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# C3 End Points watch

1. Can someone explain the end points of the inverse trig functions.

The question was the end points of y = cos -1x, I got it wrong and then thought I understood it after but now I have gone back to it and I don't understand again. I can easily find them out on my calc but would rather understand.
2. (Original post by GCSE2016Troop)
Can someone explain the end points of the inverse trig functions.

The question was the end points of y = cos -1x, I got it wrong and then thought I understood it after but now I have gone back to it and I don't understand again. I can easily find them out on my calc but would rather understand.

To obtain you must first reduce the domain of to so that it is a one-to-one function, and then you take the inverse which is just a reflection in the line . The end points of are and on this domain, hence the ends points of are and
3. (Original post by GCSE2016Troop)
Can someone explain the end points of the inverse trig functions.

The question was the end points of y = cos -1x, I got it wrong and then thought I understood it after but now I have gone back to it and I don't understand again. I can easily find them out on my calc but would rather understand.
The function is the inverse of with domain . The domain is restricted in this way to make it 1-1 so that it can have an inverse function.

Then to get from a function to its inverse, every point is reflected in which swiches y/x coordinates. The endpoints of in that domain are and . So the endpoints of are and .

A sketch will help with this to understand what I'm describing here.

Or see RDKGames post above because it's basically the same
4. (Original post by RDKGames)

To obtain you must first reduce the domain of to so that it is a one-to-one function, and then you take the inverse which is just a reflection in the line . The end points of are and on this domain, hence the ends points of are and
(Original post by Notnek)
The function is the inverse of with domain . The domain is restricted in this way to make it 1-1 so that it can have an inverse function.

Then to get from a function to its inverse, every point is reflected in which swiches y/x coordinates. The endpoints of in that domain are and . So the endpoints of are and .

A sketch will help with this to understand what I'm describing here.

Or see RDKGames post above because it's basically the same
I get it now thanks guys, it was the restricting the domain part I was messing up on so my other end point was wrong.
5. (Original post by GCSE2016Troop)
I get it now thanks guys, it was the restricting the domain part I was messing up on so my other end point was wrong.
You may also notice that you can restrict on any interval of the form , where , and it will be 1-to-1. However, the inverse would then be some shift (or reflection) of the usual along the y-axis, namely if is even (includes your question, because in your question n=0), or if is odd.

In C3 you only need to be aware of the case.
6. (Original post by RDKGames)
You may also notice that you can restrict on any interval of the form , where , and it will be 1-to-1. However, the inverse would then be some shift (or reflection) of the usual along the y-axis, namely if is even (includes your question, because in your question n=0), or if is odd.

In C3 you only need to be aware of the case.
I’m going to take a look at this tomorrow when I am on a computer so I can see the notation, looks interesting though thanks

Would we have to be aware of this for C4?
7. (Original post by GCSE2016Troop)
I’m going to take a look at this tomorrow when I am on a computer so I can see the notation, looks interesting though thanks

Would we have to be aware of this for C4?
Nah, just some extra knowledge for ya.

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