Equations with sine and cosine.

In summary, the conversation discusses solving for |θ1-θ2| using the equations cosθ1 + cosθ2 = 0 and sinθ1 + sinθ2 = 0. It is suggested to graphically solve the problem by sketching the sin(theta) and cos(theta) graphs on the same horizontal axis. It is also mentioned that setting the equations equal to each other and using trig identities can lead to the correct answer. However, there is some confusion about the proper use of trig identities and whether or not it is possible to single out components to solve for the remaining variables. Ultimately, the solution is found by setting cosθ1 and cosθ2 equal to each other and solving for the values that make them
  • #1
LogicX
181
1

Homework Statement



Solve for |θ12|

Homework Equations



cosθ1 + cosθ2 = 0
sinθ1 + sinθ2 = 0

The Attempt at a Solution



This is a silly math problem within a larger question I'm working on. I have solved for it multiple times now using different trig identities and I get different answers.

First of all, can I set them equal to each other because they both equal 0, or do I add them both together? The latter gives an extra negative sign.

Can I solve this just using one equation? i.e.:

cosθ1 + cosθ2 = 0

=2cos((θ12)/2) cos((θ12)/2)= 0

Then can I just divide both sides by 2cos((θ12)/2) and then solve for |θ12| or is that not proper math?

I've spent way too much time on this high school level question...
 
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  • #2
You can set them equal to each other as they are both equal to 0 (it works as well with every other number, variabele, and the like). Moreover your method should work fine, though you can only divide by 2cos((θ1+θ2)/2) as the cos is not equal to 0 (So you should find more solutions).

Though, because it is quite nasty when there are more variables in one sinus or cosinus, I would advise you to put first all θ_1 in one side of the equation and θ_2 on the other:

Then you will get this:
<< Solution deleted by Moderator >>
 
Last edited by a moderator:
  • #3
BishopPair said:
You can set them equal to each other as they are both equal to 0 (it works as well with every other number, variabele, and the like). Moreover your method should work fine, though you can only divide by 2cos((θ1+θ2)/2) as the cos is not equal to 0 (So you should find more solutions).

Though, because it is quite nasty when there are more variables in one sinus or cosinus, I would advise you to put first all θ_1 in one side of the equation and θ_2 on the other:

Then you will get this:
<< Solution deleted by Moderator >>

Welcome to the PF, BishopPair. Please remember the part of the Rules link at the top of the page that says the OP has to do the bulk of the work on schoolwork-type questions.
 
  • #4
LogicX said:

Homework Statement



Solve for |θ12|

Homework Equations



cosθ1 + cosθ2 = 0
sinθ1 + sinθ2 = 0

The Attempt at a Solution



This is a silly math problem within a larger question I'm working on. I have solved for it multiple times now using different trig identities and I get different answers.

First of all, can I set them equal to each other because they both equal 0, or do I add them both together? The latter gives an extra negative sign.

Can I solve this just using one equation? i.e.:

cosθ1 + cosθ2 = 0

=2cos((θ12)/2) cos((θ12)/2)= 0

Then can I just divide both sides by 2cos((θ12)/2) and then solve for |θ12| or is that not proper math?

I've spent way too much time on this high school level question...

The easiest way to solve this is graphically. Sketch the sin(theta) and cos(theta) graphs on the same horizontal axis. What difference |θ1-θ2| will always give you 0 for the two equations?

BishopPair said:
You can set them equal to each other as they are both equal to 0 (it works as well with every other number, variabele, and the like). Moreover your method should work fine, though you can only divide by 2cos((θ1+θ2)/2) as the cos is not equal to 0 (So you should find more solutions).

Though, because it is quite nasty when there are more variables in one sinus or cosinus, I would advise you to put first all θ_1 in one side of the equation and θ_2 on the other:

Then you will get this:
<< Solution deleted by Moderator >>

I don't think you can set them equal, because the two equations have "+" signs. The two quantities have to be the opposite of each other.
 
  • #5
BishopPair said:
You can set them equal to each other as they are both equal to 0 (it works as well with every other number, variabele, and the like). Moreover your method should work fine, though you can only divide by 2cos((θ1+θ2)/2) as the cos is not equal to 0 (So you should find more solutions).

Though, because it is quite nasty when there are more variables in one sinus or cosinus, I would advise you to put first all θ_1 in one side of the equation and θ_2 on the other:

cosθ1 + sinθ1 = -cos θ2 - sinθ2

There is no identity to go from here.

EDIT:

The easiest way to solve this is graphically. Sketch the sin(theta) and cos(theta) graphs on the same horizontal axis. What difference |θ1-θ2| will always give you 0 for the two equations?

But where did you get that the difference equals zero? If the difference equals zero, then that is the answer and I am done. I don't need actual numbers for theta, I just need an expression so that I can pick an arbitrary value for either of them and fix the other value.

I just used the result of θ12= 0 and I got what I know to be the correct answer for the rest of the problem. So how did you come up with that, using equations?

EDIT 2: I just used these trig identities to do it, and the method I outlined above.

I'm also an idiot and the problem is supposed to be cosθ1 - cosθ2= 0 and so on.

Using the above:

cosθ1 - cosθ2= 0

(sum to product identity)
-2sin((θ1+θ2)/2) sin((θ1-θ2)/2)= 0

sin((θ1-θ2)/2)= 0 <----- this is where I'm worried about my math. Can you just single out anyone component and get rid of it because you have a zero on the right, and then solve for what remains in the parenthases? If I get rid of the other one and solve for θ1+θ2 that also = 0 So I get conflicting answers.

(θ1-θ2)/2 = 0

(θ1-θ2) = 0

EDIT 3: Ok, this should be simple.

cosθ1 - cosθ2= 0
cosθ1 = cosθ2

Which only happens when θ1 and θ2 are equal, or negative of each other. So there are two answers. I only get the correct result in the rest of the problem when they are equal though.

Aha! But I also have sinθ1 - sinθ2= 0

sinθ1 = sinθ2

Which only happens when they are equal. Thus eliminating one solution, and θ2 and θ1 must be equal, and:

θ1-θ2=0
 
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  • #6
IF the angles are the same, then your equations are not always satisfied:

cosθ1 + cosθ2 = 0
sinθ1 + sinθ2 = 0

But there does exist a difference between the angles that does satisfy those two equations...
 
  • #7
berkeman said:
IF the angles are the same, then your equations are not always satisfied:

When? When does cosθ=/=cosθ, or sinθ=/=sinθ?

Oh sorry, maybe you missed that I updated the problem, it's actually supposed to be a minus sign instead of a plus sign, so these are the two equations: (totally my fault)

cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

But there does exist a difference between the angles that does satisfy those two equations...

For the correct above equations the two angles must be equal.

Believe it or not I passed high school trigonometry about 6 years ago... thanks for the help.
 
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  • #8
LogicX said:
cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

Well, that certainly makes it easier!

I was able to solve for [itex]\theta_1 + \theta_2[/itex] by using sum-to-product trig formulas and then using the symmetry of the unit circle to solve. Are you sure you weren't supposed to solve for [itex]\theta_1 + \theta_2[/itex]? At the moment I don't see how to solve for [itex]|\theta_1 - \theta_2|[/itex]
 
  • #9
cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

Just think about when this is true within a single period. You can then add any multiple of the period to your answers. You could think about magnitude and sign separately too. Think "if theta 1 is in quadrant 1, in what quadrant must theta 2 be?". Do this for the first equation then the second equation. Then, use whatever quadrants are in both since both equations must be satisfied. After you isolate the quadrant(s), you can think in terms of magnitudes within that quadrant.
 
  • #10
scurty said:
Well, that certainly makes it easier!

I was able to solve for [itex]\theta_1 + \theta_2[/itex] by using sum-to-product trig formulas and then using the symmetry of the unit circle to solve. Are you sure you weren't supposed to solve for [itex]\theta_1 + \theta_2[/itex]? At the moment I don't see how to solve for [itex]|\theta_1 - \theta_2|[/itex]

cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

Therefore:

cosθ1 = cosθ2
sinθ1 = sinθ2

This only happens when θ2 and θ1 are equal. Really, you could get this from only the second equation if you wanted to since there is only once solution.

θ21
So θ12=0
 
  • #11
LogicX said:
cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

Therefore:

cosθ1 = cosθ2
sinθ1 = sinθ2

This only happens when θ2 and θ1 are equal. Really, you could get this from only the second equation if you wanted to since there is only once solution.

θ21
So θ12=0

No. As for your claim that the angles must be equal, consider that sine and cosine are periodic. As for your other claim that you could arrive to this solution from only the first equation, consider that
[tex]cos(a) = cos(-a)[/tex]
 
  • #12
LogicX said:
cosθ1 - cosθ2 = 0
sinθ1 - sinθ2 = 0

Therefore:

cosθ1 = cosθ2
sinθ1 = sinθ2

This only happens when θ2 and θ1 are equal. Really, you could get this from only the second equation if you wanted to since there is only once solution.
How do you explain the following ?

sin(π/3) = sin(2π/3) = sin(7π/3) = ...
 
  • #13
RoshanBBQ said:
No.

Not helpful. When exactly does sinθ=/=sinθ?

I may as well be asking when does 2=/=2.

As for your claim that the angles must be equal, consider that sine and cosine are periodic. As for your other claim that you could arrive to this solution from only the first equation, consider that
[tex]cos(a) = cos(-a)[/tex]

Except that I said only the second equation. You get two possibilities if you only use the first equation with cosine. But sinθ1 only equals sin θ2 if they are both equal. Not if they are negatives of each other. So the second equation implies they are equal.

SammyS said:
How do you explain the following ?

sin(π/3) = sin(2π/3) = sin(7π/3) = ...

So, what are you saying, I need to add in a factor of some number n χ ∏ ?

Once again, this is irrelevant trig that is NOT THE FOCUS of an upper level science course. I need to MOVE ON ALREADY. I don't have more time to spend on this silly question.
 
  • #14
LogicX said:
Not helpful. When exactly does sinθ=/=sinθ?

I may as well be asking when does 2=/=2.



Except that I said only the second equation. You get two possibilities if you only use the first equation with cosine. But sinθ1 only equals sin θ2 if they are both equal. Not if they are negatives of each other. So the second equation implies they are equal.



So, what are you saying, I need to add in a factor of some number n χ ∏ ?

Once again, this is irrelevant trig that is NOT THE FOCUS of an upper level science course. I need to MOVE ON ALREADY. I don't have more time to spend on this silly question.
Your statement was false also for the use of the second equation. Consider that
[tex]sin(a) = sin(\pi -a)[/tex]
 
  • #15
RoshanBBQ said:
Your statement was false also for the use of the second equation. Consider that
[tex]sin(a) = sin(\pi -a)[/tex]

I get it, sine is a periodic function.

What does this tell me about θ1- θ2?

I will not be tested on this information. It is just the means to end of a homework problem. I don't need to learn trig, I need to move on. I have spent way too much time thinking about elementary trig that is not going to help me in this upper level science class.

At this point I'm just assuming that the only answer I'm going to get is my "wrong" answer(which gave me the right result) so I'm probably going to just have to go with that. I have exams coming up; the ins and outs of trig are not really my focus right now.

EDIT: I don't mean to sound angry I'm just getting kind of hyper that I have this huge assignment waiting to be done and I can't finish it because of what should be a simple problem, and now we are 15 posts in and I still don't have the answer.
 
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  • #16
LogicX said:
I get it, sine is a periodic function.

What does this tell me about θ1- θ2?

I will not be tested on this information. It is just the means to end of a homework problem. I don't need to learn trig, I need to move on. I have spent way too much time thinking about elementary trig that is not going to help me in this upper level science class.

At this point I'm just assuming that the only answer I'm going to get is my "wrong" answer(which gave me the right result) so I'm probably going to just have to go with that. I have exams coming up; the ins and outs of trig are not really my focus right now.

EDIT: I don't mean to sound angry I'm just getting kind of hyper that I have this huge assignment waiting to be done and I can't finish it because of what should be a simple problem, and now we are 15 posts in and I still don't have the answer.
The function is not periodic by a factor of pi. It is periodic by a factor of 2pi. You seem to be ignoring everything I am saying except for the short portion right before I elaborate where I say "this is wrong."

Take a look at these two pictures. When you draw one of these triangles for any given angle, you always have a portion of the triangle that lies on the x-axis and one that lies on the y-axis. The cosine of an angle gives you the portion along the x-axis. The sine gives you the portion along the y-axis. If your angle is in quadrants I or II, your y is positive, so your sine is positive. This is angles 0 + k2pi to pi +k2pi where k is an integer. Else your sign is negative. The same logic applies to cosine. If you are in quadrant I or IV, your x-axis portion is on the positive x-axis and your cosine is positive. The reason these functions are periodic is if you have an angle B (which we can assume is 45 degrees for concreteness), and you add 2pi to it, you wrapped around the entire circle back to 45 degrees in quadrant I. This line of thinking works for any multiple of 2pi. If you have 6 pi, for example, added to B, you wrapped around 3 times and ended up at B.

Can you tell me in which quadrant(s) angle 2 can be, in general, using what I've said and these pictures, if angle 1 is in quadrant I, II, III, or IV?
 

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  • #17
I know you are trying really hard but I am having trouble following you.

RoshanBBQ said:
The function is not periodic by a factor of pi. It is periodic by a factor of 2pi.

Yes.

Take a look at these two pictures. When you draw one of these triangles for any given angle, you always have a portion of the triangle that lies on the x-axis and one that lies on the y-axis. The cosine of an angle gives you the portion along the x-axis. The sine gives you the portion along the y-axis. If your angle is in quadrants I or II, your y is positive, so your sine is positive.

Yes

This is angles 0 + k2pi to pi +k2pi where k is an integer. Else your sign is negative.

Yes

The same logic applies to cosine. If you are in quadrant I or IV, your x-axis portion is on the positive x-axis and your cosine is positive.

Yes

The reason these functions are periodic is if you have an angle B (which we can assume is 45 degrees for concreteness), and you add 2pi to it, you wrapped around the entire circle back to 45 degrees in quadrant I. This line of thinking works for any multiple of 2pi. If you have 6 pi, for example, added to B, you wrapped around 3 times and ended up at B.

Yes

Can you tell me in which quadrant(s) angle 2 can be, in general, using what I've said and these pictures, if angle 1 is in quadrant I, II, III, or IV?

If angle 1 is in the first quadrant then sine can be in quadrants one or two. Cosine of angle 1 can be in quadrants one and four. The overlap happens in quadrant one.

If angle 1 is in the third quadrant then sine of angle one can be in quadrants 3 or 4. Cosine of angle 1 can be in quadrants 2 or 3. The overlap happens in quadrant 3.

And so on. The overlap happens in whatever quadrant the angle is in.Ok so you're saying if angle 1 is in quadrant 1 and cos(1)=cos(2) then the possible quadrants that angle 2 can be in is 1 and 4. If angle 1 is in quadrant 1 and sin(1)=sin(2) then the possible quadrants angle 2 can be in is 1 and 2. The overlap is only in quadrant 1, so angle 2 must also be in quadrant 1.

Now what?
 
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  • #18
LogicX said:
I know you are trying really hard but I am having trouble following you.



Yes.



Yes



Yes



Yes



Yes



If angle 1 is in the first quadrant then sine can be in quadrants one or two. Cosine of angle 1 can be in quadrants one and four. The overlap happens in quadrant one.

If angle 1 is in the third quadrant then sine of angle one can be in quadrants 3 or 4. Cosine of angle 1 can be in quadrants 2 or 3. The overlap happens in quadrant 3.

And so on. The overlap happens in whatever quadrant the angle is in.


Ok so you're saying if angle 1 is in quadrant 1 and cos(1)=cos(2) then the possible quadrants that angle 2 can be in is 1 and 4. If angle 1 is in quadrant 1 and sin(1)=sin(2) then the possible quadrants angle 2 can be in is 1 and 2. The overlap is only in quadrant 1, so angle 2 must also be in quadrant 1.

Now what?
Well, taking both equations together (both conditions), you have made the observation that angle 2 must be in whatever quadrant angle 1 is in. We have two steps to go from here. Next, we need to think magnitude. What angle must angle 2 be (which is now nicely confined to the same quadrant as angle 1) so that the cosines and sines have the same magnitude? Just ignore periodicity and think from 0 to 360 degrees. The final leap will be to take whatever answer you find for that question and say "But it can be true for any k2pi where k is an integer added to it."
 
  • #19
RoshanBBQ said:
Well, taking both equations together (both conditions), you have made the observation that angle 2 must be in whatever quadrant angle 1 is in. We have two steps to go from here. Next, we need to think magnitude. What angle must angle 2 be (which is now nicely confined to the same quadrant as angle 1) so that the cosines and sines have the same magnitude? Just ignore periodicity and think from 0 to 360 degrees. The final leap will be to take whatever answer you find for that question and say "But it can be true for any k2pi where k is an integer added to it."

sine and cosine are equal at 45, 135, 225, 315, i.e. start at 45 and add 90 to get to the next quadrant and so on.

How do I go from here to what angle 1 minus angle 2 is? I don't get what you are trying to prove. If angle 1 is 45, we have determined that angle 2 must be in the same quadrant. And sin and cosine are only equal at 45. So the angles equal each other, or rather they equal each other and either one could also have an integer +k2pi tacked on.

EDIT: At this point my final answer is θ1-θ2=k2pi where k is an integer.
 
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  • #20
LogicX said:
sine and cosine are equal at 45, 135, 225, 315, i.e. start at 45 and add 90 to get to the next quadrant and so on.

How do I go from here to what angle 1 minus angle 2 is? I don't get what you are trying to prove. If angle 1 is 45, we have determined that angle 2 must be in the same quadrant. And sin and cosine are only equal at 45. So the angles equal each other, or rather they equal each other and either one could also have an integer +k2pi tacked on.

EDIT: At this point my final answer is θ1-θ2=k2pi where k is an integer.

Yeah, that looks right to me.
 

Related to Equations with sine and cosine.

1. What is the difference between sine and cosine?

Sine and cosine are both trigonometric functions that are used to calculate the relationship between the sides and angles of a right triangle. The main difference between the two is that sine represents the ratio of the opposite side to the hypotenuse, while cosine represents the ratio of the adjacent side to the hypotenuse.

2. How do you solve equations with sine and cosine?

To solve equations with sine and cosine, you can use the basic trigonometric identities, such as the Pythagorean identity and the double angle formulas. You can also use the unit circle and inverse trigonometric functions to simplify the equations and find the solutions.

3. Can you use the sine and cosine functions to find the missing side or angle of a triangle?

Yes, you can use the sine and cosine functions to find the missing side or angle of a triangle, as long as you have enough information about the other sides and angles of the triangle. This is known as using the trigonometric ratios to solve for unknown values.

4. Why are sine and cosine important in science and engineering?

Sine and cosine are important in science and engineering because they are used to model and analyze the motion of waves, such as sound and light waves. They are also used in many physical and mathematical applications, such as signal processing, navigation, and electrical engineering.

5. How do you graph equations with sine and cosine?

To graph equations with sine and cosine, you can use the unit circle and the properties of the sine and cosine functions. You can also use transformations, such as amplitude, period, and phase shift, to graph more complex equations. Graphing calculators are also useful tools for visualizing and analyzing these equations.

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