You can put this solution on YOUR website! First calculate the arctan(-sqrt(3))
arctan(-sqrt(3)) = -60
then calculate the sin
sin(-60) = -0.86602540378443864676 is approx -0.87
The other tutor mistakenly took your word "exact"
for "approximate".
The convention for inverse tangent is:
0 < tan-1(positive_number) <
tan-1(0) = 0
< tan-1(negative_number) < 0
Therefore in your problem:
sin[tan-1(-√3)]
ee first find the tan-1(-√3).
That asks the question:
"What angle in the fourth quadrant taken in the interval (,0)
has a tangent of -√3?"
That brings to mind the special angle 60° or , whose tangent is +√3.
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 , even though by convention it's actually
measured clockwise as .
Anyway,
sin[tan-1(-√3)] = sin() = .
It's negative since the sine is negative in QIV.
Answer:
Edwin
