Download Inverse Trigonometric Functions

Precalculus - MAT 175
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Inverse Trigonometric Functions
1
Inverse Sine Function
Remember that to have an inverse of a function, the function must be 1 : 1. The
Sine function is obviously NOT a 1 : 1 function over it’s entire domain. But if we
"cut" a portion of the Sine that contained all of the range [ 1; 1] we could …nd the
inverse of the Sine over this restricted Domain. Let’s look at the full graph
y
1
x
-π -3π/4-π/2 -π/4
π/4 π/2 3π/4 π
-1
Now we want to choose a Domain
soi that the Range still hits everything between
h
1 and 1: We could choose
; .
2 2
h
i
;
The Domain of our Sine function is going to be restricted on
and the
2 2
Range will be [ 1; 1] : When we …nd
h the iInverse Sine Function it will have a
;
Domain of [ 1; 1] and a Range of
2 2
To graph the Inverse Sine Function arcsin x; or sin
1
x all we will need is a
few of the points on the restricted domain, say
; 1 , (0; 0) ; and
; 1 : To
2
2
plot them as the Inverse we just need to switch the Domain and Range:
; 1 ; (0; 0) , and
;1
2
2
becomes
1;
2
; (0; 0) , and 1;
2
Graphing we get:
π/2
y
x
-1
1
-π/2
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To work with the Sine and Inverse Sine in problems you must be very careful on
what "numbers" go where:
sin (angle) = ratio
arcsin (ratio) = angle
You must be very conscious whether a number is a Ratio or an Angle.
Example 1 Find the value of sin
1
p !
3
.
2
It is sometimes useful to write this as:
sin
1
p !
3
=
2
sin =
p
3
2
p
3
. The sign of the ratio
2
is positive so the angle must be in the First Quadrant (where the inverse is de…ned
to be positive). Draw
p this Triangle in the First Quadrant (margin). Labeling the
sides as Opposite = 3 and Hypotenuse = 2 we …nd that the third side must be 1.
The angle that must be is determined by this Special Triangle and the Opposite
p
Side. Since 3is the larger leg, the angle must be .
3
p !
3
=
sin 1
2
3
Now we need to …nd the angle whose ratio under Sine is
1.0.1
2
Inverse Cosine Function
Using the same logic as with the sine we will look at the graph of the cosine and
choose an interval such that the entire range is accounted for and none of the range
values repeat.
y
1
x
-π -3π/4-π/2 -π/4
π/4 π/2 3π/4 π
-1
Thus if we "cut" the domain from [0; ] we are left with the conditions needed to
…nd an inverse. Thus the inverse Cosine function will have a Domain of [ 1; 1] and
Precalculus - MAT 175
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a Range of [0; ] (Quad One and Two). We will denote this function as arccos x
or as cos 1 x:
π
y
3π/4
π/2
π/4
-1
x
1
-π/4
-π/2
-3π/4
-π
Example 2 Evaluate cos
1
p !
3
2
We know that the Cosine is Negative in the Second and Third Quadrants, but the
cos 1 x function only works in the First and Second Quadrants. So this angle must
lie in the Second Quad (because Cosine is negative in the Second Quad).
p !
p
p
3
3
1
Let: cos
= then we know that cos ( ) =
where 3 is the
2
2
Adjacent Side and 2 is the Hypotenuse. Since this angle lies in the Second Quadrant
we will need to use a Reference Angle, say inside the Triangle. (see margin).
p
The special triangle whose adjacent side measures 3 and has a hypotenuse of 2 is
5
the 30-60-90 and the angle must be
= (30 degrees) and angle =
so:
6
6
p !
3
5
=
cos 1
2
6
3
Tangent Inverse Function
To …nd the Inverse Tangent Function we will …rst Restrict the Domain from
;
:
2 2
Which seems natural since these endpoints are the vertical Asymptotes to the graph
of Tangent. Thus the Inverse Tangent Function will have a Domain of ( 1; 1)
and a Range of
;
(angles in the First and Fourth Quadrant)
2 2
Example 3 Evaluate
arctan
p !
3
3
The only Quadrants that Tangent is Negative in are Quad Two and Four, but the
Inverse Function is only de…ned in Quad Four so we will draw our triangle in p
this
O
3
Quadrant and label the sides according to the reference angle ; tan =
=
:
A
3
Before we go with these numbers, I don’t remember any special Triangles with a side
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p
of 3 and 3; so this might be a rationalized denominator of one of them.
"unrationalize" by rationalizing the Numerator and see what we get:
p p
3
3
3
1
p = p =p
3
3
3 3
3
: These numbers seem familiar so lets use them instead:
O
=
A
Let’s
1
p
3
p
Thus the angle is one where the Opposite Side is 1 and the Adjacent Side is 3:(see
margin) These are the sides of a 30-60-90 and = 30 = . Now is just the
6
Reference Angle. Since the angle is in the Fourth Quadrant, the solution must be
Negative (Clockwise Rotation) and we get:
p !
3
=
arctan
3
6
Review of Ideas:
1) The Inverse Sine works in the First and Fourth Quadrants.
2) The Inverse Cosine works in the First and Second Quadrants
(always use reference angle in Second).
3) The Inverse Tangent works in the First and Fourth Quadrants.
4) Always draw and label "Reference" Triangle in the appropriate Quadrant.
5) Determine the measure of the Angle. Remember sometimes
that this value could just be a Reference angle that is related to the
Angle looked for.
4
Advanced Problems Involving Inverse Trigonometry
Example 4 Evaluate: tan
1
tan
2
3
.
It sometimes is nice to realize what your answer should come out to be. The
2
is
3
2
’s answer is a Ratio, and the tan 1 (Ratio) = Angle. So
3
our answer needs to be an angle. Since we are working with the Inverse Tangent
Function, we need to only search in the First and Fourth Quadrants for an answer.
The Tangent of an angle in the Second Quadrant will be negative, so we will need
the Inverse Tangent Quadrant that is also Negative, the Fourth. Thus the answer
must be:
2
tan 1 tan
=
3
3
the angle, the tan
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Example 5 Evaluate: sin cos
3
5
1
.
The Ratio is negative, so that tells us that the Angle must be in the Second Quadrant
(where the Inverse Cosine works and is negative). Now we can draw the Triangle
A
3
in this quadrant, label the Reference Angle , and the sides according to :
=
H
5
(see margin). Using Pythagorean Theorem we …nd that the Opposite Side is 4.
Now we can …nd the Sine of this Angle:
sin cos
1
3
5
=
4
5
The answer is Positive since Sine is Positive in the Second Quadrant.
This next example is very useful in the study of Inverse Trigonometric Functions in
Calculus.
Example 6 Evaluate cos sin
1
(x)
where x 2 [0; =2]
O
x
We need a Ratio to work with, so let the Ratio x = = . Now if we draw the
1
H
…rst quadrant angle and label the sides, we …nd that the Pythagorean Theorem will
be necessary to …nd the Adjacent Side (see margin) Thus:
c2 = a2 + b2
12 = x2 + b2
1
x2 = b2
p
b= 1
x2
And the Cosine of the angle here would give the answer of:
p
p
1 x2
1
cos sin (x) =
= 1 x2
1