Results 1 to 3 of 3

Thread: why is cos(t) -1 as opposed to 1 when P = 3pi

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    12

    why is cos(t) -1 as opposed to 1 when P = 3pi

    find the coordinates of P and the exact values of the trigonomic functions at t whenever possible.

    given

    My answers
    Sin(t) = 0 csc(t)= 1/0 or Undefined
    cos(t) = 1 sec(t) = 1/1 or 1
    tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

    The book

    Sin(t) = 0 csc(t)= 1/0 or Undefined
    cos(t) =- 1 sec(t) = - 1/1 or - 1
    tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined


    Given that the the point is why is cos(t) -1 as opposed to 1?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Quote Originally Posted by rasczak View Post
    Given that the the point is why is cos(t) -1 as opposed to 1?
    Because $\displaystyle -3\pi$ is equivalent to $\displaystyle \pi$
    Two numbers, $\displaystyle t~\&~s$, are equivalent if $\displaystyle t=s+2n\pi$ for some integer $\displaystyle n$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by rasczak View Post
    find the coordinates of P and the exact values of the trigonomic functions at t whenever possible.

    given

    My answers
    Sin(t) = 0 csc(t)= 1/0 or Undefined
    cos(t) = 1 sec(t) = 1/1 or 1
    tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined

    The book

    Sin(t) = 0 csc(t)= 1/0 or Undefined
    cos(t) =- 1 sec(t) = - 1/1 or - 1
    tan(t) = 0/1 or 0 cot(t)= 1/0 or Undefined


    Given that the the point is why is cos(t) -1 as opposed to 1?
    Hi rasczak,

    A rotation of $\displaystyle \pi$ radians is an anti-clockwise rotation of $\displaystyle 180^o$ or a half circle.

    A rotation of $\displaystyle 2{\pi}$ is a full circle, back to the starting point.

    Therefore, a rotation of $\displaystyle 3{\pi}$ brings you to the same position that a rotation of $\displaystyle \pi$ does.

    This position is the extreme left of the unit circle.
    The point here is (-1,0).

    The horizontal co-ordinate is $\displaystyle Cos(angle)$ and the vertical co-ord is $\displaystyle Sin(angle)$ for any point on the circumference of the unit circle centred at (0,0).

    If the rotation is done in a clockwise direction, we refer to this using negative angles.

    Hence, a rotation of $\displaystyle 3{\pi}$ ends up at the same position that a rotation of $\displaystyle -3{\pi}$ ends up in.

    This corresponds to an angle $\displaystyle \pi$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: Aug 3rd 2010, 01:31 PM
  2. "Fix" an element, as opposed to "consider" an element?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Mar 3rd 2010, 02:10 AM
  3. Replies: 3
    Last Post: Feb 23rd 2010, 04:54 PM

Search Tags


/mathhelpforum @mathhelpforum