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Could someone please help me out here?

state the implied domain and range of

B) tan (2arccos(x))

ok, the domain of tan for which arctan exists (conventionally) is (-pi/2 , pi/2) -- really the domain of tan is R, but we're using the restricted one.

therefore, we know that the internal function must be within this:

-pi/2 < 2arccos(x)) < pi/2

dividing by two

-pi/4 < arccos(x)) < pi/4

Now, to find what x is between, we must take the cos of both sides, BUT WHY DO WE HAVE TO FLIP THE INEQUALITIES? I mean, sure, it makes no sense if one doesn't flip them -- however i presume that in maths one can't just arbitrarily change things around because they dont make sense.

Oh, crap, i forgot to consider the internal function's domain / range....

What, exactly, am i supposed to consider here? The domain or the range of the internal function?

For some questions, all one has to do is find the intersection of the domain of the external function and the range of the internal function, but this logic, i've noticed, doesn't always work. I dont really understand what i'm doing.... i guess i'm just learning it by rote without really understanding it, which irritates me.

Like with this question: y=arctan (sin x)

If one uses the Domain intersection Range logic, one gets [-1, 1] for the implied domain, whcih is wrong.

On the other hand, if one uses the other logic, ie that:

the dom of arctan is R, domain of restricted sin x is [-pi/2, pi/2], therefore for implied domain for that question is [-pi/2, pi/2].

I don't understand this. Implied ranges too, but i guess once i understand domains, ranges wont be a problem.

How does one go about doing questions like this? And could someone please explain what i'm actually doing?

Thanks

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# Finding implied domains and ranges

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