(i) Please fill the table with your preferred trigonometric function, denoted as Y = f(X) and its corresponding inverse function, denoted as f-1(Y). X    | 0 | pi/3 | 2pi/3 | pi/2 | pi | 4pi/3 | 2pi ----------------------------------------------------- Y=f(X) |   |    |     |    |   |     | ----------------------------------------------------- f^-1(Y)|   |    |     |    |   |     | (ii) Utilize the graphing tool to depict both functions Y = f(X) and f-1(Y) on the same graph. Just tell me how to do that on geogebra, preferably step by step (iii) Observe the tabular values and the graphs of Y = f(X) and f-1(Y), answer the following: Examine the periodicity of the chosen trigonometric function, Y = f(X), as presented in the table. Discuss the domain and range of both Y = f(X) and f-1(Y). Determine whether the function Y=f(X) demonstrates even, odd, or neither.

(i) Let's choose the sine function as our trigonometric function. The inverse of the sine function is the arcsine function. Here is the completed table:

X    | 0 | pi/3 | 2pi/3 | pi/2 | pi | 4pi/3 | 2pi

Y=f(X) | 0 | 1/2 | √3/2 | 1 | 0 | -√3/2 | 0

f^-1(Y)| 0 | pi/6 | pi/3 | pi/2 | 0 | -pi/3 | 0

(ii) To graph these functions in GeoGebra, follow these steps:

1. Open GeoGebra and select "Graphing" from the main menu.
2. In the input bar at the bottom, type "y = sin(x)" and press Enter. This will graph the sine function.
3. Next, type "y = asin(x)" and press Enter. This will graph the arcsine function.
4. You can adjust the view of the graph by using the zoom buttons or by dragging the graph.

(iii) Observations:

• The sine function, Y = f(X), is periodic with a period of 2π. This is evident from the table as the values of Y repeat after X increases by 2π.
• The domain of Y = f(X) is all real numbers, and the range is [-1, 1]. The domain of f^-1(Y) is [-1, 1], and the range is [-π/2, π/2].
• The sine function Y=f(X) is neither even nor odd. An even function is symmetric about the y-axis, and an odd function is symmetric about the origin. The sine function does not satisfy either of these conditions.