# Homework Help: Range of a weird function

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1. Jan 28, 2017

### Buffu

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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.

2. Jan 28, 2017

### PetSounds

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

3. Jan 28, 2017

### SammyS

Staff Emeritus
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.

4. Jan 28, 2017

### Buffu

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.
[-1,1]

Last edited: Jan 28, 2017
5. Jan 28, 2017

### PetSounds

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

6. Jan 28, 2017

### Buffu

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

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

7. Jan 28, 2017

### PetSounds

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

8. Jan 29, 2017

### Buffu

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

Last edited: Jan 29, 2017
9. Jan 29, 2017

### PetSounds

Bingo.

10. Jan 29, 2017

### Ray Vickson

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

11. Jan 29, 2017

### Buffu

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 ?

12. Feb 3, 2017

### haruspex

There are other solutions.

13. Feb 3, 2017

### Buffu

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

14. Feb 3, 2017

### haruspex

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 }

15. Feb 3, 2017

### Buffu

I did not get it. you just removed x.

16. Feb 3, 2017

### haruspex

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?

17. Feb 4, 2017
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