Difference in signs OF PHASES of -7cos(5x-3) and -cot(3x-4)?

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Discussion Overview

The discussion centers on the differences in the signs of the phases of the functions -7cos(5x-3) and -cot(3x-4). Participants explore the implications of these phase differences, the definitions of phase, and the effects of negating functions on their phases. The scope includes mathematical reasoning and conceptual clarification related to trigonometric functions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants state that the phase of -7cos(5x-3) is -3/5, while others argue it should be considered as a phase lag of 3/5 radians.
  • Participants note that the phase of -cot(3x-4) is 4/3, with some suggesting that it should be -4/3 due to its odd function nature.
  • There is a discussion about whether the phase difference should be derived from f(x) or f(-x), with some asserting that it makes no difference in the values obtained.
  • One participant emphasizes that the convention is that a positive phase indicates a lagging function, while a negative phase indicates a leading function, prompting further clarification on this point.
  • Some participants express uncertainty about how the negative sign in the cotangent function influences the phase, leading to additional questions about phase definitions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the phases and their signs. Multiple competing views remain regarding the correct representation of the phases for both functions.

Contextual Notes

There are unresolved questions about the definitions of phase difference and how negating functions affects their phases. Participants also discuss the implications of using different forms of the functions to derive phase differences.

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Homework Statement


The phase of -7cos(5x-3) is -3/5

The phase of -cot(3x-4) is 4/3


Homework Equations


3. The Attempt at a Solution
[/B]
Why the difference in the signs of phases? I think the phase of -cot(3x-4) should be -4/3. Its an odd function: -cot(3x-4)=cot(-3x+4) and even then its phase looks to be -4/3.

Also, would -7cos(5x-3) equal -7cos(-5x+3) because it's an even function?

Thanks.
 
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Try it yourself. Plug some values into a cosine function and then change the function to negative with the same values, see what you get.
 
TaxOnFear said:
Try it yourself. Plug some values into a cosine function and then change the function to negative with the same values, see what you get.

I did. cos(x)=cos(-x) holds true.

What about phases though?
 
Please, help me determine phase difference

Hi, All.

Homework Statement



1. f(x)= -7cos(5x-3)= -7cos5(x-3/5)

cosx is an even function. f(x)=f(-x)

phase difference here is -3/5

2. g(x)=- cot(3x-4)= cot(-3x+4)=cot3(-x+4/3)

cotx is an odd function g(-x)=-g(x)

phase difference is 4/3


Homework Equations


The Attempt at a Solution



I have couple questions:

1. Why did we derive the phase difference from f(x) and not from f(-x) in the first example ?
2. Why did we derive the phase difference from f(-x) and not from -f(x) in the second example ?
3. Is f(x) is preferable to work with to f(-x) and f(-x) to -f(x)?

Thanks.
 
Last edited by a moderator:


It just depends on definition of phase "difference."

If I say A-B is the difference between A and B I would be correct.
And if I say B-A is the difference between A and B I would still be correct. "Difference" by itself has no polarity.

BUT: sin(x-ψ) represents a waveform with lagging phase ψ.
If x = ωt and t is time, then you can see that sin(x-ψ) lags sin(x) by phase angle ψ or equivalently by time τ = ψ/ω.
 
(Moderator note -- multiple threads merged)
 
solve,
The phase of -7cos(5x-3) is -3/5

No, it is -7cos(5(x-(3/5)), which means that -7cos(5x-3) has a phase lag of 3/5 radians when compared to -7cos(5x).

The phase of is 4/3

Yes, it is -cot(3(x-(4/3)), which means that -cot(3x-4) has a phase lag of 4/3 radians when compared to -cot(3x). Note that this phase lag is greater than the period of -cot(3x), which is ∏/3.

Why the difference in the signs of phases?

What difference? They both show positive values, which means they both lag in phase.

I think the phase of -cot(3x-4) should be -4/3.

That would be wrong. -cot(3(x-(-4/3)) is not the equation given.

Its an odd function: -cot(3x-4)=cot(-3x+4) and even then its phase looks to be -4/3.

Its phase is 4/3 radians and it is lagging.

I did. cos(x)=cos(-x) holds true.

What about phases though?

Phase is zero for either representation.

1. f(x)= -7cos(5x-3)= -7cos5(x-3/5)

cosx is an even function. f(x)=f(-x)

phase difference here is -3/5

Which equation with respect to what equation?

2. g(x)=- cot(3x-4)= cot(-3x+4)=cot3(-x+4/3)

cotx is an odd function g(-x)=-g(x)

phase difference is 4/3

Which equation with respect to what equation?

1. Why did we derive the phase difference from f(x) and not from f(-x) in the first example ?

You could have done so, the values would not change.

2. Why did we derive the phase difference from f(-x) and not from -f(x) in the second example ?

You could have done so, the values would not change.

3. Is f(x) is preferable to work with to f(-x) and f(-x) to -f(x)?

It makes no difference, the values will still be the same.

Ratch
 
Hi, Ratch.

Ratch said:
solve,

No, it is -7cos(5(x-(3/5)), which means that -7cos(5x-3) has a phase lag of 3/5 radians when compared to -7cos(5x).
Ratch

The logic here: -7cos(5x-3)= -7cos5(x-3/5), so the phase is -3/5.

Ratch said:
solve,Yes, it is -cot(3(x-(4/3)), which means that -cot(3x-4) has a phase lag of 4/3 radians when compared to -cot(3x). Note that this phase lag is greater than the period of -cot(3x), which is ∏/3.Ratch

The logic: -cot(3x-4)= cot(-3x+4)= cot3(-x+4), so the phase looks to be 4/3, but the phase of -cot(3x-4) looks like -4/3, except I am not sure how negative cot influences the phase.
Ratch said:
solve,

Which equation with respect to what equation?
Which equation with respect to what equation?

Ratch

Not equations, but just functions:

f(x)= -7cos(5x-3)= f(-x)=-7cos5(x-3/5)

-g(x)=- cot(3x-4)= g(-x)=cot3(-x+4/3)
Ratch said:
solve,

You could have done so, the values would not change.
You could have done so, the values would not change.
It makes no difference, the values will still be the same.

Ratch

I don't know what's up with me, but asking that bunch of questions is just stupid since, for example, f(x)=f(-x) in relation to an example given above, so "it makes no difference, the values will still be the same." Thanks for that. Except, again, -cot3(x-4/3) seems to have -4/3 as its phase.
 
Last edited:
solve,

No, it is -7cos(5(x-(3/5)), which means that -7cos(5x-3) has a phase lag of 3/5 radians when compared to -7cos(5x).
Ratch

The logic here: -7cos(5x-3)= -7cos5(x-3/5), so the phase is -3/5 radians.

The convention is that a positive phase means the given term is lagging. If you say the phase is minus, then it is leading. That is not correct. See plot below.

The logic: -cot(3x-4)= cot(-3x+4)= cot3(-x+4), so the phase looks to be 4/3, but the phase of -cot(3x-4) looks like -4/3, except I am not sure how negative cot influences the phase.

cot(-3x+4)= cot(-3(x-(4/3)) , so yes the phase is 4/3 radians. A negative wave has the same phase as a positive one does.

Not equations, but just functions:

Which are defined by equations.

f(x)= -7cos(5x-3)= f(-x)=-7cos5(x-3/5)

-g(x)=- cot(3x-4)= g(-x)=cot3(-x+4/3)

What are those equations supposed to signify?

... Except, again, -cot3(x-4/3) seems to have -4/3 as its phase.
Y 09:44 PM

Put in this form, -cot3(x-4/3)==cot(3(x-(4/3)) shows it has a positive or lagging phase. See plot.

Ratch
 

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  • #10
Ratch said:
solve,

The convention is that a positive phase means the given term is lagging. If you say the phase is minus, then it is leading. That is not correct. See plot below. Ratch

I don't think I understand you correctly. For instance, y=sin(x+ pi/4) leads y=sinx by a phase of pi/4. y=sin(x- pi/4) lags behind y=sinx with a phase difference of pi/4. Would that be correct?

Ratch said:
What are those equations supposed to signify?

They have to do with the questions about the preference of use between -f(x) and f(-x) i.e -cot(3x-4) or cot(-3x+4) to find the phases. You already answered that question.

Thanks.
 
  • #11
solve,

I don't think I understand you correctly. For instance, y=sin(x+ pi/4) leads y=sinx by a phase of pi/4. y=sin(x- pi/4) lags behind y=sinx with a phase difference of pi/4. Would that be correct?

Yes, that is correct. y=sin(x-(-π/4)), shows a -π/4 (minus), which means leading. y=sin(x-(π/4)) shows a π/4 (positive), which is lagging. Isn't that what I averred? See link below.

http://www.regentsprep.org/Regents/math/algtrig/ATT7/phaseshift.htm

Ratch
 
  • #12
Ratch said:
solve,

Yes, that is correct. y=sin(x-(-π/4)), shows a -π/4 (minus), which means leading. y=sin(x-(π/4)) shows a π/4 (positive), which is lagging. Isn't that what I averred? See link below.

http://www.regentsprep.org/Regents/math/algtrig/ATT7/phaseshift.htm

Ratch

Also, is it true that cos(pi/2- x)=cos(x- pi/2)= sinx ?

Thanks, Ratch.
 
  • #13
solve,

Yes, those equations are well known trig identities.

Ratch
 
  • #14
Thanks, Ratch.
 

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