1}{4}
Now what should i do?

 

[I like Serena]

Well, isn't it starting to look more and more like your intended expression?
Which is:
C₀ + C₁ cos x + C₂ cos 2x + C₃ cos 3x + ...

You need to get rid of the remaining square and third power, and try and replace them by cos mx forms...
Any thoughts on which formulas to use for that?

 

[Saitama]

Well, isn't it starting to look more and more like your intended expression?
Which is:
C₀ + C₁ cos x + C₂ cos 2x + C₃ cos 3x + ...

You need to get rid of the remaining square and third power, and try and replace them by cos mx forms...

How can i get rid of the powers?

 

[I like Serena]

A couple of posts ago you replaced a square by some cos mx form.
Do it again?

As for the third power, perhaps you need to get some inspiration from what cos 3x would look like if you reduced it to squares and other powers.
What does it look like?

 

[Saitama]

A couple of posts ago you replaced a square by some cos mx form.
Do it again?

I didn't get it!

As for the third power, perhaps you need to get some inspiration from what cos 3x would look like if you reduced it to squares and other powers.
What does it look like?

I know cos3x = 4cos³x-3cos x. But what is its use in this question?

 

[I like Serena]

I didn't get it!

You used cos²x = (cos 2x + 1)/2
So what would cos²2x be?

I know cos3x = 4cos³x-3cos x. But what is its use in this question?

That's the one. :)
Turn it around expressing the third power into cos mx thingies?
That is, what is cos³x?

 

[Saitama]

You used cos²x = (cos 2x + 1)/2
So what would cos²2x be?

cos^22x=\frac{cos4x+1}{2}
Right..?

That's the one. :)
Turn it around expressing the third power into cos mx thingies?
That is, what is cos³x?

cos^3x=\frac{cos3x+3cosx}{4}

Now what sholud i do?

 

[SammyS]

Multiplied and it resulted to be:-
\frac{4cos2x-2cos^32x+3cos^22x+1}{4}
Now what should i do?

cos²(θ) = (cos(2θ) + 1)/2 . So, cos²(2x) = ?

Use that in the obvious place and also after splitting up cos³(2x) → cos²(2x) * cos(2x) .

I would then use the product to sum identity to split up cos(4x) cos(2x).

2 cos(A) cos(B) = cos(A+B) cos(A-B)

Added in Edit:

What you just did is fine.

Now just put your results together.

 

[I like Serena]

Substitute...?

 

[Saitama]

Substitute...?

Substituting, i get:-
\frac{8cos2x-cos3x-3cosx+3cos4x+5}{8}
Right..? What's next? :)

 

[I like Serena]

Right!
What was the question asked in the problem again?

 

[I like Serena]

Btw, I just used WolframAlpha to check if your current expression is equal to the original expression in the problem, and apparently it isn't.
So I think there is a mistake somewhere.

(Sorry, but I didn't check all your steps individually. )

 

[SammyS]

Substituting, i get:-
\frac{8cos2x-cos3x-3cosx+3cos4x+5}{8}
Right..? What's next? :)

I get something different with WolframAlpha. I'm not sure where the problem is.

 

[tiny-tim]

(just got up :zzz: …)

the question doesn't ask for the actual expansion, it only asks for the value of n, ie is it up to cos3x, or to cos4x, or …?

Multiply the round thingies away?

erm … they're not round thingies, they're curvey thingies …

don't confuse people! ​

 

[Saitama]

Ok, i did it from beginning.
I have attached my attempts. Please check them and please tell me if i am wrong somewhere.
(Pardon me for my handwriting )

 

[I like Serena]

I found a mistake where you expanded (\frac {1 - \cos 2x} 2)^2.

erm … they're not round thingies, they're curvey thingies …

don't confuse people! ​

I stand corrected! That is much less confusing!

Err, but how do you distinguish it from curly thingies?
Aren't those curvey too?

 

[Saitama]

I found a mistake where you expanded (\frac {1 - cos 2x} 2)^2.

I am very sorry for my foolishness. I would again scan my attempts and post them here.

 

[Saitama]

Here are my attempts again. :)
Hope they're correct now

 

[tiny-tim]

(Pardon me for my handwriting )

I think it's rather good handwriting …

large and clear ​

(except I think the "4" should be more angular )

Err, but how do you distinguish it from curly thingies?
Aren't those curvey too?

chunky: []

curvey: ()

curly: {}

pointy: <>

 

[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 \neq0, 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?​