Download Section 7.2 - The Inverse Trigonometric Functions

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Section 7.2 - The Inverse Trigonometric Functions
Objective 1:
Exact Value of Inverse Trigonometric Functions.
Before define the other three inverse trigonometric functions, let's practice
working more with the three inverse trigonometric functions we have
already defined.
Find the Exact Value of the following:
Ex. 1a
cos(sin – 1(
Ex. 1c
tan(sin – 1(–
3
))
2
2
))
3
Ex. 1b
sin(tan – 1(– 3))
Ex. 1d
sin(cos – 1(
2
5
))
Solution:
a)
Let θ = sin – 1(
3
2
3
2
), then sin(θ) =
π
2
[
for θ in –
,
π
2
π
3
)=
sin(θ) is positive, then θ is in quadrant I. Since sin(
then θ =
b)
π
3
. Thus, cos(sin – 1(
3
2
)) = cos(
π
3
)=
[
Let θ = tan – 1(– 3), then tan(θ) = – 3 for θ in –
1
.
2
π
π
,
2
2
]. Since
3
2
,
]. Since
tan(θ) is negative, then θ is in quadrant IV. This means the x
value is positive and the y value is negative.
3
1
Since tan(θ) = –
find r:
=
y
x
, we will let x = 1 and y = – 3. Now,
r2 = x2 + y2
r2 = (1)2 + (– 3)2 = 1 + 9 = 10
(ignore the negative root)
So, r = ± 10
Thus, the sin(tan – 1(– 3)) = sin(θ) =
c)
Let θ = sin – 1(–
2
3
), then sin(θ) = –
y
r
2
3
=
−3
10
=–
[
for θ in –
π
2
3 10
10
,
π
2
.
]. Since
sin(θ) is negative, then θ is in quadrant IV. This means the x
value is positive and the y value is negative.
Since sin(θ) = –
find x:
2
3
=
y
r
, we will let y = – 2 and r = 3. Now,
r2 = x2 + y2
(3)2 = x2 + (– 2)2
9 = x2 + 4
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5 = x2 or x = ±
2
3
Thus, the tan(sin – 1(–
d)
Let θ = cos – 1(
2
5
5
(ignore the negative root)
y
x
)) = tan(θ) =
), then cos(θ) =
2
5
=
−2
5
=–
2 5
5
.
for θ in [0, π]. Since
cos(θ) is positive, then θ is in quadrant I. This means the x
value is positive and the y value is positive.
Since cos(θ) =
find y:
2
2
5
=
2
, we will let x =
r =x +y
(5)2 = ( 2 )2 + y2
25 = 2 + y2
23 = y2 or y = ±
Thus, the sin(cos – 1(
Objective 2:
2
x
r
2
5
2 and r = 5. Now,
23 (ignore the negative root)
)) = sin(θ) =
y
r
=
23
5
.
Defining the Inverse Cosecant, Inverse Secant and
Inverse Cotangent Functions.
Let's examine the graph of f(x) = csc(x), g(x) = sec(x), and h(x) = cot(x) to
see how we would restrict the domain to develop an inverse function.
f(x) = csc(x)
– 3π
–π
π
[
Here, we will restrict the domain to –
π
2
,
π
2
3π
]. This will be consistent with
how we restricted the domain of the sine function.
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– 3π
–π
π
3π
π
3π
g(x) = sec(x)
– 3π
–π
Here, we will restrict the domain to [0, π]. This will be consistent with how
we restricted the domain of the cosine function.
– 3π
–π
π
3π
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h(x) = cot(x)
– 2π
–π
π
2π
Here, we will restrict the domain to [0, π]. This will NOT be consistent with
how we restricted the domain of the tangent function.
– 2π
–π
π
2π
Defintion
π
2
[
Let x = csc(y) be a function on the restricted domain of –
) (
, 0 U 0,
π
2
].
Then, the inverse cosecant function (arccosecant) is defined as:
y = csc – 1(x) or arccsc(x) where |x| ≥ 1 and –
π
2
≤ y < 0 or 0 < y ≤
[
Let x = sec(y) be a function on the restricted domain of 0,
π
2
π
2
) U ( 2π ,π].
Then, the inverse secant function (arcsecant) is defined as:
y = sec – 1(x) or arcsec(x) where |x| ≥ 1 and 0 ≤ y <
π
2
or
π
2
< y ≤ π.
.
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Let x = cot(y) be a function on the restricted domain of [0, π].
Then, the inverse cotangent function (arccotangent) is defined as:
y = cot – 1(x) or arccot(x) where – ∞ < x < ∞ and 0 < y < π.
Find the exact value of the following:
Ex. 2a
csc – 1(– 2)
Ex. 2b
Ex. 2c
3
3
cot – 1(–
sec – 1( 2 )
)
Solution:
a)
Let θ = csc – 1(– 2). Then, csc(θ) = – 2 for θ in the interval
[–
π
2
) (
, 0 U 0,
π
2
]. Since the cosecant is negative, then θ is in
1
1
= – 2 implies that sin(θ) = – .
sin(θ)
2
1
π
= , so θR = . Because θ is in
2
6
π
π
θR = – . Hence, csc – 1(– 2) = – .
6
6
quadrant IV. But, csc(θ) =
Now, recall that sin(
π
6
)
quadrant IV, then θ = –
b)
Let θ = sec – 1( 2 ). Then, sec(θ) =
2 for θ in the interval
[0, 2π ) U ( 2π ,π]. Since the secant is positive, then θ is in
quadrant I. But, sec(θ) =
cos(θ) =
1
2
=
2
2
1
cos(θ)
=
2 implies that
. Recall that cos(
π
4
π
4
Because, θ is in quadrant I, then θ = θR =
Hence, sec – 1( 2 ) =
c)
Let θ = cot – 1(–
3
3
π
4
2
2
)=
, so θR =
π
4
.
.
.
). Then, cot(θ) = –
3
3
for θ in (0, π).
Since cotangent is negative, then θ is in quadrant II. But, cot(θ)
=
1
tan(θ)
tan(
π
3
=–
)=
3
3
3 , so θR =
θ = π – θR = π –
Objective 3:
implies that tan(θ) = –
π
3
=
π
3
2π
3
3
3
=–
3 . Recall, that
. Because, θ is in quadrant II, then
. Hence, cot – 1(–
3
3
)=
2π
3
.
Use a Calculator to Evaluate csc – 1(x), sec – 1(x), and
cot – 1(x).
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Since there are very few calculators have all six inverse trigonometrics
functions, we will need to use the reciprocal relationship for the
trigonometric functions. For the inverse cosecant function, this does not
present a problem since the range of the inverse cosecant function
matches the range of the inverse sine function except at 0. Similarly, for the
inverse secant function, this does not present a problem since the range of
the inverse secant function matches the range of the inverse cosine
π
2
function except at
. The inverse cotangent function, however, is a
problem since the range (0, π) does not match the range of the inverse
π
2
(
tangent function –
,
π
2
). Thus, if we need to find the inverse cotangent
of a negative number, we need to remember that the angle needs to be in
quadrant II instead of quadrant IV.
Approximate the value of the following. Round to two decimal places:
Ex. 3a
sec – 1(– 5)
Ex. 3b
csc – 1(4)
Ex. 3c
1
3
cot – 1( )
cot – 1(– 1.5)
Ex. 3d
Solution:
a)
Let θ = sec – 1( – 5). Then, sec(θ) = – 5 for θ in the interval
[0, 2π ) U ( 2π ,π]. Since the secant is negative, then θ is in
quadrant II. But, sec(θ) =
cos(θ) =
1
−5
=–
1
.
5
1
cos(θ)
= – 5 implies that
So, θ = cos – 1(–
1
)
5
= 1.77215… ≈ 1.77.
This angle is in quadrant II, so sec – 1(– 5) ≈ 1.77.
b)
Let θ = csc – 1(4). Then, csc(θ) = 4 for θ in the interval
[–
π
2
) (
, 0 U 0,
π
2
]. Since the cosecant is positive, then θ is in
quadrant I. But, csc(θ) =
1
sin(θ)
= 4 implies that sin(θ) =
1
.
4
1
4
So, θ = sin – 1( ) = 0.2526802… ≈ 0.25. This angle is in
quadrant I, so csc – 1(4) ≈ 0.25.
c)
1
3
Let θ = cot – 1( ). Then, cot(θ) =
1
3
for θ in (0, π).
Since cotangent is positive, then θ is in quadrant I. But, cot(θ)
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=
1
tan(θ)
=
1
3
implies that tan(θ) = 3. So, θ = tan – 1(3)
= 1.249045… ≈ 1.25. This angle is in quadrant I, so
1
3
cot – 1( ) ≈ 1.25.
d)
Let θ = cot – 1(– 1.5). Then, cot(θ) = – 1.5 for θ in (0, π).
Since cotangent is negative, then θ is in quadrant II. But, cot(θ)
=
1
tan(θ)
= – 1.5 = –
θ = tan – 1(–
2
3
3
2
implies that tan(θ) = –
2
3
. So,
) = – 0.5880026… ≈ – 0.59. But, this angle is not
in quadrant II, but in quadrant IV. The reference angle θR is
0.5880026… Thus, the angle we need is θ = π – θR =
π – 0.5880026… = 2.55359… ≈ 2.55. Thus, cot – 1(– 1.5) ≈ 2.55
Objective 4:
Write a Trigonometric Expression as an Algebraic
Expression.
If we are asked to write a Trigonometric expression in terms of a variable,
we will use are definitions of our trigonometrc functions and inverse
trigonometric functions to get our answer. The key will be to write x, y, and r
in terms of our variable.
Write the following as an algebraic expression containing u:
Ex. 4a
cos(tan – 1(u))
Ex. 4b
sin(sec – 1(u))
Solution:
a)
π
2
(
Let θ = tan – 1(u). Then u = tan(θ) for θ in the interval –
The cosine function is positive in quadrants I & IV, so
cos(tan – 1(u)) is positive. Since tan(θ) = u =
u
1
=
y
x
, then
x = 1 and y = u. Now, find r:
r2 = x2 + y2 = (1)2 + (u)2 = 1 + u2
So, r = ±
1+u2 (ignore the negative solution)
But, cos(θ) =
x
r
=
1
1+u2
. Thus, cos(tan – 1(u)) =
1
1+u2
.
,
π
2
).
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b)
Let θ = sec – 1(u). Then u = sec(θ) for θ in the interval
[0, 2π ) U ( 2π ,π]. The sine function is positive in quadrants I &
II, so sin(sec – 1(u)) is positive. Since sec(θ) =
cos(u) =
1
u
1
cos(u)
= u, then
and x = 1 and r = u. Now, find y:
r2 = x2 + y2
(u)2 = (1)2 + y2
u2 = 1 + y2
u2 – 1 = y2
So, y = ± u2 − 1 . But the sine function is positive, so ignore
the negative solution.
But, sin(θ) =
y
r
=
u2 −1
u
. Thus, sin(sec – 1(u)) =
u2 −1
u
.