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10.6 The Inverse Trigonometric Functions 819
10.6 The Inverse Trigonometric Functions
As the title indicates, in this section we concern ourselves with finding inverses of the (circular)
trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of
the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular
functions in the same way we restricted the domain of the quadratic function in Example 5.2.3 in
Section 5.2 to obtain a one-to-one function. We first consider f (x) = cos(x). Choosing the interval
[0, π] allows us to keep the range as [− 1 , 1] as well as the properties of being smooth and continuous.
x
y
Restricting the domain of f (x) = cos(x) to [0, π].
Recall from Section 5.2 that the inverse of a function f is typically denoted f −^1. For this reason,
some textbooks use the notation f −^1 (x) = cos−^1 (x) for the inverse of f (x) = cos(x). The obvious
pitfall here is our convention of writing (cos(x))^2 as cos^2 (x), (cos(x))^3 as cos^3 (x) and so on. It
is far too easy to confuse cos−^1 (x) with
1 cos(x) = sec(x) so we will not use this notation in our
text.^1 Instead, we use the notation f −^1 (x) = arccos(x), read ‘arc-cosine of x’. To understand the
‘arc’ in ‘arccosine’, recall that an inverse function, by definition, reverses the process of the original
function. The function f (t) = cos(t) takes a real number input t, associates it with the angle
θ = t radians, and returns the value cos(θ). Digging deeper,^2 we have that cos(θ) = cos(t) is the
x-coordinate of the terminal point on the Unit Circle of an oriented arc of length |t| whose initial
point is (1, 0). Hence, we may view the inputs to f (t) = cos(t) as oriented arcs and the outputs as
x-coordinates on the Unit Circle. The function f −^1 , then, would take x-coordinates on the Unit
Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are the graphs of f (x) = cos(x)
and f −^1 (x) = arccos(x), where we obtain the latter from the former by reflecting it across the line
y = x, in accordance with Theorem 5.3.
x
y
π 2
π
− 1
1
f (x) = cos(x), 0 ≤ x ≤ π
reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates
x
y
π 2
π
− 1 1
f −^1 (x) = arccos(x).
(^1) But be aware that many books do! As always, be sure to check the context! (^2) See page 704 if you need a review of how we associate real numbers with angles in radian measure.
820 Foundations of Trigonometry
We restrict g(x) = sin(x) in a similar manner, although the interval of choice is
[
π 2 ,^
π 2
]
x
y
Restricting the domain of f (x) = sin(x) to
[
π 2 ,^
π 2
]
It should be no surprise that we call g−^1 (x) = arcsin(x), which is read ‘arc-sine of x’.
x
y
− π 2 π 2
− 1
1
g(x) = sin(x), − π 2 ≤ x ≤ π 2.
reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates
x
y
− π 2
π 2
− 1 1
g−^1 (x) = arcsin(x).
We list some important facts about the arccosine and arcsine functions in the following theorem.
Theorem 10.26. Properties of the Arccosine and Arcsine Functions
- Properties of F (x) = arccos(x)
- Domain: [− 1 , 1]
- Range: [0, π]
- arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x
- cos(arccos(x)) = x provided − 1 ≤ x ≤ 1
- arccos(cos(x)) = x provided 0 ≤ x ≤ π
- Properties of G(x) = arcsin(x)
[
π 2 ,^
π 2
]
- arcsin(x) = t if and only if − π 2 ≤ t ≤ π 2 and sin(t) = x
- sin(arcsin(x)) = x provided − 1 ≤ x ≤ 1
- arcsin(sin(x)) = x provided − π 2 ≤ x ≤ π 2
- additionally, arcsine is odd
822 Foundations of Trigonometry
(f) Since
11 π 6 does not fall between 0 and^ π, Theorem 10.26 does not apply. We are forced to
work through from the inside out starting with arccos
cos
11 π 6
= arccos
3 2
. From
the previous problem, we know arccos
3 2
π
- Hence, arccos^
cos
( (^11) π 6
π
(g) One way to simplify cos
arccos
is to use Theorem 10.26 directly. Since − 35 is
between −1 and 1, we have that cos
arccos
3 5
3 5 and we are done. However, as before, to really understand why this cancellation occurs, we let t = arccos
. Then,
by definition, cos(t) = −
3
- Hence, cos^
arccos
3 5
= cos(t) = −
3 5 , and we are finished in (nearly) the same amount of time.
(h) As in the previous example, we let t = arccos
3 5
so that cos(t) = −
3 5 for some^ t^ where 0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2 < t < π,
so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to find sin
arccos
= sin(t). Using the Pythagorean Identity cos^2 (t) + sin^2 (t) = 1, we get ( − (^35)
- sin^2 (t) = 1 or sin(t) = ± 45. Since t corresponds to a Quadrants II angle, we
choose sin(t) = 45. Hence, sin
arccos
- (a) We begin this problem in the same manner we began the previous two problems. To
help us see the forest for the trees, we let t = arccos(x), so our goal is to find a way to express tan (arccos (x)) = tan(t) in terms of x. Since t = arccos(x), we know cos(t) = x
where 0 ≤ t ≤ π, but since we are after an expression for tan(t), we know we need to throw out t = π 2 from consideration. Hence, either 0 ≤ t < π 2 or π 2 < t ≤ π so that,
geometrically, t corresponds to an angle in Quadrant I or Quadrant II. One approach^3
to finding tan(t) is to use the quotient identity tan(t) =
sin(t) cos(t). Substituting cos(t) =^ x into the Pythagorean Identity cos^2 (t) + sin^2 (t) = 1 gives x^2 + sin^2 (t) = 1, from which we get sin(t) = ±
1 − x^2. Since t corresponds to angles in Quadrants I and II, sin(t) ≥ 0,
so we choose sin(t) =
1 − x^2. Thus,
tan(t) =
sin(t)
cos(t)
1 − x^2
x
To determine the values of x for which this equivalence is valid, we consider our sub- stitution t = arccos(x). Since the domain of arccos(x) is [− 1 , 1], we know we must
restrict − 1 ≤ x ≤ 1. Additionally, since we had to discard t =
π 2 , we need to discard x = cos
π 2
= 0. Hence, tan (arccos (x)) =
√ 1 −x^2 x is valid for^ x^ in [−^1 ,^ 0)^ ∪^ (0,^ 1]. (b) We proceed as in the previous problem by writing t = arcsin(x) so that t lies in the
interval
[
− π 2 , π 2
]
with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms
of x. Since cos(2t) is defined everywhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cos(2t): cos^2 (t) − sin^2 (t),
2 cos^2 (t) − 1 and 1 − 2 sin^2 (t). Since we know x = sin(t), it is easiest to use the last form:
cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin
2 (t) = 1 − 2 x
2
(^3) Alternatively, we could use the identity: 1 + tan (^2) (t) = sec (^2) (t). Since x = cos(t), sec(t) = 1 cos(t) =^
1 x. The reader is invited to work through this approach to see what, if any, difficulties arise.
10.6 The Inverse Trigonometric Functions 823
To find the restrictions on x, we once again appeal to our substitution t = arcsin(x). Since arcsin(x) is defined only for − 1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1− 2 x^2
is valid only on [− 1 , 1].
A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in
dealing with the inverse circular functions come from the need to restrict the domains of the
original functions so that they are one-to-one. One instance of this phenomenon is the fact that
arccos
cos
( (^11) π 6
= π 6 as opposed to 116 π. This is the exact same phenomenon discussed in Section
5.2 when we saw
(−2)^2 = 2 as opposed to −2. Additionally, even though the expression we
arrived at in part 2b above, namely 1 − 2 x^2 , is defined for all real numbers, the equivalence
cos (2 arcsin(x)) = 1 − 2 x^2 is valid for only − 1 ≤ x ≤ 1. This is akin to the fact that while the
expression x is defined for all real numbers, the equivalence (
x)
2 = x is valid only for x ≥ 0. For
this reason, it pays to be careful when we determine the intervals where such equivalences are valid.
The next pair of functions we wish to discuss are the inverses of tangent and cotangent, which
are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its
fundamental cycle on
− π 2 , π 2
to obtain f −^1 (x) = arctan(x). Among other things, note that the
vertical asymptotes x = − π 2 and x = π 2 of the graph of f (x) = tan(x) become the horizontal
asymptotes y = − π 2 and y = π 2 of the graph of f −^1 (x) = arctan(x).
x
y
− π 2 − π 4 π 4 π 2 − 1
1
f (x) = tan(x), − π 2 < x < π 2.
reflect across y = x −−−−−−−−−−−−→ switch x and y coordinates
x
y
− π 4
− π 2
π 4
π 2
− 1 1
f −^1 (x) = arctan(x).
Next, we restrict g(x) = cot(x) to its fundamental cycle on (0, π) to obtain g−^1 (x) = arccot(x).
Once again, the vertical asymptotes x = 0 and x = π of the graph of g(x) = cot(x) become the
horizontal asymptotes y = 0 and y = π of the graph of g−^1 (x) = arccot(x). We show these graphs
on the next page and list some of the basic properties of the arctangent and arccotangent functions.
10.6 The Inverse Trigonometric Functions 825
Example 10.6.2.
- Find the exact values of the following.
(a) arctan(
- (b) arccot(−
(c) cot(arccot(−5)) (d) sin
arctan
- Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid.
(a) tan(2 arctan(x)) (b) cos(arccot(2x))
Solution.
- (a) We know arctan(
- is the real number t between −
π 2 and^
π 2 with tan(t) =^
- We find
t = π 3 , so arctan(
- = π 3.
(b) The real number t = arccot(−
- lies in the interval (0, π) with cot(t) = −
- We get arccot(−
- = 56 π.
(c) We can apply Theorem 10.27 directly and obtain cot(arccot(−5)) = −5. However,
working it through provides us with yet another opportunity to understand why this
is the case. Letting t = arccot(−5), we have that t belongs to the interval (0, π) and cot(t) = −5. Hence, cot(arccot(−5)) = cot(t) = −5.
(d) We start simplifying sin
arctan
by letting t = arctan
. Then tan(t) = − 34 for
some − π 2 < t < π 2. Since tan(t) < 0, we know, in fact, − π 2 < t < 0. One way to proceed is to use The Pythagorean Identity, 1+cot^2 (t) = csc^2 (t), since this relates the reciprocals
of tan(t) and sin(t) and is valid for all t under consideration.^4 From tan(t) = − 34 , we
get cot(t) = − 43. Substituting, we get 1 +
= csc^2 (t) so that csc(t) = ± 53. Since
−
π 2 < t <^ 0, we choose csc(t) =^ −^
5 3 , so sin(t) =^ −^
3
- Hence, sin^
arctan
3 4
3
(a) If we let t = arctan(x), then − π 2 < t < π 2 and tan(t) = x. We look for a way to express tan(2 arctan(x)) = tan(2t) in terms of x. Before we get started using identities, we note
that tan(2t) is undefined when 2t = π 2 + πk for integers k. Dividing both sides of this
equation by 2 tells us we need to exclude values of t where t =
π 4 +^
π 2 k, where^ k^ is an integer. The only members of this family which lie in
− π 2 , π 2
are t = ± π 4 , which
means the values of t under consideration are
π 2 ,^ −^
π 4
π 4 ,^
π 4
( (^) π 4 ,^
π 2
. Returning
to arctan(2t), we note the double angle identity tan(2t) =
2 tan(t) 1 −tan^2 (t) , is valid for all the values of t under consideration, hence we get
tan(2 arctan(x)) = tan(2t) =
2 tan(t)
1 − tan^2 (t)
2 x
1 − x^2
(^4) It’s always a good idea to make sure the identities used in these situations are valid for all values t under
consideration. Check our work back in Example 10.6.1. Were the identities we used there valid for all t under
consideration? A pedantic point, to be sure, but what else do you expect from this book?
826 Foundations of Trigonometry
To find where this equivalence is valid we check back with our substitution t = arctan(x). Since the domain of arctan(x) is all real numbers, the only exclusions come from the
values of t we discarded earlier, t = ± π 4. Since x = tan(t), this means we exclude x = tan
± π 4
= ±1. Hence, the equivalence tan(2 arctan(x)) = (^1) −^2 xx 2 holds for all x in
(−∞, −1) ∪ (− 1 , 1) ∪ (1, ∞).
(b) To get started, we let t = arccot(2x) so that cot(t) = 2x where 0 < t < π. In terms
of t, cos(arccot(2x)) = cos(t), and our goal is to express the latter in terms of x. Since
cos(t) is always defined, there are no additional restrictions on t, so we can begin using
identities to relate cot(t) to cos(t). The identity cot(t) =
cos(t) sin(t) is valid for^ t^ in (0, π), so our strategy is to obtain sin(t) in terms of x, then write cos(t) = cot(t) sin(t). The identity 1 + cot^2 (t) = csc^2 (t) holds for all t in (0, π) and relates cot(t) and csc(t) = (^) sin(^1 t).
Substituting cot(t) = 2x, we get 1 + (2x)^2 = csc^2 (t), or csc(t) = ±
4 x^2 + 1. Since t is
between 0 and π, csc(t) > 0, so csc(t) =
4 x^2 + 1 which gives sin(t) = √^1 4 x^2 +
. Hence,
cos(arccot(2x)) = cos(t) = cot(t) sin(t) =
2 x √ 4 x^2 + 1
Since arccot(2x) is defined for all real numbers x and we encountered no additional restrictions on t, we have cos (arccot(2x)) = √^2 x 4 x^2 +
for all real numbers x.
The last two functions to invert are secant and cosecant. A portion of each of their graphs, which
were first discussed in Subsection 10.5.2, are given below with the fundamental cycles highlighted.
x
y
The graph of y = sec(x).
x
y
The graph of y = csc(x).
It is clear from the graph of secant that we cannot find one single continuous piece of its graph
which covers its entire range of (−∞, −1] ∪ [1, ∞) and restricts the domain of the function so that it
is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant
functions, we must settle for a piecewise approach wherein we choose one piece to cover the top
of the range, namely [1, ∞), and another piece to cover the bottom, namely (−∞, −1]. There are
two generally accepted ways make these choices which restrict the domains of these functions so
that they are one-to-one. One approach simplifies the Trigonometry associated with the inverse
functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry
less so. We present both points of view.
828 Foundations of Trigonometry
Theorem 10.28. Properties of the Arcsecant and Arccosecant Functionsa
- Properties of F (x) = arcsec(x)
- Domain: {x : |x| ≥ 1 } = (−∞, −1] ∪ [1, ∞)
- Range:
[
0 , π 2
π 2 , π
]
π 2
; as x → ∞, arcsec(x) →
π 2
−
- arcsec(x) = t if and only if 0 ≤ t < π 2 or π 2 < t ≤ π and sec(t) = x
- arcsec(x) = arccos
1 x
provided |x| ≥ 1
- sec (arcsec(x)) = x provided |x| ≥ 1
- arcsec(sec(x)) = x provided 0 ≤ x < π 2 or π 2 < x ≤ π
- Properties of G(x) = arccsc(x)
- Domain: {x : |x| ≥ 1 } = (−∞, −1] ∪ [1, ∞)
- Range:
[
− π 2 , 0
0 , π 2
]
- as x → −∞, arccsc(x) → 0 −; as x → ∞, arccsc(x) → 0 +
- arccsc(x) = t if and only if −
π 2 ≤^ t <^ 0 or 0^ < t^ ≤^
π 2 and csc(t) =^ x
1 x
provided |x| ≥ 1
- csc (arccsc(x)) = x provided |x| ≥ 1
- arccsc(csc(x)) = x provided − π 2 ≤ x < 0 or 0 < x ≤ π 2
- additionally, arccosecant is odd
a... assuming the “Trigonometry Friendly” ranges are used.
Example 10.6.3.
- Find the exact values of the following.
(a) arcsec(2) (b) arccsc(−2) (c) arcsec
sec
( (^5) π 4
(d) cot (arccsc (−3))
- Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid.
(a) tan(arcsec(x)) (b) cos(arccsc(4x))
10.6 The Inverse Trigonometric Functions 829
Solution.
- (a) Using Theorem 10.28, we have arcsec(2) = arccos
1 2
= π 3.
(b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin
= − π 6.
(c) Since
5 π 4 doesn’t fall between 0 and^
π 2 or^
π 2 and^ π, we cannot use the inverse property stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which
yields arcsec
sec
( (^5) π 4
= arcsec(−
- = arccos
√ 2 2
3 π
(d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = − 3 and, since this is negative, we have that t lies in the interval
[
− π 2 , 0
. We are after
cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot^2 (t) = csc^2 (t). Substituting, we have 1 + cot^2 (t) = (−3)^2 , or cot(t) = ±
- Since − π 2 ≤ t < 0,
cot(t) < 0, so we get cot (arccsc (−3)) = − 2
- (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for t in [ 0 ,
π 2
( (^) π 2 , π
]
, and we seek a formula for tan(t). Since tan(t) is defined for all t values under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we
use the identity 1 + tan^2 (t) = sec^2 (t). This is valid for all values of t under consideration, and when we substitute sec(t) = x, we get 1 + tan^2 (t) = x^2. Hence, tan(t) = ±
x^2 − 1.
If t belongs to
[
0 , π 2
then tan(t) ≥ 0; if, on the the other hand, t belongs to
π 2 , π
]
then tan(t) ≤ 0. As a result, we get a piecewise defined function for tan(t)
tan(t) =
x^2 − 1 , if 0 ≤ t < π 2
x^2 − 1 , if π 2 < t ≤ π
Now we need to determine what these conditions on t mean for x. Since x = sec(t),
when 0 ≤ t <
π 2 ,^ x^ ≥^ 1, and when^
π 2 < t^ ≤^ π,^ x^ ≤ −1. Since we encountered no further restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞).
tan(arcsec(x)) =
x^2 − 1 , if x ≥ 1
x^2 − 1 , if x ≤ − 1
(b) To simplify cos(arccsc(4[ x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in − π 2 , 0
0 , π 2
]
, and we now set about finding an expression for cos(arccsc(4x)) = cos(t).
Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = (^41) x , so to find cos(t), we can make use if the identity
cos^2 (t) + sin^2 (t) = 1. Substituting sin(t) = (^41) x gives cos^2 (t) +
4 x
= 1. Solving, we get
cos(t) = ±
16 x^2 − 1
16 x^2
16 x^2 − 1
4 |x|
Since t belongs to
[
π 2 ,^0
π 2
]
, we know cos(t) ≥ 0, so we choose cos(t) =
√ 16 −x^2 4 |x|. (The absolute values here are necessary, since x could be negative.) To find the values for
10.6 The Inverse Trigonometric Functions 831
Theorem 10.29. Properties of the Arcsecant and Arccosecant Functionsa
- Properties of F (x) = arcsec(x)
- Domain: {x : |x| ≥ 1 } = (−∞, −1] ∪ [1, ∞)
- Range:
[
0 , π 2
[
π, 32 π
- as x → −∞, arcsec(x) → 32 π
− ; as x → ∞, arcsec(x) → π 2 −
- arcsec(x) = t if and only if 0 ≤ t <
π 2 or^ π^ ≤^ t <^
3 π 2 and sec(t) =^ x
1 x
for x ≥ 1 onlyb
- sec (arcsec(x)) = x provided |x| ≥ 1
- arcsec(sec(x)) = x provided 0 ≤ x < π 2 or π ≤ x < 32 π
- Properties of G(x) = arccsc(x)
- Domain: {x : |x| ≥ 1 } = (−∞, −1] ∪ [1, ∞)
- Range:
0 , π 2
]
π, 32 π
]
- as x → −∞, arccsc(x) → π+; as x → ∞, arccsc(x) → 0 +
- arccsc(x) = t if and only if 0 < t ≤ π 2 or π < t ≤ 32 π and csc(t) = x
- arccsc(x) = arcsin
x
for x ≥ 1 onlyc
- csc (arccsc(x)) = x provided |x| ≥ 1
- arccsc(csc(x)) = x provided 0 < x ≤ π 2 or π < x ≤ 32 π
a... assuming the “Calculus Friendly” ranges are used. bCompare this with the similar result in Theorem 10.28. cCompare this with the similar result in Theorem 10.28.
Our next example is a duplicate of Example 10.6.3. The interested reader is invited to compare
and contrast the solution to each.
Example 10.6.4.
- Find the exact values of the following.
(a) arcsec(2) (b) arccsc(−2) (c) arcsec
sec
( (^5) π 4
(d) cot (arccsc (−3))
- Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid.
(a) tan(arcsec(x)) (b) cos(arccsc(4x))
832 Foundations of Trigonometry
Solution.
- (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos
1 2
= π 3.
(b) Unfortunately, −2 is not greater to or equal to 1, so we cannot apply Theorem 10.29 to arccsc(−2) and convert this into an arcsine problem. Instead, we appeal to the definition.
The real number t = arccsc(−2) lies in
0 , π 2
]
π, 32 π
]
and satisfies csc(t) = −2. The t
we’re after is t =
7 π 6 , so arccsc(−2) =^
7 π
(c) Since 54 π lies between π and 32 π , we may apply Theorem 10.29 directly to simplify
arcsec
sec
5 π 4
= 54 π. We encourage the reader to work this through using the defini- tion as we have done in the previous examples to see how it goes.
(d) To simplify cot (arccsc (−3)) we let t = arccsc (−3) so that cot (arccsc (−3)) = cot(t).
We know csc(t) = −3, and since this is negative, t lies in
π,
3 π 2
]
. Using the identity
1 + cot^2 (t) = csc^2 (t), we find 1 + cot^2 (t) = (−3)^2 so that cot(t) = ±
- Since
t is in the interval
π, 32 π
]
, we know cot(t) > 0. Our answer is cot (arccsc (−3)) = 2
- (a) We begin simplifying tan(arcsec([ x)) by letting t = arcsec(x). Then, sec(t) = x for t in 0 , π 2
[
π, 32 π
, and we seek a formula for tan(t). Since tan(t) is defined for all t values
under consideration, we have no additional restrictions on t. To relate sec(t) to tan(t), we use the identity 1 + tan^2 (t) = sec^2 (t). This is valid for all values of t under consideration,
and when we substitute sec(t) = x, we get 1 + tan^2 (t) = x^2. Hence, tan(t) = ±
x^2 − 1.
Since t lies in
[
π 2
[
π,
3 π 2
, tan(t) ≥ 0, so we choose tan(t) =
x^2 − 1. Since we found
no additional restrictions on t, the equivalence tan(arcsec(x)) =
x^2 − 1 holds for all x
in the domain of t = arcsec(x), namely (−∞, −1] ∪ [1, ∞).
(b) To simplify cos(arccsc(4( x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for t in 0 , π 2
]
π, 32 π
]
, and we now set about finding an expression for cos(arccsc(4x)) = cos(t).
Since cos(t) is defined for all t, we do not encounter any additional restrictions on t. From csc(t) = 4x, we get sin(t) = (^41) x , so to find cos(t), we can make use if the identity
cos^2 (t) + sin^2 (t) = 1. Substituting sin(t) = (^41) x gives cos^2 (t) +
4 x
= 1. Solving, we get
cos(t) = ±
16 x^2 − 1
16 x^2
16 x^2 − 1
4 |x|
If t lies in
0 , π 2
]
, then cos(t) ≥ 0, and we choose cos(t) =
√ 16 x^2 − 1 4 |x|. Otherwise,^ t^ belongs
to
π, 32 π
]
in which case cos(t) ≤ 0, so, we choose cos(t) = −
√ 16 x^2 − 1 4 |x| This leads us to a (momentarily) piecewise defined function for cos(t)
cos(t) =
16 x^2 − 1
4 |x|
, if 0 ≤ t ≤ π 2
16 x^2 − 1
4 |x|
, if π < t ≤ 32 π
834 Foundations of Trigonometry
(c) Since the argument −2 is negative, we cannot directly apply Theorem 10.27 to help us
find arccot(−2). Let t = arccot(−2). Then t is a real number such that 0 < t < π and cot(t) = −2. Moreover, since cot(t) < 0, we know π 2 < t < π. Geometrically, this
means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a ‘reference angle’ approach. Consider α, the reference angle for θ, as pictured
below. By definition, α is an acute angle so 0 < α < π 2 , and the Reference Angle
Theorem, Theorem 10.2, tells us that cot(α) = 2. This means α = arccot(2) radians. Since the argument of arccotangent is now a positive 2, we can use Theorem 10.27 to get
α = arccot(2) = arctan
2
radians. Since θ = π − α = π − arctan
2
≈ 2 .6779 radians, we get arccot(−2) ≈ 2 .6779.
x
y
1
1
α
θ = arccot(−2) radians
Another way to attack the problem is to use arctan
. By definition, the real number
t = arctan
satisfies tan(t) = − 12 with − π 2 < t < π 2. Since tan(t) < 0, we know
more specifically that −
π 2 < t <^ 0, so^ t^ corresponds to an angle^ β^ in Quadrant IV. To find the value of arccot(−2), we once again visualize the angle θ = arccot(−2) radians
and note that it is a Quadrant II angle with tan(θ) = −
1
- This means it is exactly^ π units away from β, and we get θ = π + β = π + arctan
≈ 2 .6779 radians. Hence,
as before, arccot(−2) ≈ 2 .6779.
10.6 The Inverse Trigonometric Functions 835
x
y
1
1
π
β
θ = arccot(−2) radians
(d) If the range of arccosecant is taken to be
[
− π 2 , 0
0 , π 2
]
, we can use Theorem 10.28 to
get arccsc
= arcsin
≈ − 0 .7297. If, on the other hand, the range of arccosecant is taken to be
0 , π 2
]
π, 32 π
]
, then we proceed as in the previous problem by letting
t = arccsc
. Then t is a real number with csc(t) = − 32. Since csc(t) < 0, we have
that π < θ ≤
3 π 2 , so^ t^ corresponds to a Quadrant III angle,^ θ.^ As above, we let^ α^ be the reference angle for θ. Then 0 < α < π 2 and csc(α) = 32 , which means α = arccsc
3 2
radians. Since the argument of arccosecant is now positive, we may use Theorem 10. to get α = arccsc
3 2
= arcsin
2 3
radians. Since θ = π + α = π + arcsin
2 3
radians, arccsc
3 2
x
y
1
1
α
θ = arccsc
( − (^32)
) radians
10.6 The Inverse Trigonometric Functions 837
Hence, for x < 0, g(x) = π + arctan
x
x
- 2π. What about x = 0? We know g(0) = arccot(0) + π = π, and neither of the formulas for g involving arctangent
will produce this result.^6 Hence, in order to graph y = g(x) on our calculators, we need to write it as a piecewise defined function:
g(x) = arccot
x
2
arctan
x
π, if x = 0
arctan
x
We show the input and the result below.
y = g(x) in terms of arctangent y = g(x) = arccot
x 2
The inverse trigonometric functions are typically found in applications whenever the measure of an
angle is required. One such scenario is presented in the following example.
Example 10.6.6. 7 The roof on the house below has a ‘6/12 pitch’. This means that when viewed
from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination
from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded
to the nearest hundredth of a degree.
Front View Side View
Solution. If we divide the side view of the house down the middle, we find that the roof line forms
the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we
(^6) Without Calculus, of course... (^7) The authors would like to thank Dan Stitz for this problem and associated graphics.
838 Foundations of Trigonometry
find the angle of inclination, labeled θ below, satisfies tan(θ) =
6 12 =^
1
- Since^ θ^ is an acute angle, we can use the arctangent function and we find θ = arctan
1 2
radians ≈ 26. 56 ◦.
12 feet
6 feet
θ
10.6.4 Solving Equations Using the Inverse Trigonometric Functions.
In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 12 for angles θ and
tan(t) = −1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied
on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on
the other hand, we had been asked to find all angles with sin(θ) =
1 3 or solve tan(t) =^ −2 for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse
trigonometric functions, however, we are now in a position to solve these equations. A good parallel
to keep in mind is how the square root function can be used to solve certain quadratic equations.
The equation x^2 = 4 is a lot like sin(θ) = 12 in that it has friendly, ‘common value’ answers x = ±2.
The equation x^2 = 7, on the other hand, is a lot like sin(θ) = 13. We know^8 there are answers, but
we can’t express them using ‘friendly’ numbers.^9 To solve x^2 = 7, we make use of the square root
function and write x = ±
- We can certainly approximate these answers using a calculator, but
as far as exact answers go, we leave them as x = ±
- In the same way, we will use the arcsine
function to solve sin(θ) =
1 3 , as seen in the following example.
Example 10.6.7. Solve the following equations.
- Find all angles θ for which sin(θ) = 13.
- Find all real numbers t for which tan(t) = − 2
- Solve sec(x) = − 53 for x.
Solution.
- If sin(θ) = 13 , then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at y = 13. Geometrically, we see that this happens at two places: in Quadrant I
and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions
(^8) How do we know this again? (^9) This is all, of course, a matter of opinion. For the record, the authors find ±√7 just as ‘nice’ as ±2.
