SOLUTION: Sketch sinx and its inverse on the same set of axes for 0≤ x ≤ (pi/2) I sketched sinx but I don't know how to sketch it's inverse on the same axis, as for its inverse I know

Question 1183085: Sketch sinx and its inverse on the same set of axes for 0≤ x ≤ (pi/2)
I sketched sinx but I don't know how to sketch it's inverse on the same axis, as for its inverse I know that when x = 1, y = (pi/2) but I already put the radians unit on the x axis.
Found 2 solutions by Theo, math_tutor2020:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
putting them on the same axis is tricky, but it can be done.

we'll work in radians.

the graph of y = sin(x) is shown below.

the graph of y = sin^-1(x) is shown below:

the difficulty is that:

when you graph y = sin(x), x is the angle and y is the trig function value.

when you graph y = sin^-1(x), x is the trig function value and y is the angle.

to graph them on the same axes, let the x-axis remain as the angle.

the functions are y = sin(x) and y = sin^-1(sin(x)).

that graph looks like this:

the blue graph is y = sin(x).
the red graph is y = sin^-1(x).

when x = pi/2, y = sin(x) = 1 and y = sin^-1(sin(x)) = pi/2, as shown on the graph.

the coordinate point for y = sin^-1) is shown as (pi/2,pi/2)
that means the x-value is pi/2 and the y-value is pi/2).
the x-value is pi/2, which is the angle.

y = sin(x) is equal to 1.
the coordinate points are (pi/2,1)
this is what is shown on the blue graph.

y = sin^-1(sin(x)) is equal to sin^-1(1) which is equal to pi/2.
the coordinate points are (pi/2,pi/2)
this is what is shown on the red graph.

keep in mind that the function y = sin^-1(sin(x)) will always make the value of y equal to the value of x.

not matter what value of x you choose, y will always be equal to x.

for example, if x = pi/4, sin(pi/4) = .7071067812.

sin^-1(.7071067812) = .7853981634.
divide that by pi and it becomes .25 * pi which is the same as pi/4.

sin-1(sin(pi/4) is equal to pi/4.

sin^-1(sin(pi/2) is equal to pi/2 as shown on the graph.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

To be really honest, I think the tutor @Theo is greatly overcomplicating things.

We simply need to graph and on the same xy axis when

The graph of each function looks like this

The red graph is for the interval mentioned, and the blue graph is the result of reflecting the red curve over the dashed line y = x.

So we aren't considering the full curve because it's missing the left half.

Unfortunately, the two curves overlap nearly perfectly when x gets closer to x = 0. This could lead to confusion.

For the red sine curve, we have these properties:
Domain:
Range:

For the blue inverse sine curve, we have these properties
Domain:
Range:
The domain and range swap roles when going from the original function to the inverse.

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