Question 855223
<pre>

The other tutor mistakenly took your word "exact" 
for "approximate". 

The convention for inverse tangent is:

0 < tan<sup>-1</sup>(positive_number) < {{{pi/2}}}

tan<sup>-1</sup>(0) = 0

{{{-pi/2}}} < tan<sup>-1</sup>(negative_number) < 0

Therefore in your problem:

sin[tan<sup>-1</sup>(-&#8730;<span style="text-decoration: overline">3</span>)]

ee first find the tan<sup>-1</sup>(-&#8730;<span style="text-decoration: overline">3</span>).

That asks the question:

"What angle in the fourth quadrant taken in the interval ({{{-pi/2}}},0)
has a tangent of -&#8730;<span style="text-decoration: overline">3</span>?"

That brings to mind the special angle 60° or {{{pi/3}}}, whose tangent is +&#8730;<span style="text-decoration: overline">3</span>.
So we think of an angle in quadrant IV that has 60° for its referent angle,
since angles in Q4 have negative tangents.

[We may think of 300° or {{{5pi/3}}}, even though by convention it's actually
 measured clockwise as {{{-pi/3}}}.

Anyway,

sin[tan<sup>-1</sup>(-&#8730;<span style="text-decoration: overline">3</span>)] = sin({{{-pi/3}}}) = {{{-sqrt(3)/2}}}.
It's negative since the sine is negative in QIV.

Answer:  {{{-sqrt(3)/2}}}

Edwin</pre>