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Thread: Two trig equations

  1. #1
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    Two trig equations

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

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

    $\displaystyle arccosx+arccos(1-x)=arccos(-x)$
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  2. #2
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    For the first problem:

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

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

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

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

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

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

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

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

    Substitute $\displaystyle z = x^2$ yielding

    $\displaystyle 16z^2 - 16z + 3 = 0$

    and solve the quadratic for z:

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

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

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

    Hence

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

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

    Using the definition of the arccos and arcsin, the only solution that works is $\displaystyle x = \frac{1}{2}$.
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  3. #3
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    Thank you very much, i understand it now. Do you have any ideas for the second equation ?
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  4. #4
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    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:

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

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

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

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

    $\displaystyle (x(1-x) + x)^2 = (1-x^2)(2x-x^2)$

    $\displaystyle (2x-x^2)^2 = (1-x^2)(2x-x^2)$

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

    $\displaystyle 2x - x^2 = 1 - x^2$

    $\displaystyle 2x = 1$

    $\displaystyle x = \frac{1}{2}$

    and since $\displaystyle \arccos 2$ is undefined, the only solution that works is $\displaystyle x = \frac{1}{2}$.
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