Question 1099758
{{{y=csc^(-1)(x)}}} (alternatively represented as {{{y=arccsc(x)}}}
is the inverse function of {{{y=csc(x)=1/sin(x)}}} .
The graph of {{{y=csc(x)=1/sin(x)}}} looks like this:
 
{{{drawing(600,200,-4,6,-5,5,
blue(line(1.571,-5,1.571,5)),locate(1.625,-2,blue(x=pi/2)),
blue(line(-1.571,-5,-1.571,5)),locate(-1.5,3,blue(x=-pi/2)),
graph(600,200,-4,6,-5,5,1/sin(x))
)}}} .

The function {{{y=csc(x)=1/sin(x)}}} is a periodic function,
like all trigonometric functions,
so the inverse can only be defined by restricting the domains for the {{{x}}} values, so as not to repeat {{{y}}} values.
That would be taking only the part of the graph between the blue lines,
to get a function defined as
{{{y=csc(x)=1/sin(x)}}} {{{for}}} {{{-pi/2<=x<=pi/2}}} {{{only}}}
(quadrants I and IV).
 
Then we, can interchange the variables and define
{{{y=csc^(-1)(x)}}} is the angle {{{y}}} ,
between {{{-pi/2}}} and {{{pi/2}}} 
such that {{{csc(y)=x}}}
That is the same approach used to define the inverse of {{{y=sin(x)}}} .

 
So,
a. The value of the function {{{y=csc^(-1)(x)}}} is an angle {{{theta}}} in the interval {{{"["}}}{{{-pi/2}}}{{{","}}}{{{pi/2}}}{{{"]"}}} or {{{highlight(-pi/2<=x<=pi/2)}}}.
 
b. What is sin^-1 supposed to mean? {{{1/sine}}} ?
{{{csc^(-1)(2)}}} is the angle {{{theta}}} , with {{{-pi/2<=x<=pi/2}}} that has
{{{csc(theta)=1/sin(theta)=(sin(theta))^(-1)=highlight(-2)}}}
and {{{sin(theta)=highlight(-1/2)}}} .
 
NOTE:
We know that in quadrant I {{{sin(pi/6)=1/2}}} (or {{{sin(30^o)=1/2}}} .
In quadrant IV, using the symmetrical quadrant I angle as reference,
we find that {{{sin(-pi/6)=-1/2}}} ,
so {{{csc^(-1)(2)=-pi/6}}} .
However, this question wants you to crawl to the answer very, very slowly.
 
c.  Is the value of sin&#952; positive, negative, or equal to 0?
In part b. we found that we are looking for an angle whose sine is {{{-1/2}}} , so we know that {{{-1/2}}} is {{{highlight(negative)}}} .
 
d. An angle {{{theta}}} that has a negative sine,
and is in the interval we agreed to in part b. 
has to be in the interval {{{0<theta<=pi/2}}}
 
e. The terminal side of angle {{{theta}}} lies in quadrant IV.
Here it is:
{{{drawing(300,300,-1.2,1.2,-1.2,1.2,grid(0),
circle(0,0,1),arrow(0,0,1.169,-0.675),
red(arc(0,0,1.62,1.62,0,30)),locate(0.65,-0.15,red(theta)),
red(triangle(0.693,-0.4,0.743,-0.4,0.693,-0.35))
)}}} It is a negative angle because it is a counterclockwise turn from the positive x-axis.