Question 771979
<pre>
sin[cos<sup>-1</sup>({{{-1/3}}})]

First we find the angle cos<sup>-1</sup>({{{-1/3}}})

The inverse cosine of a negative number is in quadrant II
between 90° or {{{pi/2}}} and 180° or <font face="symbol">p</font>.

We'll draw the picture of the angle cos<sup>-1</sup>({{{-1/3}}}) in quadrant
II.

Since the cosine is {{{adjacent/hypotenuse}}} = {{{x/r}}} we take x
as the numerator of {{{(-1)/3}}}, x=-1, and the r as the denominator,
3, r=3:

{{{drawing(400,400,-3,3,-3,3, red(arc(0,0,1.5,-1.5,0,110)),
locate(.4,1, Cos^(-1)), locate(1.1,1.1,(-1/3)), locate(-.7,1.6,r=3),
graph(400,400,-3,3,-3,3), green(line(-1,0,-1,sqrt(8)),line(0,0,-1,sqrt(8))),locate(-.9,.3,x=-1)  )}}}

Then we calculate y by the Pythagorean theorem 

   x² + y² = r²
(-1)² + y² = 3²
    1 + y² = 9
        y² = 8
         y = &#8730;<span style="text-decoration: overline">8</span>
         y = &#8730;<span style="text-decoration: overline">4·2</span>
         y = 2&#8730;<span style="text-decoration: overline">2</span>

{{{drawing(400,400,-3,3,-3,3, red(arc(0,0,1.5,-1.5,0,110)),
locate(.4,1, Cos^(-1)), locate(1.1,1.1,(-1/3)), locate(-.7,1.6,r=3),
graph(400,400,-3,3,-3,3), green(line(-1,0,-1,sqrt(8)),line(0,0,-1,sqrt(8))),locate(-.9,.3,x=-1), locate(-1.7,1.6,y=2sqrt(2))   )}}}

Therefore sin[cos<sup>-1</sup>({{{-1/3}}})] = {{{opposite/hypotenuse}}} = {{{y/r}}} = {{{2sqrt(2)/3}}}.   

Edwin</pre>