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- TL;DR Summary
- I know we need to find out when sin(pi/x)>0.
But I can 't do it
the solution
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Because it's not just a sinx.fresh_42 said:If you know the sine function, why can't you say when ##y=\sin x## is positive?
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)##.sergey_le said:Because it's not just a sinx.
It's sin(1/x)
I know that sinx> 0 when 0+2πk<x<π+2πk.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?
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 thatsergey_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.
Thank you very much, I finally understoodfresh_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##?
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.
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.
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.
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.
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.