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Thread: 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: $\displaystyle \sin x = \pm \sqrt{1 + cos^2 x}$

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

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

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


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


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

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

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

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


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

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

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

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

    remember the basic right triangle from geometry days?



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

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

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

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

    Soroban was quicker on the draw ...
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