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

[solve]

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.

 

[TaxOnFear]

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.

 

[solve]

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?

 

[solve]

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.

 

[rude man]

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.

 

[berkeman]

(Moderator note -- multiple threads merged)

 

[Ratch]

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

 

[solve]

Hi, Ratch.

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.

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.

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)

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.

 

[Ratch]

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

 

[solve]

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?

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.

 

[Ratch]

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

 

[solve]

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.

 

[Ratch]

solve,

Yes, those equations are well known trig identities.

Ratch

 

[solve]

Thanks, Ratch.