Find the Domain for: 1) f(x)=arcsin(tg2x) and 2)f(x)=arcsin(sqrt(3)tgx)

[denis01]

Hello. I've been homeschooling myself, but I came across 2 problems I am not sure how to solve:

I have to find the domain for:
1) f(x)=arcsin(tg2x)
and
2)f(x)=arcsin(sqrt(3)tgx)

I assume I would need to know the basics of trigonometry to solve this problem, but I don't know where to start. I know how the graph is supposed to look like with tangents, and that the formula for arcsin problems goes -1<=xyz<=1, but that's about it. I really don't know how to turn the tangents into a number of some sorts

I would ask for the complete solving of both problems, along with shown work, so I have 2 examples on how to do these kinds of problems in the future, and maybe some good sources where I can learn how to do these, and other, more complicated problems.

 

[Deleted member 4993]

Hello. I've been homeschooling myself, but I came across 2 problems I am not sure how to solve:

I have to find the domain for:
1) f(x)=arcsin(tg2x)
and
2)f(x)=arcsin(sqrt(3)tgx)

I assume I would need to know the basics of trigonometry to solve this problem, but I don't know where to start. I know how the graph is supposed to look like with tangents, and that the formula for arcsin problems goes -1<=xyz<=1, but that's about it. I really don't know how to turn the tangents into a number of some sorts

I would ask for the complete solving of both problems, along with shown work, so I have 2 examples on how to do these kinds of problems in the future, and maybe some good sources where I can learn how to do these, and other, more complicated problems.

What is the range of domain function?

You already know that the domain of arcsin is -1 to +1.

So the value of tan(2x) must be -1 ≤ tan(2x) ≤ 1

At what values of 'x' do you satisfy this condition?

 

[denis01]

You add 1 to each side, and you end up with
tan2x<=2

Problem is, "Tan2x" doesn't tell me anything. What is 2 tangents, and how do I turn that into a number? I'm sorry, I must be missing a formula for this

As for the other problem, 2), I ended up with - and + sqrt(3) over 3, to cancel out the square root of 3. I know that sqrt of 3 equals one tangent, but what do I do with that information? I don't get the trigonometric part of this, over the simpler arcsin problems, which were just a formula at the beginning (-1, 1)

 

[stapel]

You add 1 to each side, and you end up with tan2x<=2

No; you would end up with:

. . . . .\(\displaystyle 0\, \leq\, \tan(2x)\, +\, 1\, \leq\, 2\)

Problem is, "Tan2x" doesn't tell me anything. What is 2 tangents...?

Um... The expression, "tan(2x)", does not mean "two tangents". It means "the tangent of the product of two and x".

Have you not studied trig functions yet...? :shock:

As for the other problem, 2), I ended up with - and + sqrt(3) over 3...

How?

...to cancel out the square root of 3.

I know that sqrt of 3 equals one tangent...

I don't get the trigonometric part of this, over the simpler arcsin problems, which were just a formula at the beginning (-1, 1)

What? :shock:

 

[ksdhart2]

You add 1 to each side, and you end up with
tan2x<=2

Problem is, "Tan2x" doesn't tell me anything. What is 2 tangents, and how do I turn that into a number? I'm sorry, I must be missing a formula for this

Well, more specifically, you'd end with \(\displaystyle 0 \le tan(2x) \le 2\). Be sure you understand why this minor change is very important. As for your confusion, the notation doesn't indicate there "two tangents" but rather the tangent of 2x. And 2x is just a number, like any other number. Specifically it's the number that is two times x. Have you learned about inverse trig functions?https://www.mathsisfun.com/algebra/trig-inverse-sin-cos-tan.html What happens if you take the inverse tangent of all three "sides" of the inequality?

\(\displaystyle tan^{-1}(0) \le tan^{-1}(tan(2x)) \le tan^{-1}(2)\)

Where does that lead you?

As for the other problem, 2), I ended up with - and + sqrt(3) over 3, to cancel out the square root of 3. I know that sqrt of 3 equals one tangent, but what do I do with that information? I don't get the trigonometric part of this, over the simpler arcsin problems, which were just a formula at the beginning (-1, 1)

I'm not sure I fully understand what you did to get here. Following the same reasoning as in the first problem, you know that arcsine must be bounded between -1 and 1, so you have:

\(\displaystyle -1 \le \sqrt{3} \cdot tan(x) \le 1\)

It sounds like you tried to divide everything through by sqrt(3):

\(\displaystyle -\dfrac{1}{\sqrt{3}} \le \tan(x) \le \dfrac{1}{\sqrt{3}}\)

And then rewrote that so the radical isn't in the denominator:

\(\displaystyle -\dfrac{\sqrt{3}}{3} \le tan(x) \le \dfrac{\sqrt{3}}{3}\)

But I don't know how any of that could have possibly led you to the conclusion that \(\displaystyle \sqrt{3}=tan(x)\).

 

[denis01]

Have you learned about inverse trig functions?

Not really. And the result I'm left with doesn't really let me know anything to be honest. I'm at a loss at what I have to do with the tangent in the equation.

I'm planning on studying inverse trig functions, among other things, but I have difficulty even knowing where to start. Hence why I asked for an example of one of those, It's easier for me to see how one of these problems was solved, so I can find out how to do the other problem, without having to spend a whole week researching a whole new subject

But that aside, I end up with a result that involves a tangent along with square rooted fractions. I know that I have to do something with the tangent, so I end up with pi divided by a number, but I have no idea how I would actually get to this number. Are there any online resources that would help me with those exact kinds of problems? What level of trigonometry do I have to know to solve this problem?

 

[denis01]

To make things simpler, I wrote down what little in the way of notes I have on the subject
https://ibb.co/kSjXyQ
This is the work I have written for f(x)=arcsin(tg2x). I don't really understand any of it, or how I am supposed to end up with any of these. There is no other work set for this problem. Is this correct? I'm probably missing something simple, due to my lack of knowledge in trigonometry