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Thread: 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:

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

    $\displaystyle \arcsin(\frac{2}{x})-\arccos(\frac{\sqrt{3}}{x})=\arccos(\frac{1}{x})$
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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:

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

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

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

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

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

    $\displaystyle \Rightarrow z = \pi -y$

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

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

    Solve this for $\displaystyle x$ 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 $\displaystyle x = 1$).

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

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

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

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

    Do you want to have another go now?

    Grandad
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