1. The problem statement, all variables and given/known data Suppose [tex]sin^3xsin3x=\sum^n_{m=0}{}^nC_mcosmx[/tex] is an identity in x, where C_{0}, C_{1}, .....C_{n} are constants, and C_{n} [itex]\neq[/itex]0, then what is the value of n? 2. Relevant equations 3. The attempt at a solution I expanded the sigma notation and got:- [tex]sin^3xsin3x={}^nC_0cos0+{}^nC_1cosx+{}^nC_2cos2x.....[/tex] I wasn't able to think what should i do next? Please help!! Thanks!!
Hi Pranav-Arora! That doesn't look right … shouldn't that be [itex]sin^3xsin3x=\sum^n_{m=0}C_mcosmx[/itex] ?
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.
No, ^{n}C_{m} means the binomial coefficient n!/m!(n-m)!, with ∑^{n}C_{m}x^{m}y^{n-m} = (x+y)^{n}. Anyway, use standard trigonometric identities to write sin^{3}x and sin3x in terms of cosx cos2x cos3x etc.
Try sin^{2}x + cos^{2}x = 1 to help break-down sin^{3}x . Write sin(3x) as sin(x + 2x) and use angle addition for the sine function. Then see what the result is & proceed further.
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 [itex](\sin^3 x)[/itex] into [itex](\sin^2 x \sin x)[/itex] 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:- [tex](1-cos^2x)(sinx)(sinxcos2x+2sinxcos^2x)[/tex] Am i right now..?
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 = 2cos^{2}x - 1 and cos 2x = 1 - 2sin^{2}x formulas, or rather use them the other way around. That is, cos^{2} x = (cos 2x + 1)/2.
Should i substitute cos^{2}x = (cos 2x + 1)/2 in this:- [tex](1-cos^2x)(sinx)(sinxcos2x+2sinxcos^2x)[/tex] Or should i go from start again?
Multiplied and it resulted to be:- [tex]\frac{4cos2x-2cos^32x+3cos^22x+1}{4}[/tex] Now what should i do?
Well, isn't it starting to look more and more like your intended expression? Which is: C_{0} + C_{1} cos x + C_{2} cos 2x + C_{3} 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?
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?