Turn on thread page Beta
    • Thread Starter
    Offline

    16
    ReputationRep:
    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.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (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.
    What about the end points?

    To obtain y= \arccos x you must first reduce the domain of y = \cos x to 0 \leq x \leq \pi so that it is a one-to-one function, and then you take the inverse which is just a reflection in the line y=x. The end points of y= \cos x are (0,1) and (\pi , -1) on this domain, hence the ends points of \arccos x are (1,0) and (-1, \pi)
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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 \cos^{-1}x is the inverse of \cos{x} with domain 0\leq x\leq\pi. 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 y=x which swiches y/x coordinates. The endpoints of \cos{x} in that domain are (0,1) and (\pi,-1). So the endpoints of \cos^{-1}x are (-1,\pi) and (1,0).

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

    Or see RDKGames post above because it's basically the same
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by RDKGames)
    What about the end points?

    To obtain y= \arccos x you must first reduce the domain of y = \cos x to 0 \leq x \leq \pi so that it is a one-to-one function, and then you take the inverse which is just a reflection in the line y=x. The end points of y= \cos x are (0,1) and (\pi , -1) on this domain, hence the ends points of \arccos x are (1,0) and (-1, \pi)
    (Original post by Notnek)
    The function \cos^{-1}x is the inverse of \cos{x} with domain 0\leq x\leq\pi. 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 y=x which swiches y/x coordinates. The endpoints of \cos{x} in that domain are (0,1) and (\pi,-1). So the endpoints of \cos^{-1}x are (-1,\pi) and (1,0).

    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.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (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 y=\cos x on any interval of the form n\pi \leq x \leq (n+1)\pi, where n \in \mathbb{Z}, and it will be 1-to-1. However, the inverse would then be some shift (or reflection) of the usual \arccos (x) along the y-axis, namely y=n \pi + \arccos x if n is even (includes your question, because in your question n=0), or y=n \pi + \arccos (-x) if n is odd.

    In C3 you only need to be aware of the n=0 case.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by RDKGames)
    You may also notice that you can restrict y=\cos x on any interval of the form n\pi \leq x \leq (n+1)\pi, where n \in \mathbb{Z}, and it will be 1-to-1. However, the inverse would then be some shift (or reflection) of the usual \arccos (x) along the y-axis, namely y=n \pi + \arccos x if n is even (includes your question, because in your question n=0), or y=n \pi + \arccos (-x) if n is odd.

    In C3 you only need to be aware of the n=0 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?
    Posted on the TSR App. Download from Apple or Google Play
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 5, 2018

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.