GRAPHING COSECANT FUNCTIONS

The graph of a reciprocal functions is related to the graph of its corresponding reciprocal trigonometric function in the following ways.

The graph of the reciprocal function has a vertical asymptote at each zero of the corresponding primary trigonometric function.

• It has the vertical asymptotes at the points where sin θ = 0
• It has the same period 2Π as y = sinθ
• It has the domain x ∈R, θ ≠ nΠ, n ∈ I
• It has the range {y ∈R, |y| ≥1}

Problem 1 :

y = csc (2(x + Π/2))

Solution :

y = csc (2(x + Π/2))

Comparing with y = a csc (k(x - c)) + d

• Amplitude a = 1
• k = 2
• Horizontal translation c = -Π/2
• Moving the graph towards the left of Π/2 units.
• No vertical shift.

Vertical asymptotes :

csc (2(x + Π/2)) = 1/sin (2(x + Π/2))

2(x + Π/2) = 0

x = -Π/2

Vertical asymptotes are -Π/2 + kΠ

Vertical asymptotes are at  -Π/2,  Π/2,  3Π/2,  5Π/2,  7Π/2,............

Period :

= 2Π/|k|

= 2Π/2

= Π

Π/4, Π/4 + Π/4, 2Π/4 + Π/4, 3Π/4 + Π/4

Inputs are Π/4, 2Π/4, 3Π/4, 4Π/4

(Π/4, -1) and (3Π/4, 1)

Problem 2 :

y = 2csc (x + Π/3)

Solution :

y = 2 csc (x + Π/3)

Comparing with y = a csc (k(x - c)) + d

• Amplitude a = 2
• k = 1
• Horizontal translation c = -Π/3
• Moving the graph towards the left of Π/3 units.
• No vertical shift.

Vertical asymptotes :

2 csc (x + Π/3) = 1/2sin (x + Π/3)

(x + Π/3) = 0

x = -Π/3

Vertical asymptotes are -Π/3 + kΠ

Vertical asymptotes are at x = -Π/3, 2Π/3, 5Π/3, 8Π/3

Period :

= 2Π/|k|

= 2Π/1

= 2Π

To find input, we divide the period by 4. 

2Π/4 ==> Π/2

-Π/3, -Π/3 + Π/2, Π/6 + Π/2, 2Π/3 + Π/2

Inputs are -Π/3, Π/6, 2Π/3, 7Π/6

Problem 3 :

y = 1 + 3 csc x

Solution :

y = 1 + 3 csc x

Comparing with y = a csc (k(x - c)) + d

• Amplitude a = 3
• k = 1
• No horizontal translation. No horizontal shift.
• Vertical shift of 1 unit up.

Vertical asymptotes :

x = Π, 2Π, 3Π,...............

Period :

= 2Π/|k|

= 2Π/1

= 2Π

To find the horizontal distance of each input, we divide period by 4. Then 

= 2Π/4

= Π/2

Π, Π + Π/2, 3Π/2 + Π/2, 2Π + Π/2, 5Π/2 + Π/2

Π, 3Π/2, 2Π, 5Π/2, 3Π

y = 1 + 3 csc x

When x = Π

y = 1 + 3 csc (Π)

Vertical asymptote is at x = Π

Problem 4 :

Graph one period of each function. Describe the transformation of the graph of its parent function. 

g(x) = csc x − 2

Solution :

g(x) = csc x − 2

Comparing with y = a csc (k(x - c)) + d

• Amplitude a = 1
• k = 1
• No horizontal translation. No horizontal shift.
• Vertical shift of 2 units down.

Vertical asymptotes :

x = Π, 2Π, 3Π,...............

Period :

= 2Π/|k|

= 2Π/1

= 2Π

To find the horizontal distance of each input, we divide period by 4. Then 

= 2Π/4

= Π/2

Π - Π/2, Π, Π + Π/2, 3Π/2 + Π/2, 2Π + Π/2, 5Π + Π/2

Π/2, Π, 3Π/2, 2Π, 5Π/2, 11Π/2