Author Topic: How to solve range for y=arccos(sin2x)? (Read 1918 times) Tweet Share

Whymme · « **on:** March 09, 2017, 06:15:51 pm »

It is given that the domain of sin is [-pie/2, pie/2], and domain of cos is [0,pie]

I understand how to solve for the domain; the range of the inner function must belong to the domain of the outer function.
But after that, how do I solve for range? Which function do I plug the domain into? Right now I'm solving for it with a calculator, but would like to know a way without.

Help would be appreciated!

Shadowxo · « **Reply #1 on:** March 09, 2017, 07:57:41 pm »

Quote from: Whymme on March 09, 2017, 06:15:51 pm

It is given that the domain of sin is [-pie/2, pie/2], and domain of cos is [0,pie]

I understand how to solve for the domain; the range of the inner function must belong to the domain of the outer function.
But after that, how do I solve for range? Which function do I plug the domain into? Right now I'm solving for it with a calculator, but would like to know a way without.

Help would be appreciated!

You know how to find the domain, and use it to find the range of the inner function. Then, sub that (range of inner function) in as the domain of the outer function to find the range.
For this, the range of sin(2x) is [-1,1], and the domain of arccos is [-1,1]. So use this to find range of the whole function, for this you can just sub in -1 and 1 to find the range of [0,π].

Whymme · « **Reply #2 on:** March 09, 2017, 08:12:00 pm »

Quote from: Shadowxo on March 09, 2017, 07:57:41 pm

You know how to find the domain, and use it to find the range of the inner function. Then, sub that (range of inner function) in as the domain of the outer function to find the range.
For this, the range of sin(2x) is [-1,1], and the domain of arccos is [-1,1]. So use this to find range of the whole function, for this you can just sub in -1 and 1 to find the range of [0,π].

Ah I see, thank you.

Sine · « **Reply #3 on:** March 20, 2017, 09:29:21 pm »

Quote from: Whymme on March 09, 2017, 06:15:51 pm

It is given that the domain of sin is [-pie/2, pie/2], and domain of cos is [0,pie]

I understand how to solve for the domain; the range of the inner function must belong to the domain of the outer function.
But after that, how do I solve for range? Which function do I plug the domain into? Right now I'm solving for it with a calculator, but would like to know a way without.

Help would be appreciated!

$y=arccos(sin(2x))$
$\text{so we are given the following initially}$
$x \in [-\frac{\pi}{2},\frac{\pi}{2}]$
$\text{since we know this we can find the range of sin(2x)}$
$sin(2x) \in [-1,1]$
$\text{Now we know the values upon which can be inserted into arccos}$
$\text{now we must find the range of the outputs possible.}$
$\text{The above restrictions still satisfies the full domain of arccos.}$
$\text{Hence the full range.}$
$arccos(sin(2x))\in[0,\pi]$
$y\in[0,\pi]$
$\text{NOTE: sketch graphs of the functions that you know to help you}$

Search the forums now!

We have moved!

Author Topic: How to solve range for y=arccos(sin2x)? (Read 1918 times) Tweet Share

Whymme

How to solve range for y=arccos(sin2x)?

Shadowxo

Re: How to solve range for y=arccos(sin2x)?

Whymme

Re: How to solve range for y=arccos(sin2x)?

Sine

Re: How to solve range for y=arccos(sin2x)?

Recent Posts

User:
Password: