6.8

Trigonometric Inverses and Their Graphs

Let us begin with a simple question:

xxf

xxf

)(

)(1

2

What is the first pair of inverse functions that pop into YOUR mind?

This may not be your pair but this is a famous pair. But something is not quite right with this pair. Do you know what is wrong?

Congratulations if you guessed that the top function does not really have an inverse because it is not 1-1 and therefore, the graph will not pass the horizontal line test.

Consider the graph of .2xy

x

y

Note the two points on the graph and also on the line y=4.

f(2) = 4 and f(-2) = 4 so what is an inverse function supposed to do with 4?

?2)4(2)4( 11 forf

By definition, a function cannot generate two different outputs for the same input, so the sad truth is that this function, as is, does not have an inverse.

So how is it that we arrange for this function to have an inverse?

We consider only one half of the graph: x > 0.

The graph now passes the horizontal line test and we do have an inverse:

xxf

xforxxf

)(

0)(1

2

Note how each graph reflects across the line y = x onto its inverse.

xy

x

4

y=x

2xy

2

A similar restriction on the domain is necessary to create an inverse function for each trig function.

Consider the sine function.

You can see right away that the sine function does not pass the horizontal line test.

But we can come up with a valid inverse function if we restrict the domain as we did with the previous function.

How would YOU restrict the domain?

x

yy = sin(x)

y = 1/2

Take a look at the piece of the graph in the red frame.

x

yWe are going to build the inverse function from this section of the sine curve because:This section picks up all the outputs of the sine from –1 to 1.This section includes the origin. Quadrant I angles generate the positive ratios and negative angles in Quadrant IV generate the negative ratios.

Lets zoom in and look at some key points in this section.

x

yy = sin(x)

12

2

3

3

2

2

4

2

1

6

002

1

6

2

2

4

2

3

3

12

)(

xfx

I have plotted the special angles on the curve and the table.

21

32

342

262

10062

142

232

32

1

)(sin 1

xx

The new table generates the graph of the inverse.

12

2

3

3

2

2

4

2

1

6

002

1

6

2

2

4

2

3

3

12

)sin(

xx

To get a good look at the graph of the inverse function, we will “turn the tables” on the sine function.

The domain of the chosen section of the sine is

So the range of the arcsin is

2,

2

2,

2

The range of the chosen section of the sine is [-1 ,1] so the domain of the arcsin is [-1, 1].

Note how each point on the original graph gets “reflected” onto the graph of the inverse.

2,11,

2

to

3

,2

3

2

3,

3

to

4

,2

2

2

2,

4

to

etc.

You will see the inverse listed as both:

)(sin)arcsin( 1 xandx

x

yy = arcsin(x)

y = sin(x)

In the tradition of inverse functions then we have:

32

3sin

32

3arcsin

2

3

3sin

2)1(sin

2)1arcsin(1

2sin

1

1

or

or

The thing to remember is that for the trig function the input is the angle and the output is the ratio, but for the inverse trig function the input is the ratio and the output is the angle.

Unless you are instructed to use degrees, you should assume that inverse trig functions will generate outputs of real numbers (in radians).

The other inverse trig functions are generated by using similar restrictions on the domain of the trig function. Consider the cosine function:

x

yy = cos(x)What do you

think would be a good domain restriction for the cosine?Congratulations if you realized that the restriction we used on the sine is not going to work on the cosine.

x

yy = cos(x)

The chosen section for the cosine is in the red frame. This section includes all outputs from –1 to 1 and all inputs in the first and second quadrants.

Since the domain and range for the section are the domain and range for the inverse cosine are

,1,1,0 and .,01,1 and

x

yy = arccos(x)

x

y

x

y

The other trig functions require similar restrictions on their domains in order to generate an inverse.

Like the sine function, the domain of the section of the

tangent that generates the arctan is .

2,

2

,

2,

2RandD

2,

2,

RandD

y=tan(x)

y=arctan(x)

The table below will summarize the parameters we have so far. Remember, the angle is the input for a trig function and the ratio is the output. For the inverse trig functions the ratio is the input and the angle is the output.

arcsin(x) arccos(x) arctan(x)

Domain

Range

11 x 11 x x

22

x

20

x

22

x

When x<0, y=arccos(x) will be in which quadrant?

When x<0, y=arcsin(x) will be in which quadrant?

When x<0, y=arctan(x) will be in which quadrant?

y<0 in IV y>0 in II

y<0 in IV

12

3

30

60

45

45 12

2

The graphs give you the big picture concerning the behavior of the inverse trig functions. Calculators are helpful with calculations (later for that). But special triangles can be very helpful with respect to the basics.

Use the special triangles above to answer the following. Try to figure it out yourself before you click.

)2(csc

2

3arccos

1

21/230csc6

30

2

330cos

630

becauseor

becauseor

12

3

30

60

45

45 12

2

OK, lets try a few more. Try them before you peek.

2

1arcsin

)3(tan

2

1arcsin

1

2

145sin)

4(45

31

360tan)

3(60

2

145sin)

4(45

becauseor

becauseor

becauseor

12

3

30

60

Negative inputs for the arccos can be a little tricky.

2

1arccos

From the triangle you can see that arccos(1/2) = 60 degrees. But negative inputs for the arccos generate angles in Quadrant II so we have to use 60 degrees as a reference angle in the second quadrant.

2

1120cos:

12060180

r

xcheckto

60

y

x

-1

23

You should be able to do inverse trig calculations without a calculator when special angles from the special triangles are involved. You should also be able to do inverse trig calculations without a calculator for quadrantal angles.

Its not that bad. Quadrantal angles are the angles between the quadrants—angles like

180,902

,00,902

orororor

To solve arccos(-1) for example, you could draw a quick sketch of the cosine section:

x

yy = cos(x)

And observe that arccos(-1) = 1,

But a lot of people feel comfortable using the following sketch and the definitions of the trig ratios.For arccos(-1) for example, you can observe that, since

the point (-1, 0) is the one we want. That point is on the terminal side of

r

xcos

.

Or for arccot(0), you can observe that, since the point (0, 1)

is the one we want. That point is on the terminal side of 90 degrees.

y

xcot

.)1arccos(

,11

1)cos(

r

xSo, since

x

y (0, 1)

(1, 0)

(-1, 0)

(-1, 0)

r = 1

.90So, arccot(0)

Good luck getting that answer off of a calculator.

Finally, we encounter the composition of trig functions with inverse trig functions. The following are pretty straightforward compositions. Try them yourself before you click to the answer.

?2

3sinsin 1

First, what do we know about

?

We know that is an angle whose sine is .2

3

so2

3sin

2

3sinsin 1

Did you suspect from the beginning that this was the answer because that is the way inverse functions are SUPPOSED to behave? If so, good instincts but….

Consider a slightly different setup:

120sinarcsin This is also the composition of two inverse functions but…

.602

3arcsin

Did you suspect the answer was going to be 120 degrees? This problem behaved differently because the first angle, 120 degrees, was outside the range of the arcsin. So use some caution when evaluating the composition of inverse trig functions.

The remainder of this presentation consists of practice problems, their answers and a few complete solutions.

Ex:

• Write the equation for the inverse of y = Arctan(2x) and then graph it.

Find each value.

2

2sinArc

22

2coscos 1

True or False for all values of x?

xxTan )(tan1