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134. In radian mode, graph both y = sin-' x and y = I - sin' x on the same coordinate-axis system. What do you notice?

User Gssi
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Answer:

Explanation:

To graph the functions y = sin^(-1) x and y = π - sin^(-1) x on the same coordinate-axis system, we can use the properties of the inverse sine function and the unit circle.

The inverse sine function, denoted as sin^(-1) x or arcsin x, is the inverse of the sine function. This means that if we input a value x into the inverse sine function, it will output the angle θ in radians that satisfies the equation sin θ = x.

The unit circle is a circle with radius 1 centered at the origin of a coordinate plane. The angles in the unit circle are measured in radians. On the unit circle, the x-coordinate of a point is equal to the cosine of the angle formed by the point and the positive x-axis, and the y-coordinate is equal to the sine of the angle.

Given this information, we can graph the functions y = sin^(-1) x and y = π - sin^(-1) x as follows:

For y = sin^(-1) x, we can input values of x into the inverse sine function and plot the resulting values of y on the coordinate plane. Since the range of the inverse sine function is -π/2 ≤ y ≤ π/2, we can start by plotting the points (-1, -π/2), (1, π/2), and (0, 0).

For y = π - sin^(-1) x, we can input values of x into the inverse sine function, subtract the resulting values from π, and plot the resulting values of y on the coordinate plane. Since the range of the inverse sine function is -π/2 ≤ y ≤ π/2, we can start by plotting the points (-1, π/2), (1, -π/2), and (0, π).

On the same coordinate-axis system, the graph of y = sin^(-1) x will be a curve that starts at (-1, -π/2), goes through (0, 0), and ends at (1, π/2). The graph of y = π - sin^(-1) x will be a curve that starts at (-1, π/2), goes through (0, π), and ends at (1, -π/2).

If we connect the points on the two curves, we will see that they intersect at (0, π/2) and (0, -π/2). This means that the two functions are reflections of each other about the x-axis.

User Ashh
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