Linearizing Cosine: Finding Derivatives w/Moivre's

[springo]

Homework Statement

I make the product [tex]\cos(x)\cos(5x)[/tex] a sum of two cosines. I think this is called linearizing but please correct me if I'm wrong.
(I need it to find the n-th derivative of that function).

Homework Equations

The Attempt at a Solution

I know the answer is:
[tex]\frac{\cos(4x)+\cos(6x)}{2}[/tex]
- Thanks HP 50g ;) -
But how do I get to this result?
I think I should be using Moivre's formula but I get both sines and cosines in my formula.

 

[Hurkyl]

But how do I get to this result?

The most straightforward way is to apply the basic trig identities. The relevant one is on the list of ones you should, if you intend to do a lot of arithmetic with trig functions, either have memorized, or know how to derive quickly.

I think I should be using Moivre's formula but I get both sines and cosines in my formula.

Did you finish simplifying the expression you got? What did you actually get?

 

[springo]

The most straightforward way is to apply the basic trig identities. The relevant one is on the list of ones you should, if you intend to do a lot of arithmetic with trig functions, either have memorized, or know how to derive quicky.Did you finish simplifying the expression you got? What did you actually get?

Well, I did get to this with Moivre's formula:
[tex]16\cos^6 x-20\cos^4 x+5\cos^2 x[/tex]

Obviously I could expand cos(4x) and cos(6x) and be like "Oh! If you sum them and divide by 2, you get that too!" but I would never think about expanding cos(4x) and cos(6x) if I didn't know the result beforehand.

 

[Hurkyl]

What were your first couple steps?

If you did what I think you did... you might consider not expanding your cis's, preferring to keep them in factored form.


(If you haven't seen that abbreviation, the function cis is defined as [itex]cis(z) = cos(z) + i sin(z)[/itex])

 

[Hurkyl]

Well, I did get to this with Moivre's formula:
[tex]16\cos^6 x-20\cos^4 x+5\cos^2 x[/tex]

Obviously I could expand cos(4x) and cos(6x) and be like "Oh! If you sum them and divide by 2, you get that too!" but I would never think about expanding cos(4x) and cos(6x) if I didn't know the result beforehand.

Actually, you might! If you knew that the expansion of [itex]cos(nx)[/itex] has [itex]cos^n x[/itex] in it, you would consider expanding [itex]cos(6x)[/itex] to kill off the leading term, and then continue going from there.

 

[springo]

What were your first couple steps?

If you did what I think you did... you might consider not expanding your cis's, preferring to keep them in factored form.(If you haven't seen that abbreviation, the function cis is defined as [itex]cis(z) = cos(z) + i sin(z)[/itex])

What I did:
[itex]cis(z) = cis(nz) cos(nz) + i sin(nz) = [cos(z) + i sin(z)]^n[/itex]
Then take real part. Then replace: [itex]sin^2 (x) = 1- cos^2 (x)[/itex]
I don't really understand what you mean by not expanding my cis(z).

Edit: That would leave me with something to the fourth power and I would think "maybe now try cos(4x)" and voilà! Yep... But I still would like to hear the explanation for cis(z) if it's possible.

 

[Hurkyl]

I had assumed you started by doing something like

[tex]
\cos x \cos 5x = \frac{cis(x) + cis(-x)}{2} \frac{cis(5x) + cis(-5x)}{2}
[/tex]

or maybe

[tex]
\cos x \cos 5x = \mathcal{R}[ cis(x) ] \mathcal{R}[ cis(5x) ]
[/tex]

(Grrr, why can't I figure out how to typeset this right? )

 

[springo]

I had assumed you started by doing something like

[tex]
\cos x \cos 5x = \frac{cis(x) + cis(-x)}{2} \frac{cis(5x) + cis(-5x)}{2}
[/tex]

or maybe

[tex]
\cos x \cos 5x = \mathcal{R}[ cis(x) ] \mathcal{R}[ cis(5x) ]
[/tex]

(Grrr, why can't I figure out how to typeset this right? )

It was the latter. The former uses formulae I have never encountered.

 

[Hurkyl]

It was the latter. The former uses formulae I have never encountered.

Do you know how to write the "real part" function in terms of complex conjugation?

 

[springo]

Do you know how to write the "real part" function in terms of complex conjugation?

[tex]\Re (z) = \frac{z + \overline{z}}{2}[/tex]
Right?

 

[Hurkyl]

[tex]\Re (z) = \frac{z + \overline{z}}{2}[/tex]
Right?

Right. That's where the other formula I wrote came from.

(Incidentally, similar ideas can be used to kill off unwanted terms in various trig identities...)