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Math Help - evaluate arccosx

  1. #1
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    evaluate arccosx

    evaluate sin (arccos x) , 0 < x< 1
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  2. #2
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    Note that: \sin x = \pm \sqrt{1 + cos^2 x}

    \sin (\arccos x) = +\sqrt{1 - \left[\cos (\arccos x)\right]^{2}} (since 0 \leq x \leq 1 \: \: \Rightarrow \: \: 0 \leq \arccos x \leq \frac{\pi}{2})

    Can you finish off?
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  3. #3
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    Hello, cruxkitty!

    {\color{red}\rlap{////////}}\text{Evaluate: }\;\sin(\arccos x),\quad 0 < x< 1 . They meant to say "Simplify".
    Recall that an inverse trig expression is an angle.


    \text{We have: }\;\sin\underbrace{(\arccos x)}_{\theta}
    . . . . . .
    angle whose cosine is x


    "The angle whose cosine is x" looks like this:
    Code:
                            *
                         *  *
                  1   *     *  ____
                   *        * √1-x
                *           *
             * θ            *
          *  *  *  *  *  *  * 
                   x

    \cos\theta \:=\:\frac{x}{1} \:=\:\frac{adj}{hyp} \quad\Rightarrow\quad adj = x,\;\;hyp = 1

    From Pythagorus, we find that: . opp = \sqrt{1-x^2}

    Hence: . \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}


    Therefore: . \boxed{\sin(\arccos x) \;=\;\sqrt{1-x^2}}

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  4. #4
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    \sin(\arccos{x})

    let ...
    \theta = \arccos{x}

    so ...
    \cos{\theta} = \frac{x}{1}

    remember the basic right triangle from geometry days?



    \cos{\theta} = \frac{x}{1} = \frac{adj}{hyp}

    the adjacent side has length x and the hypotenuse has length 1.

    using Pythagoras, the opposite side has length = \sqrt{1-x^2}

    \sin(\arccos{x}) = \sin{\theta} = \frac{opp}{hyp} = \frac{\sqrt{1-x^2}}{1}

    Soroban was quicker on the draw ...
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