01 · Secant asymptotes
List all sec x asymptotes in −2π ≤ x ≤ 2π.
Hint
Solve cos x = 0 in the interval.
Worked solution
x = −3π/2, −π/2, π/2, 3π/2. These are excluded inputs, not graph points.
Understand · explore · practise
Sec, cosec and cot graphs
Sketch sec, cosec and cot graphs with their domains, ranges, periods and asymptotes. Explore denominator zeros and practise exact intercepts and turning points.
Before you startSec, cosec and cot definitions; sine and cosine graphs in radians
01 / Sketch from the denominator
A denominator approaching zero creates a growing reciprocal.
Begin with the familiar sine or cosine graph. Where the denominator is 0, the reciprocal function is undefined. Where it is small and positive, its reciprocal is large and positive; where it is small and negative, the reciprocal is large and negative.
sec x = 1/cos x
cosec x = 1/sin x
cot x = cos x/sin x
Choose a function and move the angle yourself. A missing gold point means the selected input is excluded. The dashed vertical lines are guides: they are not part of the graph.
Follow the denominatorExplore
x = 0 rad; cos x = 1.000; sec x = 1.000.
Sec: period 2π; range y ≤ −1 or y ≥ 1; no zeros.
Blue: function. Green dashed: denominator. Gold: selected function value. Dashed vertical lines are asymptotes. The graph continues beyond the viewing window.
Watch cosine shrink while secant grows
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Sketch y = sec x
Cosine zeros become vertical asymptotes; cosine extrema give secant turning points.
Domain: x ≠ π/2 + nπ
Range: y ≤ −1 or y ≥ 1
Period: 2π; even symmetry
n is any integer.
Build one central positive branchWorked example
At x = 0, cos x = 1 so sec x = 1
This is the bottom of the positive branch.
As x approaches ±π/2 from inside this interval, cos x approaches 0 from above
Sec x grows without bound, but never reaches the asymptote.
Between π/2 and 3π/2, cosine is negative
The branch lies at or below −1, with a top at (π,−1).
Repeat every 2π
Do not join neighbouring branches through an excluded input.
The turning points are (nπ, (−1)ⁿ). The y-intercept is (0,1). There are no x-intercepts because 1/cos x cannot equal 0.
03 / Sketch y = cosec x
Sine zeros become asymptotes; the branches alternate in sign.
Domain: x ≠ nπ
Range: y ≤ −1 or y ≥ 1
Period: 2π; odd symmetry
Sketch from 0 to 2πWorked example
Asymptotes at x = 0, π and 2π
All three inputs make sine 0 and are excluded.
On 0 < x < π, sine is positive
The branch has its minimum at (π/2,1).
On π < x < 2π, sine is negative
The branch has its maximum at (3π/2,−1).
The curves continue unbounded at each end of their intervals
The top or bottom of the visible drawing is not an endpoint.
Turning points are (π/2 + nπ, (−1)ⁿ). Cosec has no x-intercepts and no y-intercept. The y-axis itself is a vertical asymptote.
04 / Sketch y = cot x
Keep its zeros even where tangent is undefined.
Domain: x ≠ nπ
Range: all real numbers
Period: π; odd symmetry
Zeros: x = π/2 + nπ
One branch on 0 < x < πWorked example
Near 0 from the right, sin x is small positive and cos x is near 1
Cot x is large positive.
At π/2, cot x = 0/1 = 0
This is an ordinary x-intercept, not an asymptote.
Near π from the left, sin x is small positive and cos x is near −1
Cot x is large negative.
The branch decreases continuously from positive to negative unbounded values
It has no turning point. Repeat this branch every π.
Use cos x/sin x when finding the domain. The expression 1/tan x is only a shortcut where tangent exists and is nonzero. Cot is defined at every odd multiple of π/2.
05 / Justify the ranges
Reciprocals of numbers between −1 and 1 cannot lie strictly between −1 and 1.
For a nonzero real number u with |u| ≤ 1, we have |1/u| ≥ 1. Sine and cosine only take values between −1 and 1, so sec and cosec lie in two separated bands.
Sec and cosec: (−∞,−1] ∪ [1,∞)
Cot: (−∞,∞)
The ±1 values are attained, so their brackets are closed. Infinity is not a real value or endpoint that can be reached. Cot takes every real value because each branch runs continuously through all heights.
Use range information before solvingWorked example
sec x = 0.7 has no real solution
0.7 falls in the forbidden gap.
cosec x = −1 is possible
It occurs when sin x = −1.
cot x = 0 is possible
It occurs when cosine is 0 and sine is ±1.
06 / Use parity and periodicity
A reflection or translation must preserve both values and excluded inputs.
sec(−x) = sec x; sec(x + 2π) = sec x
cosec(−x) = −cosec x; cosec(x + 2π) = cosec x
cot(−x) = −cot x; cot(x + π) = cot x
Sec is symmetric about the y-axis. Cosec and cot have rotational symmetry through the origin. A half-turn translation reverses sec and cosec: sec(x + π) = −sec x and cosec(x + π) = −cosec x.
Why cot repeats soonerWorked example
cos(x + π) = −cos x and sin(x + π) = −sin x
Both signs reverse.
cot(x + π) = (−cos x)/(−sin x) = cot x
The signs cancel in the quotient, with the same exclusions shifted by π.
These statements apply wherever the expressions are defined. Symmetry never creates a value at an asymptote.
07 / Find features exactly on a chosen interval
List excluded inputs before marking turning points or intercepts.
Describe sec x on −π ≤ x ≤ πWorked example
Asymptotes: x = −π/2 and π/2
These inputs lie inside the interval and are excluded.
Turning point inside a smooth branch: (0,1)
The central branch is positive.
Boundary values: sec(−π) = sec π = −1
Both endpoints are included, but they are not interior turning points of the restricted graph.
No x-intercepts; y-intercept (0,1)
Keep the negative outside branches separate from the central branch.
Describe cot x on −π ≤ x ≤ πWorked example
Exclude −π, 0 and π
Even stated closed interval endpoints must belong to the function domain.
Zeros at −π/2 and π/2
Both values are included.
Two decreasing branches; no y-intercept or turning points
The x = 0 line is an asymptote.
08 / Read each side of an asymptote
The signs on the two sides can differ.
Approach the secant asymptote x = π/2Worked example
From the left, cos x approaches 0 through positive values
Sec x grows towards +∞.
From the right, cos x approaches 0 through negative values
Sec x decreases towards −∞.
Sec(π/2) is still undefined
These are descriptions of nearby behaviour, not a value at the input.
Similarly, cot x is large negative just to the left of 0 and large positive just to the right. Drawing a vertical line between those branches would falsely give many y-values at the excluded input.
When sketching by hand, label asymptotes, exact key points and the period before drawing smooth branches. A calculator’s straight line across a gap is a plotting artefact.
09 / Your turn
Give exact positions in radians and explain the exclusions.
02 · Cosecant asymptotes
List all cosec x asymptotes in −2π ≤ x ≤ 2π.
Hint
Sine is 0 at integer multiples of π.
Worked solution
x = −2π, −π, 0, π, 2π. All are excluded, including the displayed interval boundaries.
03 · Cotangent zeros
Find the zeros of cot x for −π < x < 2π.
Hint
Cosine must be 0 and sine nonzero.
Worked solution
x = −π/2, π/2, 3π/2. Cot x = 0 at all three.
04 · Positive branch
State the minimum point of sec x on −π/2 < x < π/2.
Hint
Cosine reaches its maximum 1 at 0.
Worked solution
(0,1). For all other inputs in this branch, 0 < cos x < 1, so sec x > 1.
05 · Negative branch
State the maximum point of cosec x on π < x < 2π.
Hint
Sine reaches −1 at 3π/2.
Worked solution
(3π/2,−1). The rest of this branch lies below −1.
06 · Intercepts
Which of sec x, cosec x and cot x have a y-intercept? State any such point.
Hint
Substitute x = 0 into the definitions.
Worked solution
Only sec x has a y-intercept, (0,1). Cosec and cot are undefined because sin 0 = 0.
07 · Period and symmetry
State the least positive period and parity of sec, cosec and cot.
Hint
Check a shift by π and a reflection x → −x.
Worked solution
Sec: 2π, even. Cosec: 2π, odd. Cot: π, odd. A π shift changes the signs of sec and cosec, so it is not their period.
08 · No solutions
Without solving, decide whether sec x = −0.8 and cosec x = 0 can have real solutions.
Hint
Use the two range bands.
Worked solution
Neither can. Both right-hand sides have magnitude less than 1, outside the sec/cosec ranges.
09 · Approach an asymptote
Describe cot x just to the left and just to the right of x = 0.
Hint
Cosine is near 1; sine changes sign at 0.
Worked solution
From the left, cot x decreases without bound towards −∞. From the right, cot x is positive and grows without bound as x approaches 0. Cot 0 is undefined.
10 · Restricted domain
Give the domain and range of sec x restricted to 0 ≤ x < π/2.
Hint
The asymptote endpoint is not included; x = 0 is included.
Worked solution
Domain [0,π/2). Range [1,∞). The minimum 1 is attained at 0; arbitrarily large positive values occur nearer π/2.
11 · Compare points
The point (π/6,2) lies on cosec x. Use symmetry and periodicity to find its matching points at −π/6 and 13π/6.
Hint
Cosec is odd and repeats after 2π.
Worked solution
(−π/6,−2) and (13π/6,2). The negative input reverses the value; adding a complete period preserves it.
12 · Diagnose a sketch
A sketch of cot x has an asymptote at π/2 and excludes its zero there because tan(π/2) is undefined. Correct it.
Hint
Use cot x = cos x/sin x.
Worked solution
At π/2, cot x = 0/1 = 0: draw an x-intercept. The nearest asymptotes are x = 0 and x = π, where sine is 0.
10 / Recap
Locate excluded inputs first, then build each branch.
- Sec asymptotes occur where cosine is 0.
- Cosec and cot asymptotes occur where sine is 0.
- Sec and cosec have no zeros and |y| ≥ 1.
- Cot has zeros at odd multiples of π/2 and takes all real values.
- Sec and cosec repeat after 2π; cot repeats after π.
- Never join a curve through an asymptote.
Section 1 of 10 · Sketch from the denominator