Graphs and Inverses of Trigonometric Functions - Pre-Calculus

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.

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 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 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.

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:

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:

The two graphs help provide a visual. The domain is all possible x values, and the range is all possible y values. The arctan function exchanges all x and y values of the tangent function. However, in order to continue being a function (by passing the horizontal line test), the arctangent function's range is restricted to . It is not a coincidence that the range of the tangent function is equal to the domain of the arctangent function because of the nature of inverse relationships.

domain: , such that

range:

domain:

range:

You can see that while the range and the domain match, the domain and the range are not exact matches; this is so that can be a function (by being restricted).

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, .

If you look at a graph, you can see that the period (length of one wave) is . Without the graph, you can divide with the frequency, which in this case, is 1.

To determine the value of , solve each of the terms first.

The inverse cosine has a domain and range restriction.

The domain exists from , and the range from . The inverse cosine asks for the angle when the x-value of the existing coordinate is . The only possibility is since the coordinate can only exist in the first quadrant.

The inverse sine also has a domain and range restriction.

The domain exists from , and the range from . The inverse sine asks for the angle when the y-value of the existing coordinate is . The only possibility is since the coordinate can only exist in the first quadrant.

Therefore:

:

There is a restriction for the range of the inverse tangent function from .

The inverse tangent of a value asks for the angle where the coordinate lies on the unit circle under the condition that . For this to be valid on the unit circle, the must be very close to 1, with an value also very close to zero, but cannot equal to zero since would be undefined.

The point is located on the unit circle when , but is invalid due to the existent asymptote at this angle.

An example of a point very close to that will yield can be written as:

Therefore, the approximated rounded value of is .

Rewrite and evaluate .

The inverse sine of one-half is since is the y-value of the coordinate when the angle is .

To convert from radians to degrees, replace with 180.

For this particular problem we need to recall that the inverse cosine cancels out the cosine therefore,

.

So the expression just becomes

From here, recall the unit circle for specific angles such as .

Thus,

.

This one seems complicated, but becomes considerably easier once you implement the fact that the composite cancels out to 1 and you are left with which is equal to 1

This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which always

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of (the adjacent side over the opposite side)

This one seems complicated but becomes considerably easier once you implement the fact that the composite cancels out to and you are left with which is equal to , and so the answer is .