Download 3.1 The Inverse Sine, Cosine, and Tangent Functions

Page 1
3.1 The Inverse Sine, Cosine, and Tangent Functions
Let’s look at f(x) = sin x
The domain is all real numbers (which will represent angles).
The range is the set of real numbers where -1 ≤ sin x ≤ 1.
However, in order for the sine function to have an inverse function, it
has to be 1-to-1.
y = sin x
1.500
1.000
(π/2, 1)
(5π/2, 1)
0.500
0.000
(0,0)
Entire
domain of
sin x
does not
pass
Horizontal
Line Test
(π, 0)
(2π, 0)
-0.500
{-π/2 ≤x ≤ π/2}
-1.000
passes Horizontal Line Test
(-π/2, -1)
-1.500
(3π/2, -1)
 π π
If we restrict the domain of y = sin x to the interval  − , 
 2 2
then it will have an inverse function. The inverse sine function is
denoted f -1 (x) = sin-1 x. The input of the inverse function is a real
number between -1 and 1, and the output of inverse sine is a real
number that is an angle in radians between –π/2 and π/2.
The domain is the set of real numbers {x| -1 ≤ x ≤ 1}.
The range is the set of real numbers such that  − π2 ≤ sin x ≤ π2 


By the property of inverse functions, for any x in the domain of sin x,
sin-1(sin (x)) = x and sin(sin-1(x)) = x
Notice that the input of the sine function is an angle, and the output of
the inverse sine function is an angle (in radians).
The inverse sine function also called the arcsin function.
NOTE: sin-1 x ≠ 1 / sin x
−1
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Example 2 on p.222
Find the exact value of sin-1 ( -½ )
What this is asking is, for what angle, θ, (where -π/2 ≤ θ ≤
π/2) does sin θ = ½ ?
Let’s look at the unit circle, x2 + y2 = 1. Remember each point on
the unit circle is (cos θ, sin θ). So find a point on the unit circle
with y-coordinate = -½
sin θ = - ½
when θ= -π/6
We can’t
choose
θ = 7π/6
because that
is not within
the range of
sin-1
Therefore sin -1 (- ½ ) = -π/6
 2
You try Exercise #19 on p.230 What is sin -1  
 2 
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?
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The Inverse Cosine Function
In order for the cosine function to be one-to-one we must restrict its
domain to {x| 0 ≤ x ≤ π}.
y = cos x
1.500
{0 ≤x ≤ π} passes
1.000
0, 1.000
-π/6, 0.866 π/6, 0.866
-π/3,
0.500
0.500
Horizontal Line Test
π/3, 0.500
-π/2, 0.000
0.000
-2π/3, -0.500
2π, 1.000
11π/6, 0.866 13π/6, 0.866
-0.500
-5π/6, -0.866
-1.000
5π/3, 0.500
π/2, 0.000
3π/2, 0.000
2π/3, -0.500
4π/3, -0.500
5π/6, -0.866 7π/6, -0.866
π, -1.000
7π/3, 0.500
5π/2, 0.000
8π/3, -0.500
17π/6, -0.866
3π, -1.000
-1.500
The inverse cosine function is denoted f -1 (x) = cos-1 x. The output
of the inverse cosine function is a real number that is an angle in
radians.
The domain (input) is the set of real numbers {x| -1 ≤ x ≤ 1}.
The range (output) is the set of real numbers such that {0 ≤ cos−1 x ≤ π }
By the property of inverse functions, for any x in the domain of cos x,
cos-1(cos (x)) = x and cos(cos-1(x)) = x
The inverse cosine function is also called the arccos function.
Example 5
Find the exact value of
 2

cos 

2


−1
Look on the unit circle of an angle between 0 and π
that gives a cosine value of 2
2
Which quadrant within that range yields a positive cos value?
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The Inverse Tangent Function
In order for the tangent function to be one-to-one we must restrict
π 
 π
its domain to  − 2 < x < 2  . Notice that x cannot equal ± π
2
because tan x is undefined at those x-values.
2
{-π/2 <x < π/2} passes
Horizontal Line Test
1
(5π/4, 1)
(9π/4, 1)
(π/4, 1)
0
π/2
-π/2
(-π/4, -1)
π
(3π/4, -1)
-1
2π
3π/2
5π/2
(7π/4, -1)
-2
Notice that the range of the tangent function is all real numbers. The
graph extends to -∞ when close to –π/2 and to +∞ when close to π/2.
Therefore, for tan -1 x
The domain is all real numbers.
π
π
The range is the set of real numbers such that  − < tan −1 x < 

2
2
By the property of inverse functions, for any x in the domain of tan x,
tan-1(tan (x)) = x and tan(tan-1(x)) = x
Example 8 on p.229
Find the exact value of tan − 1 − 3
(
)
 π
π
−1
Hint: Since range is  − < tan x < , which quadrant will yield a
2
 2
negative tan value?
Now you do Exercise #17 on p.230
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3.2
More Inverse Functions
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Example 1
Find exact value of
  5π
sin  sin 
  4
−1

 

This is a “trick” problem, because we were taught that f-1(f(x))=x
However, the x in this example, 5π/4, is not in the range of
sin-1(x). So we must choose an angle within the range of
sin-1(x) {- π/2 ≤θ ≤ π/2} that yields the same value as sin(5π/4).
Use your unit circle.
− 2
2
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θ
Example 2 Find the exact value of
 −1  1  
sin  tan   
 2 

Output of tan-1
is an angle, θ,
that has a
tan θ of ½ .
What if we can’t find an angle on the unit circle that yields a tan
value of ½? That’s OK. Draw a triangle.
tan θ = opp/adj = 1/2
We are looking for
sin θ, which is opp/hyp
Hypotnuse =
2
12 + 2 2 = 5
1
sin θ = opp/hyp =
2
2 5
=
5
5
Notice that we didn’t actually have to find θ to find sin θ!
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Example 3 Find the exact value of
 −1  1  
cos sin  −  
 3 

Pay attention to the output of sin-1. The output must be an
angle in Quadrant I or IV, and -1/3 means it must be in
Quadrant IV. Will cosine be positive? Yes.
3
adjacent = 32 − ( −1) 2 = 8 = 2 2
-1
2 2
cos θ =
3
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Remember that range of sec-1 is 0≤ θ≤π, excluding π/2. [Quadrants I
and II]
Remember that range of cot-1 is 0< θ<π [Quadrants I and II]
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HOMEWORK
p. 229
Concepts and Vocab: #7-12
Exercises p.230-231 #15-57 ETP
p.236 #9-53 EOO
Extra Credit (+2 pts)
p.231 #64 (all work must be shown)
Hint for (c) (calculator graphing): Since x is the
number of feet from the screen, and the first row is 5
feet from the screen, set Xmin at 5 and set Xmax at 50
(a theater is probably that long.) Since Y represents
the angle in degrees, set Ymin =0 and Ymax = 90.
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