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! ​