Results 1 to 3 of 3

Math Help - Function

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    7

    Function

    Is: g:[-1,1]=>[0,\pi] defined by: g(x)=arccos(x) The value of g=-\frac{1}{2} is:

    a) \frac{2\pi}{3}
    b) -\frac{2\pi}{3}
    c) \frac{\pi}{3}
    d) -\frac{\pi}{3}
    e) \frac{\pi}{6}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2008
    From
    Ontario
    Posts
    80
    You're question, from what I can gather, asks: What is the value of the cosine function in radians at -1/2 if the function has a range of [-1, 1] and a domain of 0 to Pi.

    Take a look at the image below, Pi/2 occurs when y=0, and pi occurs when y= -1. Since we know that the point y=-1/2 lies between pi/2 and pi it cannot be pi/6 or pi/3. And radians are not negative since the domain is not between 0 and -pi. Therefore the answer has to be 2pi/3.

    If you're in doubt try cos of 120 or cos of 2pi/3 in your calculator, you'll get the answer -0.5 or -1/2.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,963
    Thanks
    1784
    Awards
    1
    Quote Originally Posted by boy47 View Post
    Is: g:[-1,1]=>[0,\pi] defined by: g(x)=arccos(x) The value of g=-\frac{1}{2} is:
    a) \frac{2\pi}{3}
    b) -\frac{2\pi}{3}
    c) \frac{\pi}{3}
    d) -\frac{\pi}{3}
    e) \frac{\pi}{6}
    Your question asks: \cos (?) =  - \frac{1}{2}.
    The answer to that question is: \frac{{2\pi }}{3}\,\& \, - \frac{{2\pi }}{3}.
    BUT the domain of the arccosine function is \left[ {0,\pi } \right].
    So what is the correct answer?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: November 27th 2012, 06:28 AM
  2. Replies: 0
    Last Post: October 19th 2011, 05:49 AM
  3. Replies: 4
    Last Post: October 27th 2010, 06:41 AM
  4. Replies: 3
    Last Post: September 14th 2010, 03:46 PM
  5. Replies: 2
    Last Post: September 2nd 2010, 11:28 AM

Search Tags


/mathhelpforum @mathhelpforum