We know that the parent function of  has asymptotes at  where  is any integer.  Considering  we can set the asymptotic equation equal to this one and solve for  to see if  is an integer.  If  is an integer, then there is a vertical asymptote here.

So when , then

, so the given statement is true.

The two graphs help provide a visual. The domain is all possible x values, and the range is all possible y values. The arcsin function turns the sine function on its side, but, in order to continue being a function (by passing the horizontal line test), the arcsine function's range is restricted to . It is not a coincidence that the range of the sine function is equal to the domain of the arcsin function because of the nature of inverse relationships.

y=sin x

y=arcsin x

sin(x) domain: 

sin(x) range: 

arcsin(x) domain: 

arcsin(x) range: 

This is true. Each of the inverse trig functions fails to pass the vertical line test, and is therefore not a function. For this reason, you will also see graphs of the inverse trig functions that restrict the domain such that is can be graphed as a function.

To find the inverse of a function, exchange the values of x and y, then solve for y.

Change y = sin(x) to x = sin(y). Then solve for y:

x = sin(y) 

sin^-1(x) = sin^-1(sin(y))

sin^-1(x) = y

Now graph each function:

y = sin(x)

y = sin^-1(x)

If you look at the upper graph and restrict the domain to the interval [-/2, /2], you can more clearly see the relationship between the two graphs. 

To further illustrate the relationship between the two functions, examine the inputs and outputs of the two functions on the interval [-/2, /2]: 

y = sin(x)

y = sin^-1(x)

The cotangent function is negative in quadrants II and IV, so arccot (−½) could fall in either of these quadrants. The below image shows where each function is positive. Any that are not noted are negative. Since cotangent is positive in Quadrants I and III, it is negative in Quadrants II and IV.

This is the graph of arcsec(x). Think of a few points of the function sec(x): (-, -1), (0,1), (, -1). Now, interchange the x and y values to yield: (-1, -), (1, 0), (-1, ). Notice that (-1, -) and (1, 0) both exist on this graph. (-1, ) does not exist on this graph because in order to make this a function, the domain was required to be restricted such that it would pass the vertical line test. You can see how the graph of arcsec(x) relates to the graph of sec(x) by comparing the two:

The sine function is negative in quadrants III and IV, so arcsin (−½) could fall in either of these quadrants. The below image shows where each function is positive. Any that are not noted are negative. Since sine is positive in Quadrants I and II, it is negative in Quadrants III and IV.

This is the graph of arccot(x). Think of a few characteristics of the function cot(x): it has a point at (-/2, 0), and vertical asymptotes at x=-, x=0, and x=. Notice that in the graph above, we have the point (0, -/2), which flips the x and y values of the point on cot(x), and we can also see horizontal asymptotes at both y=0 and y=. We don't also have another horizontal asymptote at y=- because we needed to restrict the domain of arccot(x) in order for it to be a function and pass the vertical line test. You can look at the graph of cot(x) below and compare the two and see their similarities:

To solve , first, let . Then, . Because this value is positive, we know that this must be in Quadrant I or Quadrant II, but given the instructions, we can assume A is in Quadrant I.

Next, recall that .

Using the Pythagorean Theorem, you can solve for the following:

x² + y² = r²

x² + 12² = 13²

x = 5

Therefore, . Now, solve: 

Therefore, 

To solve , first, let . Then, . Because this value is positive, we know that this must be in Quadrant I or Quadrant IV, but given the instructions, we can assume A is in Quadrant I.

Next, recall that . Therefore

By definition, , assuming that , since the cosine function can only output values within this range.

Therefore, .