**10.5 Graphs of the Trigonometric Functions**

**10.5.3 Graphs of the Tangent and Cotangent Functions**

Finally, we turn our attention to the graphs of the tangent and cotangent functions. When con- structing a table of values for the tangent function, we see that J(x) = tan(x) is undefined at x = π

2 and x = 3π

2 , in accordance with our findings in Section 10.3.1. As x →

π

2

−_{, sin(x)} _{→} _{1}−
and cos(x)→0+_{, so that tan(x) =} sin(x)

cos(x) → ∞producing a vertical asymptote at x=

π

2. Using a

similar analysis, we get that asx→ π_{2}+, tan(x)→ −∞; asx→ 3_{2}π−, tan(x)→ ∞; and asx→ 3_{2}π+,
tan(x)→ −∞. Plotting this information and performing the usual ‘copy and paste’ produces:

x tan(x) (x,tan(x)) 0 0 (0,0) π 4 1 π 4,1 π 2 undefined 3π 4 −1 3π 4 ,−1 π 0 (π,0) 5π 4 1 5π 4 ,1 3π 2 undefined 7π 4 −1 7π 4 ,−1 2π 0 (2π,0) x y π 4 π 2 34π π 5π 4 3π 2 7π 4 2π −1 1

The graph of y= tan(x) over [0,2π].

x y

From the graph, it appears as if the tangent function is periodic with periodπ. To prove that this is the case, we appeal to the sum formula for tangents. We have:

tan(x+π) = tan(x) + tan(π) 1−tan(x) tan(π) =

tan(x) + 0

1−(tan(x))(0) = tan(x),

which tells us the period of tan(x) is at most π. To show that it is exactly π, suppose p is a positive real number so that tan(x+p) = tan(x) for all real numbers x. For x = 0, we have tan(p) = tan(0 +p) = tan(0) = 0, which meansp is a multiple of π. The smallest positive multiple ofπ isπ itself, so we have established the result. We take as our fundamental cycle for y= tan(x) the interval −π

2,

π

2

, and use as our ‘quarter marks’x=−π

2,− π 4, 0, π 4 and π

2. From the graph, we

see confirmation of our domain and range work in Section 10.3.1.

It should be no surprise that K(x) = cot(x) behaves similarly to J(x) = tan(x). Plotting cot(x) over the interval [0,2π] results in the graph below.

x cot(x) (x,cot(x)) 0 undefined π 4 1 π 4,1 π 2 0 π 2,0 3π 4 −1 3π 4 ,−1 π undefined 5π 4 1 5π 4 ,1 3π 2 0 3π 2 ,0 7π 4 −1 7π 4 ,−1 2π undefined x y π 4 π 2 34π π 5π 4 3π 2 7π 4 2π −1 1

The graph ofy = cot(x) over [0,2π].
From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader
to prove this.14 _{We take as one fundamental cycle the interval (0, π) with quarter marks:} _{x} _{= 0,}

π

4,

π

2, 3π

4 and π. A more complete graph of y = cot(x) is below, along with the fundamental cycle

highlighted as usual. Once again, we see the domain and range of K(x) = cot(x) as read from the graph matches with what we found analytically in Section10.3.1.

14

Certainly, mimicking the proof that the period of tan(x) is an option; for another approach, consider transforming tan(x) to cot(x) using identities.

x y

The graph of y= cot(x).

The properties of the tangent and cotangent functions are summarized below. As with Theorem

10.24, each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof.

Theorem 10.25. Properties of the Tangent and Cotangent Functions

The function J(x) = tan(x)

– has domain
x:x6= π_{2} +πk, k is an integer =
∞
[
k=−∞
(2k+ 1)π
2 ,
(2k+ 3)π
2
– has range (−∞,∞)

– is continuous and smooth on its domain

– is odd

– has period π

The function K(x) = cot(x)

– has domain {x:x6=πk, k is an integer}= ∞ [

k=−∞

(kπ,(k+ 1)π)

– has range (−∞,∞)

– is continuous and smooth on its domain

– is odd

Example 10.5.5. Graph one cycle of the following functions. Find the period. 1. f(x) = 1−tan x 2 . 2. g(x) = 2 cot π 2x+π + 1. Solution.

1. We proceed as we have in all of the previous graphing examples by setting the argument of
tangent in f(x) = 1−tan x_{2}

, namely x_{2}, equal to each of the ‘quarter marks’ −π_{2},−π_{4}, 0, π_{4}
and π_{2}, and solving forx.

a x
2 =a x
−π_{2} x_{2} =−π_{2} −π
−π_{4} x_{2} =−π_{4} −π_{2}
0 x_{2} = 0 0
π
4
x
2 =
π
4
π
2
π
2
x
2 =
π
2 π

Substituting thesex-values intof(x), we find points on the graph and the vertical asymptotes.

x f(x) (x, f(x))
−π undefined
−π_{2} 2 −π_{2},2
0 1 (0,1)
π
2 0
π
2,0
π undefined
x
y
−π −π
2
π
2 π
−2
−1
1
2

One cycle ofy= 1−tan x

2

. We see that the period is π−(−π) = 2π.

2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π_{4}, π_{2}, 3_{4}π and π.
To graph g(x) = 2 cot π

2x+π

+ 1, we begin by setting π

2x+π equal to each quarter mark

a π
2x+π=a x
0 π
2x+π = 0 −2
π
4
π
2x+π=
π
4 −
3
2
π
2
π
2x+π=
π
2 −1
3π
4
π
2x+π=
3π
4 −
1
2
π π_{2}x+π =π 0
We now use thesex-values to generate our graph.

x g(x) (x, g(x))
−2 undefined
−3
2 3 −
3
2,3
−1 1 (−1,1)
−1_{2} −1 −1_{2},−1
0 undefined
x
y
−2 −1
−1
1
2
3

One cycle ofy= 2 cot π_{2}x+π
+ 1.

We find the period to be 0−(−2) = 2.

As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however.

10.5.4 Exercises

In Exercises1 -12, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function.

1. y= 3 sin(x) 2. y= sin(3x) 3. y=−2 cos(x)

4. y= cosx−π 2 5. y=−sinx+ π 3 6. y= sin(2x−π) 7. y=−1 3cos 1 2x+ π 3 8. y= cos(3x−2π) + 4 9. y= sin−x−π 4 −2 10. y= 2 3cos π 2 −4x + 1 11. y=−3 2cos 2x+π 3 −1 2 12. y= 4 sin(−2πx+π) In Exercises13 - 24, graph one cycle of the given function. State the period of the function.

13. y= tanx− π 3 14. y= 2 tan 1 4x −3 15. y= 1 3tan(−2x−π) + 1 16. y= secx−π 2 17. y=−cscx+π 3 18. y=−1 3sec 1 2x+ π 3 19. y= csc(2x−π) 20. y= sec(3x−2π) + 4 21. y= csc−x−π 4 −2 22. y= cotx+π 6 23. y=−11 cot 1 5x 24. y = 1 3cot 2x+3π 2 + 1

In Exercises 25 - 34, use Example 10.5.3 as a guide to show that the function is a sinusoid by rewriting it in the formsC(x) =Acos(ωx+φ) +B and S(x) =Asin(ωx+φ) +B forω >0 and 0≤φ <2π.

25. f(x) =√2 sin(x) +√2 cos(x) + 1 26. f(x) = 3√3 sin(3x)−3 cos(3x)

27. f(x) =−sin(x) + cos(x)−2 28. f(x) =−1

2sin(2x)− √

3

2 cos(2x)

29. f(x) = 2√3 cos(x)−2 sin(x) 30. f(x) = 3_{2}cos(2x)− 3
√
3
2 sin(2x) + 6
31. f(x) =−1
2cos(5x)−
√
3
2 sin(5x) 32. f(x) =−6
√
3 cos(3x)−6 sin(3x)−3

33. f(x) = 5 √ 2 2 sin(x)− 5√2 2 cos(x) 34. f(x) = 3 sin x 6 −3√3 cosx 6

35. In Exercises25-34, you should have noticed a relationship between the phasesφfor theS(x) and C(x). Show that if f(x) = Asin(ωx+α) +B, then f(x) = Acos(ωx+β) +B where β =α−π

2.

36. Let φ be an angle measured in radians and let P(a, b) be a point on the terminal side of φ
when it is drawn in standard position. Use Theorem 10.3 and the sum identity for sine in
Theorem10.15to show thatf(x) =asin(ωx) +bcos(ωx) +B (withω >0) can be rewritten
asf(x) =√a2_{+}_{b}2_{sin(ωx}_{+}_{φ) +}_{B.}

37. With the help of your classmates, express the domains of the functions in Examples 10.5.4

and 10.5.5using extended interval notation. (We will revisit this in Section 10.7.)

In Exercises38 - 43, verify the identity by graphing the right and left hand sides on a calculator.
38. sin2(x) + cos2_{(x) = 1} _{39. sec}2_{(x)}_{−}_{tan}2_{(x) = 1} _{40. cos(x) = sin}π

2 −x

41. tan(x+π) = tan(x) 42. sin(2x) = 2 sin(x) cos(x) 43. tanx 2

= sin(x) 1 + cos(x) In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates.

44. f(x) = cos(3x) + sin(x). Is this function periodic? If so, what is the period?
45. f(x) = sin(_{x}x). What appears to be the horizontal asymptote of the graph?

46. f(x) =xsin(x). Graph y=±x on the same set of axes and describe the behavior of f. 47. f(x) = sin 1

x

. What’s happening asx→0?

48. f(x) =x−tan(x). Graph y=x on the same set of axes and describe the behavior of f.
49. f(x) = e−0.1x_{(cos(2x) + sin(2x)). Graph} _{y} _{=} _{±e}−0.1x _{on the same set of axes and describe}

the behavior of f.

50. f(x) =e−0.1x_{(cos(2x) + 2 sin(x)). Graph} _{y} _{=}_{±e}−0.1x _{on the same set of axes and describe}

the behavior of f.

51. Show that a constant function f is periodic by showing that f(x+ 117) = f(x) for all real numbers x. Then show that f has no period by showing that you cannot find a smallest number p such that f(x+p) = f(x) for all real numbers x. Said another way, show that f(x+p) =f(x) for all real numbersx for ALL values of p >0, so no smallest value exists to satisfy the definition of ‘period’.

10.5.5 Answers
1. y= 3 sin(x)
Period: 2π
Amplitude: 3
Phase Shift: 0
Vertical Shift: 0 x
y
π
2 π 3_{2}π 2π
−3
3
2. y= sin(3x)
Period: 2π
3
Amplitude: 1
Phase Shift: 0
Vertical Shift: 0 _{x}
y
π
6
π
3
π
2 2_{3}π
−1
1
3. y=−2 cos(x)
Period: 2π
Amplitude: 2
Phase Shift: 0
Vertical Shift: 0
x
y
π
2 π 32π 2π
−2
2
4. y= cosx−π
2
Period: 2π
Amplitude: 1
Phase Shift: π
2
Vertical Shift: 0 _{x}
y
π
2 π 3_{2}π 2π 5_{2}π
−1
1

5. y=−sinx+ π
3
Period: 2π
Amplitude: 1
Phase Shift: −π
3
Vertical Shift: 0 _{x}
y
−π_{3} π_{6} 2π
3
7π
6
5π
3
−1
1
6. y= sin(2x−π)
Period: π
Amplitude: 1
Phase Shift: π
2
Vertical Shift: 0
x
y
π
2 34π π
5π
4
3π
2
−1
1
7. y=−1
3cos
1
2x+
π
3
Period: 4π
Amplitude: 1
3
Phase Shift: −2π
3
Vertical Shift: 0
x
y
−2π
3
π
3 4_{3}π 7_{3}π 10_{3}π
−1
3
1
3
8. y= cos(3x−2π) + 4
Period: 2π
3
Amplitude: 1
Phase Shift: 2π
3
Vertical Shift: 4
x
y
2π
3
5π
6 π
7π
6
4π
3
3
4
5

9. y= sin−x−π 4 −2 Period: 2π Amplitude: 1 Phase Shift: −π

4 (You need to use
y =−sinx+π
4
−2 to find this.)15
Vertical Shift: −2
x
y
−9π
4 −
7π
4 −
5π
4 −
3π
4 −
π
4
π
4 3_{4}π 5_{4}π 7_{4}π
−3
−2
−1
10. y= 2
3cos
π
2 −4x
+ 1
Period: π
2
Amplitude: 2
3
Phase Shift: π

8 (You need to use
y= 2
3cos
4x− π
2
+ 1 to find this.)16
Vertical Shift: 1
x
y
−3π
8 −
π
4 −
π
8
π
8
π
4 3_{8}π
π
2 5_{8}π
1
3
1
5
3
11. y=−3
2cos
2x+π
3
− 1
2
Period: π
Amplitude: 3
2
Phase Shift: −π
6
Vertical Shift: −1
2
x
y
−π_{6} _{12}π π_{3} 7π
12
5π
6
−2
−1_{2}
1
12. y= 4 sin(−2πx+π)
Period: 1
Amplitude: 4
Phase Shift: 1

2 (You need to use y=−4 sin(2πx−π) to find this.)17

Vertical Shift: 0 x y −1 2 − 1 4 1 4 1 2 3 4 1 5 4 3 2 −4 4

15_{Two cycles of the graph are shown to illustrate the discrepancy discussed on page}_{796}_{.}
16

Again, we graph two cycles to illustrate the discrepancy discussed on page796.

13. y= tanx− π
3
Period: π
x
y
−π_{6} _{12}π π_{3} 7π
12
5π
6
−1
1
14. y= 2 tan
1
4x
−3
Period: 4π
x
y
−2π −π π _{2π}
−5
−3
−1
15. y= 1
3tan(−2x−π) + 1
is equivalent to
y=−1
3tan(2x+π) + 1

via the Even / Odd identity for tangent. Period: π 2 x y −3π 4 − 5π 8 − π 2 −3π8 − π 4 4 3 1 2 3

16. y= sec x−π_{2}

Start withy= cos x−π

2
Period: 2π
x
y
π
2 π 3_{2}π 2π 5_{2}π
−1
1
17. y=−cscx+π
3

Start withy=−sinx+π
3
Period: 2π
x
y
−π_{3} π_{6} 2π
3
7π
6
5π
3
−1
1
18. y=−1
3sec
1
2x+
π
3
Start withy=−1
3cos
1
2x+
π
3
Period: 4π
x
y
−2_{3}π π_{3} 4_{3}π 7_{3}π 10_{3}π
−1_{3}
1
3

19. y= csc(2x−π)

Start withy= sin(2x−π)
Period: π
x
y
π
2 3_{4}π π 5_{4}π 3_{2}π
−1
1
20. y= sec(3x−2π) + 4

Start withy= cos(3x−2π) + 4 Period: 2π 3 x y 2π 3 5π 6 π 7π 6 4π 3 3 4 5 21. y= csc−x−π 4 −2 Start withy= sin−x−π

4
−2
Period: 2π _{x}
y
−π_{4} π_{4} 3π
4
5π
4
7π
4
−3
−2
−1

22. y= cotx+π
6
Period: π
x
y
−π
6
π
12
π
3 7_{12}π 5_{6}π
−1
1
23. y=−11 cot
1
5x
Period: 5π
x
y
5π
4
5π
2
15π
4 5π
−11
11
24. y= 1
3cot
2x+3π
2
+ 1
Period: π
2
x
y
−3π_{4} −5π_{8} −π
2 −3π8 −
π
4
4
3
1
2
3

25. f(x) =√2 sin(x) +√2 cos(x) + 1 = 2 sinx+π 4 + 1 = 2 cos x+7π 4 + 1

26. f(x) = 3√3 sin(3x)−3 cos(3x) = 6 sin 3x+ 11π 6 = 6 cos 3x+ 4π 3

27. f(x) =−sin(x) + cos(x)−2 =√2 sin x+3π 4 −2 =√2 cosx+π 4 −2 28. f(x) =−1 2sin(2x)− √ 3 2 cos(2x) = sin 2x+4π 3 = cos 2x+5π 6

29. f(x) = 2√3 cos(x)−2 sin(x) = 4 sin x+2π 3 = 4 cosx+π 6 30. f(x) = 3 2cos(2x)− 3√3 2 sin(2x) + 6 = 3 sin 2x+5π 6 + 6 = 3 cos2x+ π 3 + 6 31. f(x) =−1 2cos(5x)− √ 3 2 sin(5x) = sin 5x+7π 6 = cos 5x+2π 3

32. f(x) =−6√3 cos(3x)−6 sin(3x)−3 = 12 sin 3x+4π 3 −3 = 12 cos 3x+5π 6 −3 33. f(x) = 5 √ 2 2 sin(x)− 5√2 2 cos(x) = 5 sin x+7π 4 = 5 cos x+ 5π 4 34. f(x) = 3 sinx 6 −3√3 cosx 6 = 6 sin x 6 + 5π 3 = 6 cos x 6 + 7π 6