Sketch Trigonometric Functions - 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 = 2π
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 = π/2 + kπ , k = 0 , ~+mn~1, ~+mn~2, ...
Vertical asymptotes of y = csc(x) = 1 / sin(x) at the zeros of sin(x) given by x = kπ , k = 0 , ~+mn~1, ~+mn~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 secant and cosecant Functions: Examples with Detailed Solutions

  1. Sketch the graph of y = sec(2x - π/3) over one period.
    Solution
    Graphing Parameters
    range: (-∞ , - 1) ∪ (1, +∞)
    Period = 2π/2 = π
    Vertical asymptotes given by the soltuion to the equation: 2x - π/3 = π/2 + kπ which gives: x = 5π/12 + kπ/2, , k = 0 , ~+mn~1, ~+mn~2, ...
    Horizontal Shift: Because of the term - π/3, the graph is shifted horizontally. We first rewrite the given function as: y = sec[2(x - π/6)] and we can now write the shift as being equal to π/6 to the right.
    We sketch y = sec(2x - π/3) translating the graph of y = sec(2x) by π/6 to the right (red graph below) so that the sketched period starts at π/6 and ends at π/6 + π = 7π/6 which is one period equal to π.
    Graph of y = sec(2x - π/3)

  2. Sketch the graph of y = - 3 csc(x/2 + π/2) over one period.
    Solution
    Graphing Parameters
    range: (-∞ , -3) ∪ (3, +∞)
    Period = 2π/|k| = 2 π / (1/2) = 4 π
    Vertical asymptotes given by the solution to the equation: x/2 + π/2 = kπ which gives: x = (2k-1)π, , k = 0 , ~+mn~1, ~+mn~2, ...
    Horizontal Shift: Because of the term π/2, the graph is shifted horizontally. We first rewrite the given function as: y = - 3 csc[(1/2)(x + π)] and we can now write the shift as being equal to π to the left.
    We sketch - 3 csc(x/2 + π/2) by translating the graph of y = - 3 csc(x/2) to the left by π (red graph below) so that the sketched period starts at -π and ends at π + 4 π = 3π which is an interval equal to one period.
    Graph of  y = - 3 csc(x/2 + π/2)


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