Inverse Trig Functions2 (1)

7/28/2019 Inverse Trig Functions2 (1)

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Inverse Trig Functions

By

Richard Gill

Supported in Part by a Grant from

VCCS LearningWare


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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 thisis 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.


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Consider the graph of .2

xy

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.


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

inverse?

We consider only one half ofthe 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


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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 aswe did with the previous

function.

How would YOU restrict the domain?

x

y

y = sin(x)

y = 1/2


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

x

yWe are going to build

the inverse functionfrom this section of the

sine curve because:

This section picks upall the outputs of the

sine from1 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.


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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.


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2

1

32

342

2

62

10062

142

2

32

321

)(sin 1

xx

The new table generates the graph of the inverse.

1

2

2

3

3

22

4

2

1

6

002

1

6

2

2

4

2

3

3

12

)sin(

xx

To get a goodlook at the

graph of the

inverse

function, wewill turn the

tables on the

sine function.

The domain of

the chosen

section of thesine 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].


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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,

22

22,

4

to

etc.

You will see the

inverse listed as

both:

)(sin)arcsin( 1 xandx

x

yy = arcsin(x)

y = sin(x)


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In the tradition of inverse functions then we have:

32

3sin

32

3arcsin

2

3

3sin

2)1(sin2)1arcsin(12sin

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 areinstructed to use

degrees, you

should assume

that inverse trig

functions willgenerate outputs

of real numbers

(in radians).


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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 forthe cosine?

Congratulations if

you realized that

the restriction weused on the sine is

not going to work

on the cosine.


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x

yy = cos(x)

The chosen section for the cosine is in the red frame. This

section includes all outputs from1 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)


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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

,2

RandD

2,

2,

RandD

y=tan(x)y=arctan(x)


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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


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1

2

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

21/230csc630

2

330cos

630

becauseor

becauseor


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1

2

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


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1

2

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


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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

quadrantsangles like

180,902

,00,902

orororor

To solve arccos(-1) for example, youcould draw a quick sketch of the

cosine section:

x

yy = cos(x)

And observe that arccos(-1) =

1,


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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.


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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

so

2

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.


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Consider a slightly different setup:

120sinarcsin This is also thecomposition of two inversefunctions 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.


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First, some calculator problems. On most calculators, you

access the inverse trig functions by using the 2nd function

option on the corresponding trig functions. The mode buttonallows you to choose whether your work will be in degrees or

in radians.

You have to stay on top of this because the answer is not in

a format that tells you which mode you are in.

Answers and selected complete solutions can be found

after the exercises.


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Find the exact

value of each

expression

without usinga calculator.

When your

answer is an

angle, express

it in radians.

Work out the

answers

yourself before

you click.

2

3

cos.9

1sin.8

3arctan.7

2

1cos.6

0arcsin.5

3

1arctan.4

1tan.3

1arccos.2

2

1sin.1

1

1

1

1

1

2

1cossin.16

3cosarccos.15

2

1arccostan.14

270sinarcsin.13

2sinarcsin.12

2

1arccos.11

2sec.10

1

1


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Use a calculator. For 17-20,round to the nearest tenth of a

degree.

)8787.arccos(.20

)1234.arcsin(.19

)345.2arctan(.18)6666(.cos.17

1

Use a calculator. For 21-24,

express your answers inradians rounded to the nearest

hundredth.

)7878arctan(..24

2345.cos.23

)7878arcsin(..22585.3tan.21

1

1

On most calculators, you access the inverse trig functions by

using the 2nd function option on the corresponding trig functions.

The mode button allows you to choose whether your work will

be in degrees or in radians. You have to stay on top of thisbecause the answer is not in a format that tells you which mode

you are in.


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Answers for problems 1 9


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21sin.8

33arctan.7

42

1cos.6

00arcsin.5

63

1

arctan.4

41tan.3

1arccos.2

62

1sin.1

1

1

1

1

6

5

2

3cos.9 1

Answers for problems 1 9.

Negative ratios for arccos generate

angles in Quadrant II.

y

x

1 2

3

The reference angle is

so the answer is

6

6

5

66

6

6

2/1210 11


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2

3

3

2sin

2

1cossin.16

32

1arccos

3cosarccos.15

33

2tan

2

1arccostan.14

2901arcsin270sinarcsin.13

22sinarcsin.12

4

3

2

1arccos.11

32/1cos2sec.10

1

11

60

y

x

-1

23

14.

x

3

1

2

y

15.


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5.151)8787.arccos(.20

1.7)1234.arcsin(.19

9.66)345.2arctan(.18

2.48)6666(.cos.17 1

67.0)7878arctan(..24

81.12345.cos.23

91.0)7878arcsin(..22

30.1585.3tan.21

1

1

1455.4...3341.1tan2345.arccostan.308175.0...6136.0cos5758.sincos.29

5758.5758.arccoscos.28

62.0...7184.0arctan)34.2arctan(sin.27

57.0...5403.0arcsin1cosarcsin.26

44.0...4245.0arcsin58.3sinarcsin.25

1

Answers for 17 30.

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