Multiplying Cosines: Learn How to Represent x(t)

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The discussion centers on representing the product of cos(ω1t) and cos(ω2t) using trigonometric identities. The key realization comes from applying the Product to Sum Identity, which allows the expression to be rewritten as x(t) = 1/2 (cos(ω2 - ω1)t) + 1/2 (cos(ω2 + ω1)t). Participants clarify the steps involving the expansion of cos(A+B) and cos(A-B) to eliminate sine components. Ultimately, the solution is confirmed, and the original poster expresses gratitude for the assistance. Understanding these identities is crucial for manipulating cosine functions in physics applications.
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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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