Understanding Arccos and Its Undefined Value for π/2

[opus]

Homework Statement

State the exact value of $y=arccos\left(\frac π 2 \right)$

Homework Equations

The Attempt at a Solution

In a "typical" question, such as $y=arccos\left(-1\right)$, I look at it like "The cosine of what angle gives the value -1?" I then see that it's π.
This question however is giving an x value rather than a y value which isn't making sense to me. So I look at it as "the cosine of what value will give me $\left(\frac π 2\right)$ and that would be zero. But the solution to the question is undefined, and I am not seeing how this is the case. The range of arccos is [0,π] and the domain is [-1,1]. So I'm not seeing where the undefined comes into play here. Any tips?

 

[phinds]

I don't see how x vs y has any bearing. There is NO answer to the question "what angle has a cosine of pi/2"

 

[phyzguy]

π/2 is about 1.57. What angle has a cos equal to 1.57?

 

[Mark44]

Homework Statement

State the exact value of $y=arccos\left(\frac π 2 \right)$

Homework Equations

The Attempt at a Solution

In a "typical" question, such as $y=arccos\left(-1\right)$, I look at it like "The cosine of what angle gives the value -1?" I then see that it's π.
This question however is giving an x value rather than a y value which isn't making sense to me. So I look at it as "the cosine of what value will give me $\left(\frac π 2\right)$ and that would be zero.

No. The range of the cosine function is [-1, 1], so there aren't any angles that produce a value > 1.6, which is approximately equal to $\pi/2$.

But the solution to the question is undefined, and I am not seeing how this is the case. The range of arccos is [0,π] and the domain is [-1,1].

Right, the domain of arccos() is [-1, 1], so $\pi/2$ isn't in the domain.

So I'm not seeing where the undefined comes into play here. Any tips?

 

[opus]

I don't see how x vs y has any bearing. There is NO answer to the question "what angle has a cosine of pi/2"

Which I suppose is why I am confused. Some of the questions have "x values", that is radian measures in the parentheses. Some have "y values", that is, real numbers in the parentheses. So in my normal way of wording the question, stated in my OP, makes no sense now. So what is the question asking me?

π/2 is about 1.57. What angle has a cos equal to 1.57?

So you divided 2 into pi to get a real number approximation, and now we can ask the question like a "normal" question? Is this what we would do in every case where pi is included as the input the an inverse trigonometric function?
And to answer your question, I don't believe any angle can have a cosine equal to more than 1, as the unit circle has radius 1.

No. The range of the cosine function is [-1, 1], so there aren't any angles that produce a value > 1.6, which is approximately equal to $\pi/2$.
Right, the domain of arccos() is [-1, 1], so $\pi/2$ isn't in the domain.

Ok I see. Getting that ~1.6 is what I did not do. So if it asked for something like $\frac π 8$, that would not be undefined, as the value is less than 1, and in the domain?

And as a side question, I don't understand how it is "undefined". I see that it's not in the domain, because we limited ourselves to [-1,1] so that we have a monotone function. But I've always thought something to be undefined as a zero in the denominator. Is the true definition of an undefined value to be something that isn't in the domain? Even if it's doesn't necessarily have a zero in the denominator?

 

[Mark44]

Ok I see. Getting that ~1.6 is what I did not do. So if it asked for something like $\frac π 8$, that would not be undefined, as the value is less than 1, and in the domain?

$\arccos(\frac \pi 8)$ would be defined, because $\frac \pi 8$ is in the domain of this function.

[-1,1] so that we have a monotone function. But I've always thought something to be undefined as a zero in the denominator. Is the true definition of an undefined value to be something that isn't in the domain?

Yes. Division by zero isn't the only operation that causes something to be undefined. For example, ln(0) is undefined, because 0 isn't in the domain of this function. $\sqrt{-1}$ is undefined, if you're talking about the real-valued square root function.

Even if it's doesn't necessarily have a zero in the denominator?

 

[opus]

Makes sense! Thanks for the responses everyone.

 

[Mark44]

@opus, our question is related to a thread you started a week or so ago, about inverses.

Since y = cos(x) isn't one-to-one, it doesn't have an inverse. However, if we restrict the domain enough, we can make a function that is one-to-one. The usual restricted domain is $[0, \pi]$.

So, $y = \cos(x) \text{ and } x \in [0, \pi] \Leftrightarrow x = \cos^{-1}(y)$. Instead of cos^-1, arccos is sometimes used.

For the restricted cosine function, the domain is $[0, \pi]$ and the range is [-1, 1].
For its inverse, the domain is [-1, 1] and its range is $[0, \pi]$.

The two equations y = cos(x) (with the restriction) and x = cos^-1(y), have exactly the same graphs.

 

[opus]

I figured the concept was the same. What has been throwing me for a loop is the interchange between x and y. For example, for your plain vanilla cosine function, x values are associated with the angle, and y values are associated with the cosine value at that angle listed on the y axis. My book changes this for the inverse cosine function, having angle values as y listed on the y axis, and cosine values as x listed on the x axis. My instructor does the opposite and keeps angle values as x and cosine values as y. This flip flopping has been causing me some confusion although I do understand the concept thanks to your explanation in the other thread. Would you suggest looking at it as the x-axis always has the angled values and the y-axis has the cosine values or vice versa?

 

[opus]

An example of what I'm talking about.

 

[Mark44]

I figured the concept was the same. What has been throwing me for a loop is the interchange between x and y. For example, for your plain vanilla cosine function, x values are associated with the angle, and y values are associated with the cosine value at that angle listed on the y axis. My book changes this for the inverse cosine function, having angle values as y listed on the y axis, and cosine values as x listed on the x axis. My instructor does the opposite and keeps angle values as x and cosine values as y. This flip flopping has been causing me some confusion although I do understand the concept thanks to your explanation in the other thread. Would you suggest looking at it as the x-axis always has the angled values and the y-axis has the cosine values or vice versa?

On the graph in your next post, the red graph is y = cos^-1(x) AKA arccos(x). This is also the graph of x = cos(y).

Normally we have x as the independent variable (on the horizontal axis) and y as the dependent variable. I don't know about always keeping x as angles and y as cosine values. For example if you have w = 3cos(t), just remember that w has to be within the interval [-1, 1], and t can be any real value (unless we're talking about the restricted cosine function, sometimes written as Cos(x).

 

[SammyS]

The range of arccos is [0,π] and the domain is [-1,1]. So I'm not seeing where the undefined comes into play here. Any tips?

That phrase highlighted in red is all that is needed to answer your question.

 

[opus]

Ok thanks guys!