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Section 6.5 – Graphs of the Tangent, Cotangent, Cosecant, and
Secant Functions
In this section, we will look at the graphs of the other four trigonometric
functions. We will start by examining the tangent function. Recall that the
period of the tangent function is π. Since tan(x) =
sin(x)
,
cos( x)
then the tangent
function is undefined when cos(x) = 0 which is when
x = …, –
3π
2
between –
,–
π
2
π
2
,
and
π 3π
, 2 , … So, we will first sketch the graph for values of x
2
π
and then use the fact that the graph is periodic to draw
2
the rest of the graph. We will need to look at the behavior of the function near
–
π
2
and
As x → –
π
2
π
2
and make a table of values to graph the tangent function:
in the fourth quadrant, cos(x) → a very small positive number
going to zero and sin(x) → – 1. Thus, tan(x) =
π
2
Likewise, As x →
sin(x)
cos( x)
≈
−1
small+ #
= – ∞.
in the first quadrant, cos(x) → a very small positive
number going to zero and sin(x) → 1. Thus, tan(x) =
sin(x)
cos( x)
Thus, the tangent function has vertical asymptotes at x = –
≈
π
2
1
small+ #
= ∞.
and x =
π
2
.
Now, let's make a table of values.
x
–
π
2
–
tan(x) undef. –
π
3
3
–
π
4
–1
–
–
π
6
3
3
Now, will put the information on a graph:
0
0
π
6
3
3
π
4
1
π
3
3
π
2
undef.
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–π
–
π
2
π
2
π
π
2
π
Now, draw a smooth curve:
–π
–
π
2
This is the graph of y = tan(x) on the interval [–
π
2
,
π
2
]. Since it is periodic,
this shape repeats and thus we get the following graph:
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y = tan(x)
– 2π
–π
π
2π
Properties of the Tangent Function
(2n+ 1)π
2
1)
The domain is {x| x ≠
where n is an integer}.
2)
3)
4)
The range is (– ∞, ∞).
The tangent function is odd so it is symmetric with respect to the origin.
The tangent function is periodic with a period of π.
The x-intercepts are {(nπ, 0) | n is an integer}. The y-intercept is (0, 0).ß
5)
The vertical asymptotes are { x = a | a =
(2n+ 1)π
2
where n is an integer}.
We will now examine the cotangent function. Recall that the period of the
cotangent function is π. Since cot(x) =
cos(x)
sin(x)
, then the cotangent function is
undefined when sin(x) = 0 which is when
x = …, – 2π, – π, 0, π, 2π, … So, we will first sketch the graph for values of x
between 0 and π and then use the fact that the graph is periodic to draw the
rest of the graph. We will need to look at the behavior of the function near 0
and π and make a table of values to graph the cotangent function:
As x → 0 in the first quadrant, sin(x) → a very small positive number going to
zero and cos(x) → 1. Thus, cot(x) =
cos(x)
sin(x)
≈
1
small+ #
= ∞. Likewise, As x → π
in the second quadrant, sin(x) → a very small positive number going to zero
and cos(x) → – 1. Thus, cot(x) =
cos(x)
sin(x)
≈
−1
small+ #
= – ∞. Thus, the cotangent
function has vertical asymptotes at x = 0 and x = π.
Now, let's make a table of values.
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x
0
cot(x) undef.
π
6
π
4
3
1
π
3
3
3
π
2
0
–
2π
3
3
3
3π
4
–1
5π
6
–
3
π
2
undef.
Now, will put the information on a graph:
–
π
2
π
2
π
3π
2
π
2
π
3π
2
Now, draw a smooth curve:
–
π
2
This is the graph of y = cot(x) on the interval [0, π]. Since it is periodic, this
shape repeats and thus we get the following graph:
y = cot(x)
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– 2π
–π
π
2π
Properties of the Cotangent Function
1)
The domain is {x| x ≠ nπ where n is an integer}.
The range is (– ∞, ∞).
2)
The cotangent function is odd so it is symmetric with respect to the
origin.
3)
The cotangent function is periodic with a period of π.
(2n+ 1)π
,
2
0) | n is an integer}.
4)
The x-intercepts are {(
5)
There is no y-intercept.
The vertical asymptotes are { x = a | a = nπ where n is an integer}.
Objective 1:
Graph Functions of the Form y = Atan(ωx) and y = Acot(ωx).
Just with the sine and cosine function, the amplitude |A| is a vertical
stretch/compression and
1
ω
is a horizontal stretch/compression. The main
difference is the period. Since the tangent and cotangent functions have a
period of π, a tangent or cotangent function with a horizontal
stretch/compression of
1
ω
will has a period of T =
π
ω
.
Use transformations to sketch the graph of the following:
Ex. 1a
y=–
Solution:
1
tan(3x)
4
+2
Ex. 1b
π
y = 2cot( 4 x) – 1
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a)
The amplitude is |–
1
|
4
1
4
=
and the period is T =
π
3
. The graph is
also reflected across the x-axis, vertically compressed by a factor of
1
,
4
and shifted up 2 units. Since the period is 1/3 the size of the period for
π
2
tan(x), instead of going from –
–
π
2
–π
b)
÷ 3=–
π
6
to
–
π
2
π
2
÷3=
π
6
to
π
2
, we will graph from
.
π
2
The amplitude is |2| = 2 and the period is T =
π
π
π
4
= 4. The graph is
not reflected across the x-axis but is vertically stretched by a factor of 2,
and shifted down 1 unit.
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The next function we want to explore is the cosecant function. The period for
the cosecant function is 2π. Since csc(x) =
1
sin(x)
then the cosecant function is
undefined when sin(x) = 0 which is when x = …, – 2π, – π, 0, π, 2π, …
These will act as the vertical asymptotes. To obtain the graph of the
cosecant, we can first draw the sine function and then use the reciprocal
relationship to "turn the graph inside out:"
– 3π
–π
π
3π
π
3π
Now, erase the sine function:
– 3π
–π
Properties of the Cosecant Function:
1)
The domain is {x| x ≠ nπ where n is an integer}.
The range is (– ∞,– 1] U [1, ∞).
2)
The cosecant function is odd so it is symmetric with respect to the
origin.
3)
The cosecant function is periodic with a period of 2π.
4)
There are no intercepts.
5)
The vertical asymptotes are { x = a | a = nπ where n is an integer}.
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The final function we want to explore is the secant function. The period for the
secant function is 2π. Since sec(x) =
1
cos(x)
then the cosecant function is
undefined when sin(x) = 0 which is when x = …, –
3π
2
,–
π
2
,
π
2
,
3π
2
,…
These will act as the vertical asymptotes. To obtain the graph of the secant,
we can first draw the cosine function and then use the reciprocal relationship
to "turn the graph inside out:"
– 3π
–π
π
3π
π
3π
Now, erase the cosine function:
– 3π
–π
Properties of the Secant Function:
1)
The domain is {x| x ≠
(2n+ 1)π
2
where n is an integer}.
3)
4)
The range is (– ∞,– 1] U [1, ∞).
The secant function is even so it is symmetric with respect to the
y-axis.
The secant function is periodic with a period of 2π.
There are no x-intercepts. The y-intercept is (0, 1).
5)
The vertical asymptotes are { x = a | a =
2)
(2n+ 1)π
2
where n is an integer}.
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Objective 2:
Graph Functions of the Form y = Acsc(ωx) and
y = Asec(ωx).
Just with the sine and cosine function, the amplitude |A| is a vertical
stretch/compression and
1
ω
is a horizontal stretch/compression. Since the
cosecant and secant functions have a period of 2π, a cosecant or secant
function with a horizontal stretch/compression of
T=
2π
ω
1
ω
will has a period of
.
Use transformations to sketch the graph of the following:
Ex. 2a
y = – csc(πx) – 2
Ex. 2b
y = 2sec(
π
3
x) + 5
Solution:
a)
2π
π
The amplitude is |– 1| = 1. The period is T =
= 2. The function
is reflected across the x-axis, compressed by a factor of
1
π
and shifted
down 2 units.
b)
The amplitude is |2| = 2. The period is T =
2π
π
3
= 6. The function
is reflected across the x-axis, compressed by a factor of
up 5 units.
3
π
and shifted
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Sometimes it might be easier to first draw the sine or cosine function first. We
would draw it reflected, stretched/compressed, and shifted in the same
fashion as the corresponding cosecant or secant function and then turn the
graph inside out. So, in part b, we could have drawn the corresponding
cosine function and use that to obtain the secant graph: