# Trigonometry sigma problem

1. Jul 22, 2011

### Saitama

1. The problem statement, all variables and given/known data
Suppose $$sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx$$ is an identity in x, where C0, C1, .....Cn are constants, and Cn $\neq$0, then what is the value of n?

2. Relevant equations

3. 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?

Thanks!!

2. Jul 22, 2011

### tiny-tim

Hi Pranav-Arora!
That doesn't look right

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

3. Jul 22, 2011

### Saitama

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.

4. Jul 22, 2011

### tiny-tim

No, nCm means the binomial coefficient n!/m!(n-m)!, with ∑nCmxmyn-m = (x+y)n.

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

5. Jul 22, 2011

### Saitama

Which identity should i use?

6. Jul 22, 2011

### SammyS

Staff Emeritus
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.

7. Jul 22, 2011

### Saitama

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

Am i right? What should i do next?

8. Jul 22, 2011

### 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?

9. Jul 22, 2011

### Saitama

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

10. Jul 22, 2011

### I like Serena

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

11. Jul 22, 2011

### Saitama

How??
I still have a "cos2x".

12. Jul 22, 2011

### I like Serena

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.

13. Jul 22, 2011

### Saitama

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

14. Jul 22, 2011

### I like Serena

At this stage it doesn't matter much.

15. Jul 22, 2011

### Saitama

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

16. Jul 22, 2011

### I like Serena

Multiply the round thingies away?

17. Jul 22, 2011

### Saitama

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

18. Jul 22, 2011

### 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?

19. Jul 22, 2011

### Saitama

How can i get rid of the powers?

20. Jul 22, 2011

### 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?