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).
y = csc(x) = 1 / sin(x)
All zeros of sin(x) (which is in the denominator) are vertical asymptotes of the 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 ?.
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.
