Sketch secant cosecant and cotangent with separate branches
Use the zeros and signs of the original ratios to locate asymptotes, turning points, periods and ranges.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.3 and 3.3 cover all six trigonometric functions, reciprocal squared identities, compound and double angles, and amplitude-phase forms used in simplification, exact evaluation and equations. State angle units, keep original domains and find every solution in the supplied interval. Written practice is self-checked.
Before you begin
- Sketch sine, cosine and tangent and use their zeros.
- Evaluate reciprocal ratios and recognise an asymptote.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which range belongs to sec x?
What you will learn
- Sketch the reciprocal functions over a stated radian interval without joining excluded inputs.
- Compare the domains, ranges and periods of all six trigonometric functions.
Build reciprocal graphs from their denominators
Sine and cosine have all-real input domains, range [−1, 1] and period 2π. Their reciprocal secant and cosecant both have range y ≤ −1 or y ≥ 1 and period 2π. Neither reciprocal can be zero. Secant is undefined where cosine vanishes: on 0 ≤ x ≤ 2π these inputs are π/2 and 3π/2. Its labelled extrema are (0, 1), (π, −1) and (2π, 1), with separate branches tending to vertical asymptotes.
Cosecant is undefined where sine vanishes: x = 0, π, 2π in this interval. On (0, π) it is positive with minimum (π/2, 1); on (π, 2π) it is negative with maximum (3π/2, −1). Near a denominator zero, the magnitude grows without bound. The endpoint asymptotes remain excluded even when the drawing window includes 0 and 2π. A graph clipped to a finite vertical window has continuing branches, not extra endpoints.
Cotangent changes the zero and asymptote positions
Tangent and cotangent have range all real numbers and period π. Tangent is undefined where cosine is zero and has zeros where sine is zero. Cotangent reverses those roles: its denominator sine excludes 0, π and 2π here, while its zeros occur at π/2 and 3π/2. Each cotangent branch decreases from positive unbounded values to negative unbounded values as x runs across (0, π) or (π, 2π). It is not a tangent graph with unchanged asymptotes.
For y = sec 2x, the factor 2 halves the period to π. On 0 ≤ x ≤ π, the asymptotes are π/4 and 3π/4, and key points are (0, 1), (π/2, −1) and (π, 1). Its range still excludes values strictly between −1 and 1. Use the transformed denominator cos 2x to locate forbidden inputs before solving sec 2x = 2 or connecting any branches.
Pause and explain
Which statement correctly describes cot x on 0 < x < π?
Worked example
Sketch y = sec x for 0 ≤ x ≤ 2π, giving its asymptotes, key extrema, period and range.
Show the worked solution
- The zeros of cos x locate excluded asymptotes at π/2 and 3π/2.
- Reciprocate cos values 1, −1, 1 at 0, π, 2π to get the three key extrema (0, 1), (π, −1), (2π, 1).
- Draw separate signed branches approaching the asymptotes; the period is 2π and range is y ≤ −1 or y ≥ 1.
Answer Asymptotes π/2, 3π/2; extrema (0, 1), (π, −1), (2π, 1); period 2π; range y ≤ −1 or y ≥ 1.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Sketch cosec x for 0 ≤ x ≤ 2π, stating the asymptotes, signed extrema and period.
Give me a hint
Use sine's zeros for the excluded inputs and its ±1 values for the reciprocal extrema.
Compare my reasoning
- Sine vanishes at 0, π and 2π, so these are excluded vertical asymptotes.
- The positive branch has minimum (π/2, 1); the negative branch has maximum (3π/2, −1).
- The period is 2π, with range y ≤ −1 or y ≥ 1; neither branch crosses zero.
Asymptotes 0, π, 2π; minimum (π/2, 1), maximum (3π/2, −1); period 2π; range y ≤ −1 or y ≥ 1.
Look for these in your work
- Locate asymptotes from the original denominator.
- Keep signs and branch extrema distinct.
Sketch cot x over 0 ≤ x ≤ 2π and compare its period and asymptotes with tan x.
Give me a hint
Use cos x/sin x, so sine controls forbidden inputs and cosine controls the zeros.
Compare my reasoning
- Cotangent excludes 0, π and 2π and has zeros at π/2 and 3π/2.
- Draw a decreasing branch on each interval (0, π) and (π, 2π), with range ℝ and period π.
- Tangent also has period π and range ℝ, but its asymptotes are π/2 and 3π/2 here and its branches increase.
cot: asymptotes 0, π, 2π; zeros π/2, 3π/2; decreasing branches; range ℝ; period π. Tangent's asymptotes are π/2, 3π/2.
Look for these in your work
- Use the correct denominator rather than copy tangent's graph.
- State separate branches, range and period.
For y = sec 2x on 0 ≤ x ≤ π, state the asymptotes and solve y = 2 exactly.
Give me a hint
First find the denominator zeros, then convert the valid reciprocal equation to cos 2x = 1/2.
Compare my reasoning
- cos 2x = 0 at x = π/4 and 3π/4, giving the two excluded asymptotes.
- The equation sec 2x = 2 gives cos 2x = 1/2; 0 ≤ 2x ≤ 2π supplies angles π/3 and 5π/3.
- Divide by 2 to obtain x = π/6 and 5π/6; both lie in the interval and have non-zero cosine denominator.
Asymptotes x = π/4, 3π/4; y = 2 at x = π/6 or 5π/6.
Look for these in your work
- Apply the input scaling to both asymptotes and equation angles.
- Check the original reciprocal domain for every solution.
Common mistakes
- Secant and cosecant have no zeros and no finite amplitude.
- Mark cotangent's sine zeros as asymptotes rather than use tangent's cosine zeros.
- An excluded asymptote is not an attained endpoint; separate branches must stay separate.
What can you explain now?
What are cotangent's zeros and excluded asymptotes in 0 ≤ x ≤ 2π?
Compare with the explanation
Zeros π/2 and 3π/2; excluded asymptotes 0, π and 2π.
Cosine zeros give cotangent zeros when sine is non-zero; sine zeros make its defining quotient undefined.
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Make it stick
Sketch all six functions for one repeated cycle. Explain why the reciprocal functions' zeros, ranges and excluded inputs differ from their original ratios.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.