Results 1 to 6 of 6

Math Help - Inverse trigonometric equation

  1. #1
    Junior Member Hardwork's Avatar
    Joined
    Sep 2010
    Posts
    36

    Inverse trigonometric equation

    How many \displaystyle x\in\mathbb{R} exist such that \displaystyle \arcsin{x} = \arccos{x} ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Just one.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Hardwork's Avatar
    Joined
    Sep 2010
    Posts
    36
    Quote Originally Posted by Also sprach Zarathustra View Post
    Just one.
    How do you show that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    There is a simple way:

    Draw the graphs for the arcsin(x) & arccos(x),
    The x-coordinate of the point of intersection is the solution.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Hardwork's Avatar
    Joined
    Sep 2010
    Posts
    36
    Quote Originally Posted by General View Post
    Draw the graphs for the arcsin(x) & arccos(x),
    The x-coordinate of the point of intersection is the solution.
    Ok. I know that much. But let's say that we want to find x. How do we go about that?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,115
    Thanks
    992
    Quote Originally Posted by Hardwork View Post
    Ok. I know that much. But let's say that we want to find x. How do we go about that?
    it's just a matter of being familiar with both functions.

    y = \arcsin(x) has range -\frac{\pi}{2} \le y \le \frac{\pi}{2}

    y = \arccos(x) has range 0 \le y \le \pi

    the ranges intersect in quad I ... the angle in quad I that has the same sine and cosine value is y = \frac{\pi}{4}


    if you insist on doing it algebraically ...

    \arcsin(x) = \arccos(x)

    \sin[\arcsin(x)] = \sin[\arccos(x)]

    x = \sqrt{1 - x^2}

    x^2 = 1 - x^2

    2x^2 - 1 = 0

    (\sqrt{2} x - 1)(\sqrt{2} x + 1) = 0

    x = \frac{1}{\sqrt{2}} ... the negative solution is invalid since

    y = \arcsin\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}

    and

    y = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}


    \arcsin\left(\frac{1}{\sqrt{2}}\right) = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: August 29th 2010, 06:23 PM
  2. Replies: 2
    Last Post: April 28th 2009, 07:42 AM
  3. Inverse Trigonometric Functions
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 20th 2009, 03:16 AM
  4. Inverse trigonometric function
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 15th 2008, 06:08 AM
  5. Solve the inverse trigonometric equation?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 24th 2008, 08:29 AM

Search Tags


/mathhelpforum @mathhelpforum