How do you solve cos(α+θ)=0 for different values of α?

[PainterGuy]

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

\theta_{1}=\frac{3}{2}(\pi-2)
\theta_{2}=\frac{1}{2}(5 \pi-6)

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

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

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 \cos ^{-1} y is used, and please note that α is not given any specific value:

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

Could you please help me? It's not homework but I still thought I better post it here nonetheless.

 

[anuttarasammyak]

\alpha+\theta=\pi(n+\frac{1}{2})
\theta=\pi(n+\frac{1}{2})-\alpha
integer n should be chosen properly so that
0 < \theta < 2\pi
You may make use of floor function to show solutions explicitly.

 

[PeroK]

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

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

What happens if you add $\alpha$ to those solutions for $\theta$?

Do you know how the function $\cos^{-1}$ is defined?

 

[PainterGuy]

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.

 

[PeroK]

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.

 

[PainterGuy]

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

 

[PainterGuy]

\alpha+\theta=\pi(n+\frac{1}{2})
\theta=\pi(n+\frac{1}{2})-\alpha
integer n should be chosen properly so that
0 < \theta < 2\pi
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.

 

[PeroK]

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 $y$?

 

[PainterGuy]

Can you see the relationship between the two values of $y$?

I think both values of y should be equal.

 

[anuttarasammyak]

Also, I don't see how the floor function could be used here.

Say
\frac{1}{4}-\frac{\alpha}{2\pi}=123456.789
\frac{3}{4}-\frac{\alpha}{2\pi}=123457.289
You will be able to reduce them to 0.789, 0.289 using floor function.

.

 

[PeroK]

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 $[0, 2\pi]$.

 

[pbuk]

Whoah, this seems to have gone WAY off track. Assuming this is the question:

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?

 

[PeroK]

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.

 

[pbuk]

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 $\cos (\alpha + \theta) = 0$.

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

\theta_{1}=\frac{3}{2}(\pi-2)
\theta_{2}=\frac{1}{2}(5 \pi-6)

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

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

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

 

[PeroK]

@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!

 

[PainterGuy]

@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.

 

[PeroK]

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?

 

[pbuk]

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:

for what values of $x$ is $\cos x = 0$ true?

 

[pbuk]

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.

 

[PainterGuy]

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.

 

[PainterGuy]

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.

 

[pbuk]

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 $\alpha$.