What Is the Range of \( y = \sqrt{\ln(\cos(\sin(x)))} \)?

[Buffu]

Homework Statement

Find the range $y = \sqrt{\ln({\cos(\sin (x)}))}$

Homework Equations

The Attempt at a Solution

[/B]
https://www.desmos.com/calculator

I used a graphing calculator to find the intersection between $y = e^{x^2}$ and $y = \cos(\sin(x))$.
Which I get as $(0,1)$. So the range is $\{0\}$.

But I want to find the range without graphs and by analytical methods.
Thanks for help.

 

[PetSounds]

What is the range of $y = cos (x)$ ?

 

[SammyS]

Homework Statement

Find the range $y = \sqrt{\ln({\cos(\sin (x)}))}$

Homework Equations

The Attempt at a Solution

[/B]
https://www.desmos.com/calculator

I used a graphing calculator to find the intersection between $y = e^{x^2}$ and $y = \cos(\sin(x))$.
Which I get as $(0,1)$. So the range is $\{0\}$.

But I want to find the range without graphs and by analytical methods.
Thanks for help.

All that you actually found here is that if $\ x=0\,,\$ then $\ y=1\,.\$ Therefore, 1 is in the range of your function.

I suggest the first thing to do is to determine the (implied) domain of your function.

 

[Buffu]

All that you actually found here is that if $\ x=0\,,\$ then $\ y=1\,.\$ Therefore, 1 is in the range of your function.

I suggest the first thing to do is to determine the (implied) domain of your function.

Putting x = 0 $y = \sqrt{\ln(\cos(\sin(0)))} =\sqrt{\ln(\cos 0))} = \sqrt{\ln(1)} = 0$, So y = 0 is also in range.
So the range is {0,1}.

Domain of function is (0 to pi/2) + 2n*pi.

What is the range of $y = cos (x)$ ?

[-1,1]

 

[PetSounds]

[-1,1]

And how does that overlap with the domain of $y = ln (x)$?

 

[Buffu]

And how does that overlap with the domain of $y = ln (x)$?

domain of ln x is (0, $\infty$) .

So $(0, 1]$ part of cos x domain is only useful in this problem

 

[PetSounds]

domain of ln x is (0, $\infty$) .

So $(0, 1]$ part of cos x domain is only useful in this problem

And what is the range of $ln (x)$ for $0 < x \leq 1$ ?

 

[Buffu]

And what is the range of $ln (x)$ for $0 < x \leq 1$ ?

less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.

 

[PetSounds]

less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.

Bingo.

 

[Ray Vickson]

less than 0 but we cannot have less than zero because of square root. So only 1 is left; Thus range is {0}.

Yes. And the domain of $f$ is also very limited in the real line. What would it (the domain) be?

 

[Buffu]

Yes. And the domain of $f$ is also very limited in the real line. What would it (the domain) be?

Domain of my original function would be when sin x is 0, that is 2pi or for general solution 2* pi *n. So my domain would be {x : x = 2pi * n $\forall n \in \mathbb Z$}. Right ?

 

[haruspex]

when sin x is 0, that is 2pi

There are other solutions.

 

[Buffu]

There are other solutions.

Oh yes sin x is also zero at π So the domain should be {x : x = π * n ∀n ∈ ℤ}

 

[haruspex]

Oh yes sin x is also zero at π So the domain should be {x : x = π * n ∀n ∈ ℤ}

Looks right. Your use of the predicates is a little inaccurate. There does not exist an x such that it equals π * n for all integers n. You mean {π * n : n∈ ℤ }

 

[Buffu]

Looks right. Your use of the predicates is a little inaccurate. There does not exist an x such that it equals π * n for all integers n. You mean {π * n : n∈ ℤ }

I did not get it. you just removed x.

 

[haruspex]

I did not get it. you just removed x.

What you had posted said :
"the set of things x such that x equals πn for all integers n".
There is no number that can equal πn for two different integers n, let alone all infinity of them.
If you want to use x and n then I suggest using ∃n. Maybe {x:∃n∈ℕ:x=πn}. But why not omit x and write it my way?

 

[Buffu]

But why not omit x and write it my way?

Your way is better.

"the set of things x such that x equals πn for all integers n".
There is no number that can equal πn for two different integers n, let alone all infinity of them.

Oh I understand what you mean.