Ma 113: The graph of y=csc(x)
Ma 113: The graph of y=csc(x)

Graph secant and cosecant

The sketching of the secant and
cosecant
functions of the form

y = a sec[ k ( x – d) ] and y = a csc[ k ( x – d)] are discussed with detailed examples.

Graphing Parameters of y = sec(x) and y = csc(x)

range: (-? , -1) ? (1 , +?)

Period = 2?

Horizontal Shift (translation) = d , to the left if (- d) is positive and to the right if (- d) is negative.

Vertical asymptotes of y = sec(x) = 1 / cos(x) at the zeros of cos(x) given by x = ?/2 + k? , k = 0 , �1, �2, …

Vertical asymptotes of y = csc(x) = 1 / sin(x) at the zeros of sin(x) given by x = k? , k = 0 , �1, �2, …

We need to know how to sketch basic secant and cosecant functions using the identities y = sec(x) = 1 / cos(x) and y = csc(x) = 1 / sin(x) to understand the vertical asymptotes.

y = sec(x) = 1 / cos(x)

All zeros of cos(x) (which is in the denominator) are vertical asymptotes of the sec(x).
graph of y = sec(x)

y = csc(x) = 1 / sin(x)

All zeros of sin(x) (which is in the denominator) are vertical asymptotes of the csc(x).
graph of y = csc(x)

Sketching and Graphing secant and cosecant Functions: Examples with Detailed Solutions

Example 1

Sketch the graph of y = sec(2x – ?/3) over one period.

solution

Graphing Parameters

range: (-? , – 1) ? (1, +?)

Period = 2?/2 = ?

Vertical asymptotes given by the soltuion to the equation: 2x – ?/3 = ?/2 + k? which gives: x = 5?/12 + k?/2, , k = 0 , �1, �2, …

Horizontal Shift: Because of the term – ?/3, the graph is shifted horizontally. We first rewrite the given function as: y = sec[2(x – ?/6)] and we can now write the shift as being equal to ?/6 to the right.

We sketch y = sec(2x – ?/3) translating the graph of y = sec(2x) by ?/6 to the right (red graph below) so that the sketched period starts at ?/6 and ends at ?/6 + ? = 7?/6 which is one period equal to ?.
Graph of y = sec(2x - ?/3)

Example 2

Sketch the graph of y = – 3 csc(x/2 + ?/2) over one period.

solution

Graphing Parameters

range: (-? , -3) ? (3, +?)

Period = 2?/|k| = 2 ? / (1/2) = 4 ?

Vertical asymptotes given by the solution to the equation: x/2 + ?/2 = k? which gives: x = (2k-1)?, , k = 0 , �1, �2, …

Horizontal Shift: Because of the term ?/2, the graph is shifted horizontally. We first rewrite the given function as: y = – 3 csc[(1/2)(x + ?)] and we can now write the shift as being equal to ? to the left.

We sketch – 3 csc(x/2 + ?/2) by translating the graph of y = – 3 csc(x/2) to the left by ? (red graph below) so that the sketched period starts at -? and ends at ? + 4 ? = 3? which is an interval equal to one period.
Graph of y = - 3 csc(x/2 + ?/2)

More References and links

secant functions

Cosecant Function

High School Maths (Grades 10, 11 and 12) – Free Questions and Problems With Answers

Middle School Maths (Grades 6, 7, 8, 9) – Free Questions and Problems With Answers

e-mail

Home Page

You are watching: Graph secant and cosecant. Info created by Bút Chì Xanh selection and synthesis along with other related topics.