DarkHorse's Spesh questions

[DarkHorse]

Hi guys,

I know that this is a really noob question, but how would you go about working out the range of cos(sqrt(1-x^2))? And could you please explain your reasoning behind it. 

[paulsterio]

Firstly, think about the range of the function sqrt(1-x^2) - that's essentially the top half of a circle, so it will output between 0 and 1.

We know that the cos graph will "max" at 0 - so the maximum will be 1.

The minimum is a little tougher, we know that the cos grapgh will "min" at pi, but of course, pi is greater than 1, so in this case, it will actually never get to pi, so the "min" will actually be cos(1)

[kensan]

If we look at that, we can see that there is a maximal domain of

Then you could differentiate to get,





   solve for x

 so x=0

Therefore if you sub in 0 and 1 into the original function, you should get the range, which turns out to be

That's how I would do it, Paul's way is probably quicker

[DarkHorse]

Thanks for that guys!

[DarkHorse]

Okay I have another one - how is the domain of cos(2sin^-1(x)) = [-1,1]? I thought it would be [0,1]....

[Jenny_2108]

Okay I have another one - how is the domain of cos(2sin^-1(x)) = [-1,1]? I thought it would be [0,1]....

Because its composite function and domain of 2sin^-1(x)=[-1,1].
Why do you think its [0,1]?

[kensan]

The range of     is and the domain

Since the range of the inside function fits in the domain of the outside function, the domain of the inside function exists.
Hope that makes sense

Because its composite function and domain of 2sin^-1(x)=[-1,1].

Got to be careful, if the range of the inside doesn't 'fit' with the domain of the outside, it won't work. It would only work if you restricted the domain. But in this case it's ok

[DarkHorse]

Hmmm...but I thought that it wouldn't be defined because we would have to take the principal cos which has the domain [0,pi] ?

[kensan]

By principal cos do you mean cos^-1?   Haven't come across that terminology
And I'm not sure where you are pulling that domain from sorry

[Jenny_2108]

Because its composite function and domain of 2sin^-1(x)=[-1,1].

Got to be careful, if the range of the inside doesn't 'fit' with the domain of the outside, it won't work. It would only work if you restricted the domain. But in this case it's ok

The question asks domain of cos(2sin^-1(x)) => I assume this composite function is defined already so ran of inside should be considered as "fit" with dom of the outside.

Hmmm...but I thought that it wouldn't be defined because we would have to take the principal cos which has the domain [0,pi] ?

why does cos have domain [0,pi]?
You mean range of cos^-1 is [0,pi]?

[DarkHorse]

When we take the inverse of any circular function we have to restrict its domain so that it becomes a one-to-one function, thus these new functions with restricted domains are identified with the prefix 'principal'.

[/quote]

The question asks domain of cos(2sin^-1(x)) => I assume this composite function is defined already so ran of inside should be considered as "fit" with dom of the outside.
[/quote]

In what cases would you assume this?

[Jenny_2108]

Nah when we take the inverse of any circular function we have to restrict its domain so that it becomes a one-to-one function, thus these new functions with restricted domains are identified with the prefix 'principal'.

I got what you mean. I've never heard about "principal" cos before 
but in this question, they dont ask about inverse of cos, so why do we have to care about "principal cos"?

[Jenny_2108]

In what cases would you assume this?

I assume because if ran of inside doesnt fit dom outside, this composite function doesnt exist, so whats the point of finding the dom of the function?

P/S: You edit your post so I have to post another one, not spam though

[DarkHorse]

Yeah I realised that I had a somewhat trivial issue with conventions for identifying circular functions with restricted domains.