Solve cos(a+θ)=0: Exploring Solutions & Understanding Results

  • Thread starter PainterGuy
  • Start date
In summary, The solutions to the equation y=cos(α+θ) where α is a constant and θ has range [0,2π) are given by θ1=cos−1(y)−α and θ2=2π−(cos−1(y))−α. However, it is important to note that the inverse cosine function is not a one-to-one function, meaning that for every value of y there are two possible values of θ that satisfy the equation. This is due to the symmetry of the cosine function about θ=π. To find a specific value for θ, the floor function can be used to reduce the solution to a value within the range of [0,2π
  • #1
PainterGuy
940
69
Homework Statement
How do I solve cos(a+θ)=0?
Relevant Equations
Please check the posting.
Hi,

I was trying to solve y=cos(α+θ) where α is a constant and θ has range [0,2π).

If α=3 is assumed, the solution is as shown below and it makes sense

[itex]\theta_{1}=\frac{3}{2}(\pi-2)[/itex]
[itex]\theta_{2}=\frac{1}{2}(5 \pi-6)[/itex]

But if I don't use any specific value for α, the solution doesn't make sense to me.

[itex]\pi n-\frac{5 \pi}{2} \leq \alpha \leq \pi n-\frac{\pi}{2}[/itex] and [itex]\theta=\pi\left(n-\frac{1}{2}\right)-\alpha[/itex] and [itex]n \in \mathbb{Z}[/itex]

Source: https://www.wolframalpha.com/input/?i=solve+cos(α+θ)=0+for+θ+from+0+to+2pi

Somewhere I found the solution to be as shown below but don't understand why [itex]\cos ^{-1} y[/itex] is used, and please note that α is not given any specific value:

[itex]\theta_{1}=\left(\cos ^{-1} \mathrm{y}\right)-\alpha[/itex]
[itex]\theta_{2}=2 \pi-\left(\cos ^{-1} \mathrm{y}\right)-\alpha[/itex]

Could you please help me? It's not homework but I still thought I better post it here nonetheless.
 
Last edited:
Physics news on Phys.org
  • #2
[tex]\alpha+\theta=\pi(n+\frac{1}{2})[/tex]
[tex]\theta=\pi(n+\frac{1}{2})-\alpha[/tex]
integer n should be chosen properly so that
[tex]0 < \theta < 2\pi[/tex]
You may make use of floor function to show solutions explicitly.
 
Last edited:
  • Like
Likes PainterGuy
  • #3
PainterGuy said:
Somewhere I found the solution to be as shown below but don't understand why [itex]\cos ^{-1} y[/itex] is used, and please note that α is not given any specific value:

[itex]\theta_{1}=\left(\cos ^{-1} \mathrm{y}\right)-\alpha[/itex]
[itex]\theta_{2}=2 \pi-\left(\cos ^{-1} \mathrm{y}\right)-\alpha[/itex]
What happens if you add ##\alpha## to those solutions for ##\theta##?

Do you know how the function ##\cos^{-1}## is defined?
 
  • Like
Likes PainterGuy
  • #4
PeroK said:
What happens if you add ##\alpha## to those solutions for ##\theta##?

Do you know how the function ##\cos^{-1}## is defined?
Yes, I'd say I know how arccosine is defined. Cosine of an angle, cos(angle), gives you the ratio of base/hypotenuse and arccosine of the ratio, arccosine(base/hypotenuse), gives the corresponding angle.
 
  • #5
PainterGuy said:
Yes, I'd say I know how arccosine is defined. Cosine of an angle, cos(angle), gives you the ratio of base/hypotenuse and arccosine of the ratio, arccosine(base/hypotenuse), gives the corresponding angle.
At an elementary level, the cosine is defined in terms of a right-angled triangle, but a question like this demands a more general definition.

More generally, ##\cos \theta## it is the x-coordinate on the unit circle corresponding to an angle ##\theta## as measured anti-clockwise from the positive x-axis.

You've missed the key point that the cosine is not a one-to-one function. E.g. ##\cos \theta = \cos (-\theta)##. It is not necessarily the case, therefore, that ##\cos^{-1}(cos \theta) = \theta##

This means you need a slightly more sophisticated definition of the inverse cosine. I'll let you look that up.
 
  • Like
Likes PainterGuy
  • #6
PeroK said:
This means you need a slightly more sophisticated definition of the inverse cosine. I'll let you look that up.

Thank you for pointing this out.

On a unit circle the cosine is the length of the adjacent side (which is the x-coordinate) divided by the length of the hypotenuse (which is 1). So, as you said, the cosine is just the x-coordinate.

The domain of the inverse cosine function is [−1,1] and the range is [0,π]. That means a positive value will yield a 1st quadrant angle and a negative value will yield a 2nd quadrant angle.Helpful links:
https://www.mathwords.com/c/cosine_inverse.htm
/watch?v=ZEAgtXTIxk0 (add youtube.com in front)
/watch?v=xEGZsWgCn-Q
https://www.khanacademy.org/math/al...0c9fb89:unit-circle/a/trig-unit-circle-review
 
  • #7
anuttarasammyak said:
[tex]\alpha+\theta=\pi(n+\frac{1}{2})[/tex]
[tex]\theta=\pi(n+\frac{1}{2})-\alpha[/tex]
integer n should be chosen properly so that
[tex]0 < \theta < 2\pi[/tex]
You may make use of floor function to show solutions explicitly.

Thanks!

But it also involves α in addition to n; two variables.

Also, I don't see how the floor function could be used here.
 
  • #8
PainterGuy said:
Thank you for pointing this out.

On a unit circle the cosine is the length of the adjacent side (which is the x-coordinate) divided by the length of the hypotenuse (which is 1). So, as you said, the cosine is just the x-coordinate.

The domain of the inverse cosine function is [−1,1] and the range is [0,π]. That means a positive value will yield a 1st quadrant angle and a negative value will yield a 2nd quadrant angle.Helpful links:
https://www.mathwords.com/c/cosine_inverse.htm
/watch?v=ZEAgtXTIxk0 (add youtube.com in front)
/watch?v=xEGZsWgCn-Q
https://www.khanacademy.org/math/al...0c9fb89:unit-circle/a/trig-unit-circle-review
Okay. So for every point ##y## in ##[-1,1]## there are two values of ##\theta## for which ##\cos \theta = y##.

Can you see the relationship between the two values of ##y##?
 
  • Like
Likes PainterGuy
  • #9
PeroK said:
Can you see the relationship between the two values of ##y##?
I think both values of y should be equal.
 
  • #10
PainterGuy said:
Also, I don't see how the floor function could be used here.
Say
[tex]\frac{1}{4}-\frac{\alpha}{2\pi}=123456.789[/tex]
[tex]\frac{3}{4}-\frac{\alpha}{2\pi}=123457.289[/tex]
You will be able to reduce them to 0.789, 0.289 using floor function.

.
 
Last edited:
  • Like
Likes PainterGuy
  • #11
PainterGuy said:
I think both values of y should be equal.
They are not. The graph of ##\cos \theta## is symmetric about ##\theta = \pi##. For every value of ##y## there are two values of ##\theta## for which ##\cos \theta = y##: one in the interval ##[0, \pi]## and one in the interval ##[\pi, 2\pi]##. The only exception is ##y = -1##, for which there is only ##\theta = \pi##.

This symmetry may be expressed in two ways (these are common trig identities):$$\cos (\pi - \theta) = \cos (\pi + \theta)$$ or $$\cos \theta = \cos(2\pi - \theta)$$You should convince yourself of these by looking at the graph of ##\cos \theta## on the interval ##[0, 2\pi]##.
 
  • Like
Likes PainterGuy
  • #12
Whoah, this seems to have gone WAY off track. Assuming this is the question:
PainterGuy said:
How do I solve cos(a+θ)=0?
Then you are going about this completely the wrong way.

Here is the first step along the right way: for what values of ## x ## is ## \cos x = 0 ## true?
 
  • Like
Likes SammyS
  • #13
pbuk said:
Whoah, this seems to have gone WAY off track. Assuming this is the question:

Then you are going about this completely the wrong way.

Here is the first step along the right way: for what values of ## x ## is ## \cos x = 0 ## true?
If you read the OP that ##0## should be a general ##y##. Given the solution, I assume the first line is a typo.
 
  • Like
Likes pbuk
  • #14
PeroK said:
If you read the OP that ##0## should be a general ##y##. Given the solution, I assume the first line is a typo.
That is also a possibility, but it doesn't fit with this part of the OP which is clearly referring to ## \cos (\alpha + \theta) = 0 ##.
PainterGuy said:
If α=3 is assumed, the solution is as shown below and it makes sense

[itex]\theta_{1}=\frac{3}{2}(\pi-2)[/itex]
[itex]\theta_{2}=\frac{1}{2}(5 \pi-6)[/itex]

But if I don't use any specific value for α, the solution doesn't make sense to me.

[itex]\pi n-\frac{5 \pi}{2} \leq \alpha \leq \pi n-\frac{\pi}{2}[/itex] and [itex]\theta=\pi\left(n-\frac{1}{2}\right)-\alpha[/itex] and [itex]n \in \mathbb{Z}[/itex]

Source: https://www.wolframalpha.com/input/?i=solve+cos(α+θ)=0+for+θ+from+0+to+2pi
 
Last edited:
  • Like
Likes PeroK
  • #15
@PainterGuy please clarify whether you are only looking at the special case where ##y = 0##. That said, this is as good an opportunity as any to understand the inverse cosine function generally!
 
  • Like
Likes SammyS
  • #16
PeroK said:
@PainterGuy please clarify whether you are only looking at the special case where ##y = 0##. That said, this is as good an opportunity as any to understand the inverse cosine function generally!
Yes, I was trying to find the roots of y=cos(α+θ) which would require y=0. α is a constant and I'd assume that it's not zero.
 
  • #17
PainterGuy said:
Yes, I was trying to find the roots of y=cos(α+θ) which would require y=0. α is a constant and I'd assume that it's not zero.
Okay, so we have an answer in post #2. Without knowing a range for ##\alpha## it looks a bit messy to try to keep ##\theta## in a certain range. Is that what you want to do?
 
  • Like
Likes PainterGuy
  • #18
PainterGuy said:
Yes, I was trying to find the roots of y=cos(α+θ) which would require y=0.
Why are you writing this in such a complicated way? You don't need to introduce ## y ##, you are simply trying to find values of ## \theta ## that satisfy ## cos (\alpha + \theta) = 0 ##.

The first step to answer that question is answering:
pbuk said:
for what values of ## x ## is ## \cos x = 0 ## true?
 
  • Like
Likes PainterGuy
  • #19
Note that to get the answer from Wolfram Alpha you simply need to enter "solve cos(\alpha + \theta) = 0 for \theta", but in order to use a tool like Wolfram Alpha you need to have some understanding of the basics.
 
  • #20
PeroK said:
Okay, so we have an answer in post #2. Without knowing a range for ##\alpha## it looks a bit messy to try to keep ##\theta## in a certain range. Is that what you want to do?
Yes, I was trying to keep the theta in the given range without knowing the range for alpha.
 
  • #21
pbuk said:
Why are you writing this in such a complicated way? You don't need to introduce ## y ##, you are simply trying to find values of ## \theta ## that satisfy ## cos (\alpha + \theta) = 0 ##.

The first step to answer that question is answering:

cos(x)=0 for x=pi/2 and x=3pi/2 if x is restricted to 0 to 2pi.
 
Last edited:
  • #22
PainterGuy said:
cos(x)=0 for x=pi/2 and x=3pi/2 if x is restricted to 0 to 2pi.

Right, let's just look at the first solution, ## x = \dfrac \pi 2 ##.

Look back at the original problem ## \cos(\alpha + \theta) = 0 ##. What I have done is made a substitution ## x = \alpha + \theta ##. Using substitutions is an important tool in solving mathematical problems; choosing the right substitution makes the problem easier to solve. We can now write an equation with the substitution I made on the LHS and the solution you found for x on the RHS:
$$ \alpha + \theta = \dfrac \pi 2 $$
and we can simplify this to
$$ \theta = \dfrac \pi 2 - \alpha $$
Note that this will only give ## 0 \le \theta \lt 2 \pi ## for certain values of ## \alpha ## (what are these?)

You can do something similar with your other solution for ## x ##, and again look at how this works for different values of ## \alpha ##.
 

1. What is the meaning of cos(a+θ)?

Cos(a+θ) is a mathematical function that calculates the cosine of the sum of two angles, a and θ. It is a trigonometric function commonly used in geometry and physics.

2. How do you solve cos(a+θ)=0?

To solve cos(a+θ)=0, you can use the inverse cosine function (arccos) to find the value of (a+θ) that results in a cosine of 0. Then, you can solve for the individual values of a and θ by using basic algebraic equations.

3. What are the possible solutions for cos(a+θ)=0?

The possible solutions for cos(a+θ)=0 depend on the values of a and θ. In general, there are an infinite number of solutions, as cosine is a periodic function. However, in most cases, there are only a few distinct solutions within a given range of values for a and θ.

4. How does solving cos(a+θ)=0 relate to real-world applications?

Solving cos(a+θ)=0 is useful in many real-world applications, such as calculating the position of objects in space, determining the direction and magnitude of forces, and analyzing the behavior of waves and oscillations. It is also used in fields such as engineering, navigation, and astronomy.

5. What are some tips for understanding the results of solving cos(a+θ)=0?

When solving cos(a+θ)=0, it is important to remember that the solutions will depend on the values of a and θ, and may not always result in whole numbers. It is also helpful to graph the function to visualize the solutions and check your work. Additionally, understanding the unit circle and the properties of cosine can aid in understanding the results.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
7
Views
543
  • Precalculus Mathematics Homework Help
Replies
2
Views
985
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
502
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
546
  • Precalculus Mathematics Homework Help
Replies
7
Views
125
  • Introductory Physics Homework Help
Replies
5
Views
384
  • Calculus and Beyond Homework Help
Replies
1
Views
220
  • Precalculus Mathematics Homework Help
Replies
17
Views
2K
Back
Top