The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation of the graph?
Y=cos^-1 x
Y=cot^-1 x
Y=sin^-1 x
Y=tan^-1 x
The answer is going to be y=sin^-1 x. I used desmos for this
well, sin(pi/2) = csc(pi/2) = 1
The correct equation for the graph passing through the point (1, pi/2) is Y = sin^-1 x.
Here's how you can determine this:
1. Recall that the inverse sine function, sin^(-1)(x) also written as asin(x) or arcsin(x), maps values between -1 and 1 to angles between -π/2 and π/2.
2. The point (1, pi/2) means that when x = 1, y = pi/2 or 90 degrees.
3. The equation Y = sin^-1 x takes a value of x (between -1 and 1) and gives the corresponding angle y (between -π/2 to π/2).
4. In this case, when x = 1, the only possible value for y is π/2.
5. Therefore, the equation Y = sin^-1 x fits the given condition and the graph passes through the point (1, pi/2).
Hence, the correct equation for the graph is Y = sin^-1 x.
To determine which of the given equations could represent the graph passing through the point (1, pi/2), we need to consider the behavior of the inverse trigonometric functions.
Let's go through each option:
1. Y = cos^(-1)(x):
The inverse cosine function has a range from 0 to pi, which means it can output values between 0 and pi. However, the given point (1, pi/2) has a y-value of pi/2, which is not in the range of the inverse cosine function. Hence, this equation cannot represent the graph passing through the given point.
2. Y = cot^(-1)(x):
The inverse cotangent function has a range from 0 to pi. The cotangent function is undefined at x = 0. Given that the point (1, pi/2) lies on the graph, which has an x-value of 1, this equation cannot represent the graph passing through the given point.
3. Y = sin^(-1)(x):
The inverse sine function has a range from -pi/2 to pi/2, which means it can output values between -pi/2 and pi/2. The point (1, pi/2) lies within this range, making it a possible candidate for the equation. However, it's important to note that the inverse sine function would return pi/2 when the input is 1, not the other way around. So, this equation does not represent the graph passing through the given point.
4. Y = tan^(-1)(x):
The inverse tangent function has a range from -pi/2 to pi/2, similar to the inverse sine function. Additionally, the tangent function has a period of pi, which means the graph passes through the point (1, pi/2) at multiple locations. Therefore, the equation Y = tan^(-1)(x) is a possible representation of the graph passing through the given point.
In summary, the equation that could represent the graph passing through the point (1, pi/2) is Y = tan^(-1)(x).