Graphing Inverse Trigonometric Functions
Key Questions

Throughout the following answer, I will assume that you are asking about trigonometry restricted to real numbers.
Using Domain of
#arc sin x# Find
#arc sin (3)# .#3# is not in the domain of#arc sin# , (#3# is not in the range of#sin# ) so#arc sin (3)# does not exist.Using Range of
#arc sin x# Find
#arc sin (1/2)# .Although there are infinitely many
#t# with#sin t = 1/2# , the range of#arc sin# restricts the value to those#t# with#(pi)/2 <= t <= pi/2# , So the value we want is#pi/6# .Using the quadrant
This is the same as using the range, but it involves thinking about the problem more geometrically.Find
#arc sin (1/2)# .#arc sin (1/2)# is a number (an angle) in quadrant I or IV. It is the#t# with smallest absolute value (the shortest path from the initial side).Again,
#arc sin (1/2) = pi/6# . 
For a trig function, the range is called "Period"
For example, the function
#f(x) = cos x# has a period of#2pi# ; the function#f(x) = tan x# has a period of#pi# . Solving or graphing a trig function must cover a whole period.The range depends on each specific trig function.
For example, the inverse function#f(x) = 1/(cos x) = sec x# has as period#2pi# .Its range varies from (+infinity) to Minimum
#1# then back to (+infinity), between (#pi/2# and#pi/2# ).Its range also varies from (infinity) to Max 1 then back to to (infinity), between (
#pi/2# and#3pi/2# ). 
Since the graphs of
#f(x)# and#f'(x)# are symmetric about the line#y=x# , start with the graph of a trigonometric function with an appropriate restricted domain, then reflect it about the line#y=x# .(Caution: Their domains must be restricted to an appropriate interval so that their inverses exist.)
Let us sketch the graph of
#y=sin^{1}x# .The graph of
#y=sinx# on#[pi/2, pi/2]# looks like:By reflecting the graph above about the line
#y=x# ,The curve in purple is the graph of
#y=sin^{1}x# .The graphs of other inverse trigonometric functions can be obtained similarly.
I hope that this was helpful.