Multiplying Cosines: Learn How to Represent x(t)

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Homework Help Overview

The discussion revolves around the mathematical representation of the product of two cosine functions, specifically how to express cos(ω1t) times cos(ω2t) using trigonometric identities. The subject area is trigonometry within the context of physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of trigonometric identities, particularly the Product to Sum Identity, to transform the product of cosines into a sum. Questions arise regarding the presence of sine components in the identities and their relevance to the problem at hand.

Discussion Status

There is an ongoing exploration of the identities involved, with some participants suggesting specific steps to take, such as adding the cosine identities to eliminate sine components. The original poster expresses uncertainty about the relevance of sine terms and appears to be moving towards a resolution.

Contextual Notes

Participants express concerns about clarity and understanding of the identities being discussed, indicating a need for further exploration of the assumptions underlying the problem.

SMOF
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Hello,

I hope everyone is well.

Also, I hope someone can help me understand something. I am trying to understand how cos(ω1t) times cos(ω2t) can be represented as

x(t) 1/2 (cos(ω2 - ω1)t) + 1/2 (cos(ω2 + ω1)t)

Thanks in advance for any help.

Seán
 
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Try expanding cos(A+B) and cos(A-B) :wink:
 
Hello,

Thanks for the reply.

So, cos(A+B) = cosA cosB - sinA sinB, and cos(A-B) = cosA cosB + sinA sinB.

But I don't see how this gets me to the solution I need, since it doesn't have a sin component.

But maybe I am just being stupid here.

Seán
 
Oh, I may have got it with the Product to Sum Identity.

Seán
 
Sorry about the late reply, I am on holidays.
Ok notice that the cos(A+B) has a sinAsinB component and cos(A-B) has a -sinAsinB component, so why not add these two cosines sums to get rid of the sinAsinB?
 
Hello.

That will just leave me with 2(cosAcosB) ...yea?

Seán
 
SMOF said:
Hello,

Thanks for the reply.

So, cos(A+B) = cosA cosB - sinA sinB, and cos(A-B) = cosA cosB + sinA sinB.

But I don't see how this gets me to the solution I need, since it doesn't have a sin component.

But maybe I am just being stupid here.

Seán
So add them!
 
SMOF said:
Hello.

That will just leave me with 2(cosAcosB) ...yea?

Seán

Right, and you want cosAcosB, so...
 
Hello.

Yea, so, I divide by 2 ...or, multiply by a half ...I think I have it from here.

Many thanks for your help :)

Seán
 
  • #10
SMOF said:
Hello.

Yea, so, I divide by 2 ...or, multiply by a half ...I think I have it from here.

Many thanks for your help :)

Seán

Yep that's it :smile:

You're welcome, and good luck with your physics!
 

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