1. ## evaluate arccosx

evaluate sin (arccos x) , 0 < x< 1

2. 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?

3. 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-x²
*           *
* 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: . $\boxed{\sin(\arccos x) \;=\;\sqrt{1-x^2}}$

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

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 ...