Inverse Trig Functions

Remember that the inverse of a relationship is found by interchanging the x’s and the y’s. Given a series of points in a relation, you can arrive at the inverse by simply switching the x’s and the y’s. If a relation is a function, its inverse is not necessarily a function and vice versa. (1,2), (3,4), (5,6), (7,8) (2,1), (4,3), (6,5), (8,7) function inverse (1,2), (2,4), (4,2), (6,8) function (2,1), (4,2), (2,4), (8,6) inverse Not a function

If you are given an equation in algebraic form, the procedure to find the inverse is outlined below. Find the inverse of 2) Switch the x’s and the y’s. 3) Solve for y. Is the inverse of 1) Changeintoy

If x is the angle and y is the value of the function at that angle then:

Are all symbols for angles just as  is a symbol for angle.  A means “the angle whose vertex is A” “means the angle whose sine is n”

We know that the graph of is a function because it passes the vertical line test.

If you are given the graph of a function, you can predict whether or not its inverse will be a function by the horizontal line test. If a horizontal line intersects the graph in more than one point, the inverse of the function WILL NOT BE A FUNCTION Clearly the inverse sine function will not be a function.

However, if we limit the domain of the sine function, we can insure that its inverse will be a function. Although there are clearly multiple intervals we could choose, by convention we choose the interval

x y 1

Graphically, the inverse of a function can be found by reflecting the graph over the y = x line.

Notice that the x and y values of the coordinates have been interchanged.

Essentially, for The input (x) is any angle and the output (y) is a value between -1 and 1 Domain: Range: The input (x) is a value between -1 and 1, and the output is an angle in the restricted range of Domain: Range:

This means find an angle whose sine is 1, within the limitations of the range of the inverse sine function. Since the limits for the range of sine are There is only one angle in this range that has a sine of 1 and that is

This means find an angle whose sine is ½, within the limitations of the range of the inverse sine function. Since the limits for the range of sine are There is only one angle in this range that has a sine of ½.

Use the calculator The calculator is in radian mode in these examples. The calculator always returns the restricted values when using the inverse functions.

Here the angle is within the restricted domain so the sin and the undo each other. Answer is Here the angle is NOT within the restricted domain. If possible, replace the angle with a different angle that has the same reference angle within the restricted domain. Now the sin and the undo each other. Answer is

Since 0.5 is between - 1 and 1, the sin undoes the inverse, the answer is 0.5. Since 1.8 is NOT between -1 and 1, the inverse of sine does not exist and the expression is undefined.

Essentially, for The input (x) is any angle and the output (y) is a value between -1 and 1 Domain: Range: The input (x) is a value between -1 and 1, and the output is an angle in the restricted range of Domain: Range:

Read: “Find me an angle whose cosine is 0 in the range of

Cosine is negative in the quadrants II and III, but only quadrant II contains angle in the restricted range. The answer is an angle in the second quadrant with a reference angle Means we are dealing with a reference angle of

a) b) c) Not in domain In domain Answer: d.  = 3.14, not in domain, undefined.

Several periods of tangent fail the horizontal line test.

We limit the function to one period, we get With limited domain as follows. Note that x is not = to b/c of asymptotes.

Essentially, for The input (x) is any angle and the output (y) is a value between -  and . Domain: Range: The input (x) is a value between -  and , and the output is an angle in the restricted range of Domain: Range: Excluding odd multiples of

It is assumed when you find the inverse sine, cosine or tangent, you find the principal value, that is the value within the limited range of the inverse function. Your calculator computes principal values only. That is, it will only return angles within the limited range of the inverse function. Examples:

A) B) C) D)

A) B) C) D)

A) B) C) D)

A) B) C) D)

OBJECTIVE 4

OBJECTIVE 5