How do you graph #y=sin^-1(x-2)# over the interval #1<=x<=3#?

1 Answer
Dec 22, 2017

Please see below.

Explanation:

As #y=sin^(-1)x# means #x=siny#, the graph is similar to graph of #y=sinx# but wave is formed along #y#-axis. The range of #y=sin^(-1)x# is, however, #[-pi/2,pi/2]# and hence its graph is limited between #[-pi/2,pi/2]# along #y#-axis and corresponding values of #x# ranges from #[-1,1]#,

other values are #(-1,-pi/2),(-sqrt3/2,-pi/3),(-1/sqrt2,-pi/4),(-1/2,-pi/6),(0,0),(1/2,pi/6),(1/sqrt2,pi/4),(sqrt3/2,pi/3),(1,pi/2)#.

The graph appears as follows:

graph{arcsinx [-10, 10, -5, 5]}

However, in #y=sin^(-1)(x-2)# #x# can take values from #[1,3]# and hence the graph is similar to that of #y=sin^(-1)x# but shifted #2# units to right. The graph appears as follows:

graph{arcsin(x-2) [-10, 10, -5, 5]}