Question

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

Answer 1

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]}