# Find the domain of a function of the form: ln(sin(1/x))

• I
• sergey_le
In summary, when you know the sine function, you can't say when ##y=\sin x## is positive. You need to know the intervals of x for which the function is positive.

#### sergey_le

TL;DR Summary
I know we need to find out when sin(pi/x)>0.
But I can 't do it

the solution

Last edited by a moderator:
If you know the sine function, why can't you say when ##y=\sin x## is positive?

fresh_42 said:
If you know the sine function, why can't you say when ##y=\sin x## is positive?
Because it's not just a sinx.
It's sin(1/x)

sergey_le said:
Because it's not just a sinx.
It's sin(1/x)
Proceed step by step. When is ##y=\sin x > 0##? Which are the intervals of ##x## for which this is true? Let's change the variable names. When is ##Y=\sin X > 0##? If we know that, we get e.g. ##X\in (0,\pi)##.

With that we have now ##X=\dfrac{\pi}{x}##. That is ##0<X=\dfrac{\pi}{x} < \pi##. Can you restructure that in a way, that ##a<x<b## is the result?

fresh_42 said:
Proceed step by step. When is ##y=\sin x > 0##? Which are the intervals of ##x## for which this is true? Let's change the variable names. When is ##Y=\sin X > 0##? If we know that, we get e.g. ##X\in (0,\pi)##.

With that we have now ##X=\dfrac{\pi}{x}##. That is ##0<X=\dfrac{\pi}{x} < \pi##. Can you restructure that in a way, that ##a<x<b## is the result?
I know that sinx> 0 when 0+2πk<x<π+2πk.

I do not understand why it is okay to say that 0+2πk<π/x<π+2πk.

sergey_le said:
I know that sinx> 0 when 0+2πk<x<π+2πk.

I do not understand why it is okay to say that 0+2πk<π/x<π+2πk.
Let us write the sine function as ##\varphi = \sin \alpha##. Then you just said that ##\varphi >0## when ##2k\pi < \alpha < (2k+1)\pi## for ##k\in \mathbb{Z}##. Now how can we compare ##\sin \alpha## and ##\sin (\pi/x)##? We can do this by setting ##\alpha:=\pi/x.## Thus we get from the above inequality that
$$\varphi =\sin \alpha = \sin \left(\dfrac{\pi}{x}\right) > 0 \text{ when } 2k\pi < \alpha = \dfrac{\pi}{x} < (2k+1)\pi \quad (k\in \mathbb{Z})$$
Next question is: Can you transform ##2k\pi < \dfrac{\pi}{x} < (2k+1)\pi## into something like ##A < x < B##?

sergey_le
fresh_42 said:
Let us write the sine function as ##\varphi = \sin \alpha##. Then you just said that ##\varphi >0## when ##2k\pi < \alpha < (2k+1)\pi## for ##k\in \mathbb{Z}##. Now how can we compare ##\sin \alpha## and ##\sin (\pi/x)##? We can do this by setting ##\alpha:=\pi/x.## Thus we get from the above inequality that
$$\varphi =\sin \alpha = \sin \left(\dfrac{\pi}{x}\right) > 0 \text{ when } 2k\pi < \alpha = \dfrac{\pi}{x} < (2k+1)\pi \quad (k\in \mathbb{Z})$$
Next question is: Can you transform ##2k\pi < \dfrac{\pi}{x} < (2k+1)\pi## into something like ##A < x < B##?
Thank you very much, I finally understood

berkeman

## 1. What is the domain of a function of the form ln(sin(1/x))?

The domain of a function is the set of all possible input values that the function can take. In this case, the domain of ln(sin(1/x)) is all real numbers except for 0 and any value that makes sin(1/x) negative or 0. This is because the natural logarithm function is undefined for non-positive values.

## 2. Why is the domain of ln(sin(1/x)) restricted to exclude certain values?

The domain of a function is restricted in order to avoid undefined or imaginary outputs. In this case, the natural logarithm function is undefined for non-positive values, and the sine function can only take values between -1 and 1. Therefore, the domain of ln(sin(1/x)) is restricted to avoid these values.

## 3. How do I find the domain of a function of the form ln(sin(1/x))?

To find the domain of a function, you need to consider any restrictions on the input values that would result in undefined or imaginary outputs. In this case, you need to exclude 0 and any value that makes sin(1/x) negative or 0.

## 4. Can the domain of ln(sin(1/x)) be extended beyond the restrictions?

No, the domain of a function is determined by its definition and cannot be extended beyond the restrictions. In this case, the restrictions are necessary to avoid undefined or imaginary outputs.

## 5. Are there any other ways to write the domain of ln(sin(1/x))?

Yes, the domain can also be written as the set of all real numbers except for 0 and any value that makes 1/x undefined, since the sine function will also be undefined for those values. This can be expressed as x ≠ 0 and x ≠ kπ, where k is any integer.

• Calculus
Replies
3
Views
1K
• Calculus
Replies
12
Views
1K
• Calculus
Replies
28
Views
2K
• Calculus
Replies
11
Views
1K
• Calculus
Replies
10
Views
737
• Calculus
Replies
3
Views
904
• Calculus
Replies
4
Views
1K
• Calculus
Replies
1
Views
685
• Calculus
Replies
1
Views
1K
• Precalculus Mathematics Homework Help
Replies
15
Views
441