7. inverse trig functions and linear trig equations-x

–π/2
π/2
The Graphs of Inverse Trig–Functions
π
0
π/2
–π/2
π
0

A function f(x) = y takes an input x and produces
one output y.
Inverse Functions

one output y. Often we represent a function by the
following figure.
Inverse Functions
domian rangex y=f(x)
f

following figure.
Inverse Functions
We like to reverse the operation, i.e., if we know
the output y, what was (were) the input x?
f

following figure.
Inverse Functions
This procedure of associating the output y to the
input x may or may not be a function.
f

following figure.
Inverse Functions
f
If it is a function, it is called
the inverse function of f(x)
and it is denoted as f –1(y).

following figure.
Inverse Functions
f
x y=f(x)
f

following figure.
Inverse Functions
f
x=f–1(y) y=f(x)
f
f –1

following figure.
Inverse Functions
f
We say f(x) and f –1(y) are
the inverse of each other.
x=f–1(y) y=f(x)
f
f –1

Example A.
a. The function y = f(x) = 2x takes the input x and
doubles it to get the output y.
Inverse Functions

Example A.
doubles it to get the output y. To reverse the
operation, take an output y,
Inverse Functions

Example A.
operation, take an output y, divided it by 2 and we
get back to the x.
Inverse Functions

Example A.
get back to the x. In other words f –1(y) = y/2.
Inverse Functions

Example A.
So, for example, f –1(6) = 3 because f(3) = 6.
Inverse Functions

Example A.
b. Given y = f(x) = x2 and y = 9,
Inverse Functions

Example A.
b. Given y = f(x) = x2 and y = 9, there are two inputs
namely x = 3 and x = –3 that give the output 9.
x=3
y=9
f(x)=x2
x=–3
Inverse Functions

Example A.
Therefore, the reverse procedure is not a function
x=3
y=9
f(x)=x2
x=–3
not a function
Inverse Functions

Example A.
Inverse Functions
x=3
y=9
f(x)=x2
x=–3
not a function
Unless we intentionally select an
answer, f –1(9) is not well defined
Because there are two choices.

Example A.
so f –1(y) does not exist.
Inverse Functions
x=3
y=9
f(x)=x2
x=–3
not a function
Unless we intentionally select an
answer, f –1(9) is not well defined
Because there are two choices.

Example A.
Inverse Functions
But g –1(y) exists for
y = g(x) = x2 with x ≥ 0,
and that g –1(9) = 3. In fact g –1(y) = √y.
x=3
y=9
f(x)=x2
x=–3
not a function

Example A.
Inverse Functions
But g –1(y) exists for
y = g(x) = x2 with x ≥ 0,
and that g –1(9) = 3. In fact g –1(y) = √y. Hence the
existence of the inverse depends on the domain.
x=3
y=9
f(x)=x2
x=–3
not a function

Inverse Functions
u
v
u = v
a one-to-one function
A function is one-to-one if different inputs produce
different outputs or that two inputs give two outputs.

Inverse Functions
That is, f(x) is said to be one-to-one if for every two
inputs u and v such that u  v, then f(u)  f(v).

Inverse Functions
u
v
u = v

Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)

Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
v
u = v
not a one-to-one function

Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v

Example B.
a. g(x) = 2x is one-to-one
Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v

Example B.
because if u  v, then 2u  2v.
Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v

Example B.
b. f(x) = x2 is not one-to-one
because 3  –3, but f(3) = f(–3) = 9.
Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v

Example B.
e.g. 3  –3, but f(3) = f(–3) = 9.
Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v
Note:
To justify a function is 1–1,
we have to show that for every
pair of u  v that f(u)  f(v).

Example B.
e.g. 3  –3, but f(3) = f(–3) = 9.
Inverse Functions
u f(u)
v f(v)
u = v f(u) = f(v)
u
f(u)=f(v)
v
u = v
Note:
To justify a function is 1–1,
we have to show that for every
pair of u  v that f(u)  f(v).
To justify a function is not 1–1,
all we need to do is to produce
one pair of u  v but f(u) = f(v).

Inverse Functions
If y = f(x) is one-to-one, then each output y came
from one input x so that f –1(y) = x is well defined.

Inverse Functions
Given y = f(x), to find f –1(y), just solve the equation
y = f(x) for x in terms of y.

Inverse Functions
Example C.
Find the inverse function of y = f(x) = x – 53
4

Inverse Functions
Example C.
Given y = x – 5 and solve for x.
3
4
3
4

Inverse Functions
Example C.
Clear denominator: 4y = 3x – 20
3
4
3
4

Inverse Functions
Example C.
4y + 20 = 3x
3
4
3
4

Inverse Functions
Example C.
4y + 20 = 3x
x =
3
4
3
4
4y + 20
3

Inverse Functions
Example C.
4y + 20 = 3x
x =
3
4
3
4
4y + 20
3
Hence f –1(y) =
4y + 20
3

Inverse Trigonometric Functions
Trig–functions assign multiple inputs to the same
output, e.g. sin(π/6) = sin(5π/6) = sin(–7π/6) = .. = ½.

Hence their inverse association are not functions
because there are multiple answers to the question
such as “which angle has sine value ½?”.

In such cases, we choose one specific angle as the
principal answer to restrict the inverse procedures
as functions.

as functions. For example, if sin() = ½,
we select  = π/6 as the sine-inverse of ½,
we write that sin–1(½) = π/6.

we write that sin–1(–½) = –π/6. In general,
given a number a such that –1 < a <1,
π/2
–π/2
a
–1
+1

we define sin–1(a) =  where
sin() = a and –π/2 <  < π/2.
π/2
–π/2
aa

–1
+1

we define sin–1(a) =  where
sin() = a and –π/2 <  < π/2. Sin–1(a) is also
written as arcsin(a) or asin(a).
π/2
–π/2
aa

–1
+1

In a similar manner, we define the inverse cosine.
Given a number a such that –1 < a <1,
cos–1(a) =  where cos() = a and 0 <  < π
cos–1(a) is read as "cosine–inverse of a"
and it is also written as arccos(a) or acos(a).

0π
a

–1 +1
cos–1(a) = 
with 0 ≤  ≤ π

Tangent outputs all real numbers, so the
inverse tangent is defined for all a’s.
0π
a

–1 +1
cos–1(a) = 
with 0 ≤  ≤ π

Given any number a, define tan–1(a) = ,
where –π/2 <  < π/2 and that tan() = a.
0π
a

–1 +1
cos–1(a) = 
with 0 ≤  ≤ π
tan–1(a) =  with
–π/2 <  < π/2
π/2
–π/2
a
–1
+1

–π/2 <  < π/2
0π
a

–1 +1
π/2
–π/2
a

–1
+1
cos–1(a) = 
with 0 ≤  ≤ π

tan–1(a) is also written as arctan(a) or atan(a).
–π/2 <  < π/2
0π
a

–1 +1
π/2
–π/2
a

–1
+1
cos–1(a) = 
with 0 ≤  ≤ π

Example D.
a. sin–1(1/2) = π/6 b. cos–1(–1/2) = 2π/3
c. cos–1(–1) = π d. sin–1(2) = UND
e. cos–1(0.787)  38.1o f. tan–1(3) = π/3
For –1≤ a ≤1,
cos–1(a) = ,
0 ≤  ≤ π
Here are the inverse trig-functions again.
For –1≤ a ≤1,
sin–1(a) = ,
–π/2 ≤  ≤ π/2
All a’s,
tan–1(a) =  ,
–π/2<  < π/2

Example D.
a. sin–1(1/2) = π/6 b. cos–1(–1/2) = 2π/3
c. cos–1(–1) = π d. sin–1(2) = UDF
e. cos–1(0.787)  38.1o f. tan–1(3) = π/3
For –1≤ a ≤1,
cos–1(a) = ,
0 ≤  ≤ π
Here are the inverse trig-functions again.
For –1≤ a ≤1,
sin–1(a) = ,
–π/2 ≤  ≤ π/2
All a’s,
tan–1(a) =  ,
–π/2<  < π/2

Example E.
Solve the triangle with
a=29, <B=16o, c=74.

From the cosine law we have that
b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.

b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.
We can't solve for C using sine–inverse directly
because it’s more than 90o so we solve for A first.

b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.
sin(A) sin(16)
29 46.8
=

b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.
sin(A) sin(16)
29 46.8
=
sin(A)
29sin(16)
46.8=  0.171

b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.
sin(A) sin(16)
29 46.8
=
sin(A)
29sin(16)
46.8=  0.171
Hence A = sin–1(0.171)  9.83o

b2 = 742 + 292 – 2ac(cos 16o) or that b  46.8.
Example E.
a=29, <B=16o, c=74.
sin(A) sin(16)
29 46.8
=
sin(A)
29sin(16)
46.8=  0.171
Hence A = sin–1(0.171)  9.83o
and C = 180 – 16 – 9.83  154.2o

Using the SOACAHTOA relations, given an inverse
trig–value, we may draw a generic right triangle that
reflects the relations between the values.

Example F. Draw the angle cos–1(2/a) and find
tan(cos–1(2/a) ).

tan(cos–1(2/a) ).
Let α = cos–1(2/a) so cos(α) = 2/a = Adj/Hyp,

tan(cos–1(2/a) ).
then the following triangle represents α.

α
2 = Adj.
a = Hyp.
tan(cos–1(2/a) ).

a2 – 4
α
2 = Adj.
a = Hyp.
tan(cos–1(2/a) ).
By Pythagorean Theorem the third side is a2 – 4.

a2 – 4
Therefore,
tan(cos–1(2/a))
= tan(α)
α
2 = Adj.
a = Hyp.
tan(cos–1(2/a) ).

a2 – 4
Therefore,
tan(cos–1(2/a))
= tan(α)
=
α
2 = Adj.
a = Hyp.
tan(cos–1(2/a) ).
a2 – 4
2

and the sine–inverse
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We will use the variable x and y for the functions
y = sin(x): [–π/2, π/2] [–1, 1].

y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].
As in the case for all inverse pairs of functions,
the graph of y = sin–1(x) may be
obtained by reflecting
y = sin(x) across y = x.

–1
y = sin–1(x)
y = x
y = sin(x)
y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].
–π/2 π/2
1
–1

–1
–π/2
y = sin–1(x)
1
–1
π/2
y = x
y = sin(x)
y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].

–1
–π/2
1
–1
–π/2
π/2
1–1 π/2
y = x
y = sin(x)
y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].
y = sin–1(x)

–1
–π/2
1
–1
–π/2
π/2
1–1 π/2
y = x
y = sin(x)
y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].
(π/6 ½)
For example, the point
(π/6, ½) on y = sin(x)
y = sin–1(x)

–1
–π/2
1
–1
–π/2
π/2
1–1 π/2
y = x
y = sin(x)
y = sin–1(x) : [–1, 1] [–π/2, π/2].
y = sin(x): [–π/2, π/2] [–1, 1].
(½, π/6)
(π/6 ½)
For example, the point
(π/6, ½) on y = sin(x)
corresponds to (½, π/6)
on y = sin–1(x) since
(, a) and (a, ) are the
reflection of each other.
y = sin–1(x)

The graph of y = cos–1(x) is the reflection of the graph
of y = cos (x) where y = cos(x): [0, π] [–1, 1]
and y = cos–1(x): [–1, 1] [0, π]

y = x
π
0
1
–1
π
y = cos(x)
y = cos–1(x)
and y = cos–1(x): [–1, 1] [0, π]

y = x
π
0
1
–1
Here are some specific cosine inverse values:
π
(1, 0)
(–1, π)
(0, π/2)
cos–1(1) = 0,
cos–1(0) = π/2, and
cos–1(–1) = π.
y = cos(x)
y = cos–1(x)
and y = cos–1(x): [–1, 1] [0, π]

y = cos(x)
y = x
π
0
1
–1
Here are some specific cosine inverse values:
π
(1, 0)
(–1, π)
(0, π/2)
cos–1(1) = 0,
cos–1(0) = π/2, and
cos–1(–1) = π.
P  (0.74, 0.74) intersection point
P  (0.74, 0.74)
in radian.
By solving cos(x) = x
numerically, we've the
.y = cos–1(x)
and y = cos–1(x): [–1, 1] [0, π]

The domain of y = tan–1(x) is all real numbers and
the output y is restricted to –π/2 < y < π/2,

x
tan–1(x)
0 1/3 1 3 ∞–1/3–1–3–∞

x
tan–1(x)
0 1/3 1 3 ∞–1/3–1–3–∞
π/60 π/4 π/3 π/2

x
tan–1(x) π/60 π/4 π/3
0 1/3 1 3 ∞–1/3–1–3
π/2
–∞
y = π/2
(0,0)
(1,π/4)
(√3, π/3)

x
tan–1(x) π/60 π/4 π/3–π/2
0 1/3 1 3 ∞
–π/6
–1/3
–π/4
–1
–π/3
–3
π/2
–∞
y = π/2
(0,0)
(1,π/4)
(√3, π/3)

x
tan–1(x) π/60 π/4 π/3–π/2
0 1/3 1 3 ∞
–π/6
–1/3
–π/4
–1
–π/3
–3
π/2
–∞
y = π/2
y = –π/2
(0,0)
(1,π/4)
(–1,–π/4)
(√3, π/3)
(–√3, –π/3)

Here is the graph of y = tan–1(x)
x
tan–1(x) π/60 π/4 π/3–π/2
0 1/3 1 3 ∞
–π/6
–1/3
–π/4
–1
–π/3
–3
π/2
–∞
y = π/2
y = –π/2
(0,0)
(1,π/4)
(–1,–π/4)
(√3, π/3)
(–√3, –π/3)

Here is the graph of y = tan–1(x)
x
tan–1(x) π/60 π/4 π/3–π/2
0 1/3 1 3 ∞
–π/6
–1/3
–π/4
–1
–π/3
–3
π/2
–∞
y = π/2
y = –π/2
(0,0)
(1,π/4)
Remark: y =tan–1(x) has two horizontal asymptotes.
(–1,–π/4)
(√3, π/3)
(–√3, –π/3)

Cotangent–inverse is similar to tangent–inverse.
Given any number a, we define cot–1(a) = ,
where 0 <  < π and that cot() = a.
Cot–1(a) is also written as arccot(a) or acot(a).

Cotangent–inverse is similar to tangent–inverse.
Given any number a, we define cot–1(a) = ,
where 0 <  < π and that cot() = a.
Cot–1(a) is also written as arccot(a) or acot(a).
Here is the graph of y = cot–1(x)
(0, π/2)
(1, π/4)
x
y
π
(–1, 3π/4)
(√3, π/6)
(–√3, 5π/6)

For |a| ≥ 1, we define sec–1(a) =  with sec() = a,
where 0 <  < π,  ≠ π/2.
Sec–1(a) is also noted as arcsec(a) or asec(a).
Sin() and cos() vary between –1 and 1,
so their reciprocals |csc()| and |sec()| ≥ 1.

Using x and y, here is the graph of y = sec–1(x).
(1, 0) x
y
(√2, π/4)
(2, π/3)
(–1, π)
(–√2, 3π/4)
(–2, 2π/3)
For |a| ≥ 1, we define sec–1(a) =  with sec() = a,
where 0 <  < π,  ≠ π/2.
Sec–1(a) is also noted as arcsec(a) or asec(a).
Sin() and cos() vary between –1 and 1,
so their reciprocals |csc()| and |sec()| ≥ 1.
π

Given a number a such that |a| ≥ 1,
we define csc–1(a) =  with csc() = a
where –π/2 <  < π/2,  ≠ 0.
Csc–1(a) is also noted as arccsc (a) or acsc (a).

Given a number a such that |a| ≥ 1,
we define csc–1(a) =  with csc() = a
where –π/2 <  < π/2,  ≠ 0.
Csc–1(a) is also noted as arccsc (a) or acsc (a).
Using x and y, here is the graph of y = csc–1(x).
(1, π/2)
x
y
(–√2, –π/4)
(2, π/6)
(–1, –π/2)
(√2, π/4)
(–2, –π/6)
–π/2
π/2

One of the reasons for creating inverse functions is
for solving equations.
Linear Trig-Equations

The function 3x = y is different from x2 or sin(x)
in that if we know the output y = 1/2, then we know
the precise input x, in this case, 3x = ½, so x = 1/6.

Such is not the case for x2 or sin(x).

If the output y = ½ so that x2 = ½ or sin(x) = ½,
we won’t be able to determine a precise input,
instead we have multiple or even infinite possibilities.

In these situations, we utilize trig-inverse functions to
obtain their solutions.

In these situations, we utilize trig-inverse functions to
obtain their solutions.
Specifically, we will solve linear trig-equations,
that is, equations of the form aX + b = 0
where X is a trig-function.

Important Trigonometric Values
Solving linear trig-equations is to solve for positions
on the unit circle.

on the unit circle. From the position-solutions,
we apply the inverse trig-functions to obtain angular-
solutions and the list of all solutions, if necessary.

Example g. a. Draw the two locations on the unit
circle corresponding to  where 5sin() – 3 = 0.

5sin() – 3 = 0 or
sin() = 3/5 = y-coordinate of the
point on the circle.

5sin() – 3 = 0 or
It’s positive so both points are at
the top-half of the circle as shown.
(–4/5, 3/5) (4/5, 3/5)

By the Pythagorean Theorem we find
x = ±4/5 so the points are (4/5, 3/5) or (–4/5, 3/5).
5sin() – 3 = 0 or
It’s positive so both points are at
the top-half of the circle as shown.
(–4/5, 3/5) (4/5, 3/5)

b. Find the values of between 0 and π for the ’s.
(–4/5, 3/5) (4/5, 3/5)
A
Since sin() = 3/5, from the figure,
angle A is sin–1(3/5)  36.9o,

and angle B  180 – 36.9o = 143.1o
(–4/5, 3/5) (4/5, 3/5)
A
B

and angle B  180 – 36.9o = 143.1o
(–4/5, 3/5) (4/5, 3/5)
A
B
c. List all solutions of 5sin() – 3 = 0.

and angle B  180 – 36.9o = 143.1o
(–4/5, 3/5) (4/5, 3/5)
A
B
Each 360o corresponds to one cycle,
or that n360o corresponds to going around n times,
where n is an integer.

and angle B  180 – 36.9o = 143.1o
(–4/5, 3/5) (4/5, 3/5)
A
B
Hence the following is a complete list of solutions:
{36.9o+ n360o or 143.1o + n360o}

and angle B  180 – 36.9o = 143.1o
(–4/5, 3/5) (4/5, 3/5)
A
B
Hence the following is a complete list of solutions:
{36.9o+ n360o or 143.1o + n360o}
Sometimes we use the argument to π or 2π in the
equation to obtain answers in integers or fractions,
instead of answers in multiples of π.

Example H.
a. Draw the two locations corresponding to angle α
on the unit circle where 2cos(α) – 1 = 0.

Example H.
We have 2cos(α) – 1 = 0
or cos(α) = ½ .

Example H.
on the unit circle where 2cos(α) – 1 = 0. (1/2, √2/2)
(1/2, –√2/2)
or cos(α) = ½ .
Hence the points are (1/2, √2/2)
and(1/2, –√2/2).

Example H.
b. List two angles for α that correspond
to the two points then list all the solutions for α.
or cos(α) = ½ .
and(1/2, –√2/2).
(1/2, √2/2)
(1/2, –√2/2)

Example H.
(1/2, √2/2)
(1/2, –√2/2)
or cos(α) = ½ .
and(1/2, –√2/2).
–π/3
π/3
Two angles for the positions are α = ±π/3.

Example H.
(1/2, √2/2)
(1/2, –√2/2)
or cos(α) = ½ .
and(1/2, –√2/2).
Hence a list for all the solutions for α is {±π/3 + 2nπ}.
–π/3
π/3

Example H.
(1/2, √2/2)
(1/2, –√2/2)
or cos(α) = ½ .
and(1/2, –√2/2).
Hence a list for all the solutions for α is {2nπ±π/3}.
c. Find all the solutions for  where 2cos(2π) – 1 = 0.
–π/3
π/3

Example H.
(1/2, √2/2)
(1/2, –√2/2)
or cos(α) = ½ .
and(1/2, –√2/2).
We have that cos(2π) = ½ so 2π = 2nπ±π/3,
–π/3
π/3

Example H.
(1/2, √2/2)
A
B
(1/2, –√2/2)
or cos(α) = ½ .
and(1/2, –√2/2).
We have that cos(2π) = ½ so 2π = 2nπ±π/3,
or  = (2nπ±π/3)/2π = n±1/6 where n is any integer.

7. inverse trig functions and linear trig equations-x

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