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Math Help - inverse trig function

  1. #1
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    inverse trig function

    i'm trying to study for a test and i'm having trouble understanding how to solve these, my textbook doesn't really cover this very well. could someone please help me?

    solve for x:

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

    \arcsin(\frac{2}{x})-\arccos(\frac{\sqrt{3}}{x})=\arccos(\frac{1}{x})
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  2. #2
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    Hello yaykittyeee
    Quote Originally Posted by yaykittyeee View Post
    i'm trying to study for a test and i'm having trouble understanding how to solve these, my textbook doesn't really cover this very well. could someone please help me?

    solve for x:

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

    \arcsin(\frac{2}{x})-\arccos(\frac{\sqrt{3}}{x})=\arccos(\frac{1}{x})
    Inverse trig functions are always a pain. Get rid of them as soon as you can. Here's the first one:

    Let y = \arcsin\left(\frac{x}{2}\right). So \sin y = \frac{x}{2}

    Let z = \arccos\left(\frac{-x\sqrt3}{2}\right). So \cos z = \frac{-x\sqrt3}{2}

    And the original equation becomes:
    y+z=\pi

    \Rightarrow z = \pi -y

    \Rightarrow \cos z = -\frac{x\sqrt3}{2}= \cos(\pi-y) = -\cos y
    But \cos y = \sqrt{1 - \sin^2 y}=\sqrt{1-\frac{x^2}{4}}

    So \frac{x\sqrt3}{2}=\sqrt{1-\frac{x^2}{4}}

    Solve this for x by squaring both sides. You get a quadratic, one of whose roots is not valid. The other is the answer to the question (which is x = 1).

    Number 2 is a bit trickier, but start it in the same way. Let:
    y = \arcsin\left(\frac{2}{x}\right) i.e. \sin y = ...

    z = \arccos\left(\frac{\sqrt3}{x}\right) i.e. \cos z = ...

    w = \arccos\left(\frac{1}{x}\right) i.e. \cos w = ...
    Then say:
    y-z=w
    and:
    \frac{1}{x}=\cos w = \cos(y-z)
    =\cos y \cos z + \sin y \sin z

    = ... (substitute to get everything in terms of x)
    Then solve the equation in x.

    Do you want to have another go now?

    Grandad
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