Results 1 to 4 of 4

Math Help - Two trig equations

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    18

    Two trig equations

    I ran into problems with these two trig equations, can someone help me ?

    arccosx-arcsinx=arccos\frac{\sqrt{3}}{2}

    arccosx+arccos(1-x)=arccos(-x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    For the first problem:

    \arccos x - \arcsin x = \arccos \frac{\sqrt{3}}{2}

    \cos (\arccos x - \arcsin x) = \frac{\sqrt{3}}{2}

    \cos (\arccos x) \cos (\arcsin x) + \sin (\arccos x) \sin (\arcsin x) = \frac{\sqrt{3}}{2}

    x (\sqrt{1 - x^2}) + x (\sqrt{1 - x^2}) = \frac{\sqrt{3}}{2}

    2x(\sqrt{1 - x^2}) = \frac{\sqrt{3}}{2}

    4x^2(1 - x^2) = \frac{3}{4}

    -4x^4 + 4x^2 = \frac{3}{4}

    4x^4 - 4x^2 + \frac{3}{4} = 0

    Substitute z = x^2 yielding

    16z^2 - 16z + 3 = 0

    and solve the quadratic for z:

    z = \frac{16 \pm \sqrt{256 - 4(16)(3)}}{32}

    z = \frac{16 \pm \sqrt{64}}{32}

    z = \frac{16 \pm 8}{32} = \frac{1}{4}, \frac{3}{4}

    Hence

    x^2 = \frac{1}{4}, \frac{3}{4}

    x = \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}

    Using the definition of the arccos and arcsin, the only solution that works is x = \frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    18
    Thank you very much, i understand it now. Do you have any ideas for the second equation ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by ldx2 View Post
    Thank you very much, i understand it now. Do you have any ideas for the second equation ?
    It's basically the same drill. Take the cosine of both sides:

    \cos(\arccos x + \arccos (1-x)) = -x

    \cos(\arccos x)\cos(\arccos(1-x)) - \sin(\arccos x)\sin(\arccos(1-x)) = -x

    x(1-x) - \sqrt{(1-x^2)(2x-x^2)} = -x

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

    (x(1-x) + x)^2 = (1-x^2)(2x-x^2)

    (2x-x^2)^2 = (1-x^2)(2x-x^2)

    Now x = 2 satisfies this equation. We can look for other solutions by dividing by 2x - x^2 if x \neq 2:

    2x - x^2 = 1 - x^2

    2x = 1

    x = \frac{1}{2}

    and since \arccos 2 is undefined, the only solution that works is x = \frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Equations
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: December 17th 2008, 11:10 PM
  2. Trig Equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 3rd 2008, 06:26 PM
  3. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 05:50 PM
  4. Trig Equations with Multiple Trig Functions help
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 6th 2008, 05:48 PM
  5. Trig Equations with Multiple Trig Functions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 6th 2008, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum