Graphs of Reciprocal Trig Functions
You already know $\sin x$, $\cos x$, and $\tan x$ — but their reciprocals $\csc x$, $\sec x$, and $\cot x$ behave very differently. Wherever the original function hits zero, the reciprocal explodes to infinity, creating vertical asymptotes. In this lesson you'll see exactly where those asymptotes appear, what the graphs look like, and how period and symmetry carry over from the original functions.
Where does $y = \csc x$ have vertical asymptotes? What happens to $\csc x$ when $\sin x = 1$? When $\sin x = -1$? Write your gut answer before reading on.
Key facts
- The key features of $y = \csc x$, $y = \sec x$, and $y = \cot x$
- The period of each reciprocal trig function
- The equations of all vertical asymptotes for each function
Concepts
- How the reciprocal relationship creates asymptotes at the zeros of the original function
- Why local minima and maxima of the reciprocal occur at the extrema of the original
- Why $\cot x$ has period $\pi$ while $\csc x$ and $\sec x$ have period $2\pi$
Skills
- Sketch $\csc x$, $\sec x$, and $\cot x$ over any specified interval
- State domains, ranges, and asymptote equations
- Determine the sign of a reciprocal trig function in each quadrant
Since $\csc x = \dfrac{1}{\sin x}$, the graph of $y = \csc x$ is determined entirely by $y = \sin x$:
- Period: $2\pi$ (same as $\sin x$)
- Vertical asymptotes where $\sin x = 0$: at $x = n\pi$, $n \in \mathbb{Z}$
- Local minima at $x = \dfrac{\pi}{2} + 2n\pi$ where $\csc x = 1$ (because $\sin x = 1$ is its maximum)
- Local maxima at $x = \dfrac{3\pi}{2} + 2n\pi$ where $\csc x = -1$ (because $\sin x = -1$ is its minimum)
- Range: $(-\infty, -1] \cup [1, +\infty)$ — the function never takes values between $-1$ and $1$
- Sign: positive where $\sin x > 0$ (Q1 and Q2), negative where $\sin x < 0$ (Q3 and Q4)
Each branch of $y = \csc x$ is a U-shaped (or inverted-U) curve that opens away from the $x$-axis between consecutive asymptotes.
Dashed curve: $y = \sin x$ (guide). Solid curve: $y = \csc x$. Red dashed lines: vertical asymptotes. Teal lines: $y = \pm 1$.
x = 1{ x}; period = 2; range = (-,-1] [1,); Vertical asymptotes at x = n (where x = 0)
Pause — copy the key features of $y = \csc x$ into your book: period $2\pi$, range $(-\infty,-1]\cup[1,+\infty)$, vertical asymptotes at $x = n\pi$, branch minimum at $(\pi/2,\,1)$.
Quick check: What are the equations of the vertical asymptotes of $y = \csc x$ in the interval $[0, 2\pi]$?
We just saw that $y = \csc x$ has vertical asymptotes at $x = n\pi$ and range $(-\infty,-1]\cup[1,+\infty)$. That raises a question: does $y = \sec x$ share the same range and period, or are its asymptotes in different positions? This card answers it → $\sec x = 1/\cos x$ has the same range but asymptotes shifted to $x = \frac{\pi}{2}+n\pi$ because cosine is zero there, not sine.
Since $\sec x = \dfrac{1}{\cos x}$, the graph is determined by $y = \cos x$. Note that the asymptote positions are shifted by $\dfrac{\pi}{2}$ compared to $\csc x$:
- Period: $2\pi$
- Vertical asymptotes where $\cos x = 0$: at $x = \dfrac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$
- Local minima at $x = 2n\pi$ where $\sec x = 1$ (because $\cos x = 1$ is its maximum)
- Local maxima at $x = \pi + 2n\pi$ where $\sec x = -1$ (because $\cos x = -1$ is its minimum)
- Range: $(-\infty, -1] \cup [1, +\infty)$ — same as $\csc x$
- Sign: positive where $\cos x > 0$ (Q1 and Q4), negative where $\cos x < 0$ (Q2 and Q3)
The shape of each branch is the same as $\csc x$ — it is simply shifted $\dfrac{\pi}{2}$ to the left, because $\sec x = \csc\!\left(x + \dfrac{\pi}{2}\right)$.
x = 1{ x}; period = 2; range = (-,-1] [1,); Vertical asymptotes at x = {2} + n (where x = 0)
Pause — copy the key features of $y = \sec x$ into your book: period $2\pi$, range $(-\infty,-1]\cup[1,+\infty)$, vertical asymptotes at $x = \frac{\pi}{2}+n\pi$, key points $(0,1)$, $(\pi,-1)$, $(2\pi,1)$.
Did you get this? True or false: $y = \sec x$ has a vertical asymptote at $x = \pi$.
Worked examples · 3 in a row, reveal as you go
Sketch $y = \sec x$ for $0 \le x \le 2\pi$, marking asymptotes and key points clearly.
For $\dfrac{\pi}{2} < x \le \pi$: $\cos x$ decreases from 0 to $-1$, so $\sec x$ increases from $-\infty$ to $-1$ → left branch opens downward.
Repeat symmetrically for $\pi < x \le 2\pi$.
We just saw that both $\csc x$ and $\sec x$ have range $(-\infty,-1]\cup[1,+\infty)$ and period $2\pi$. That raises a question: does $y = \cot x$ behave the same way, or is it fundamentally different in range and period? This card answers it → $\cot x$ has range $\mathbb{R}$ (all real values), period $\pi$, and is strictly decreasing on each branch — unlike the U-shaped branches of $\csc$ and $\sec$.
$\cot x = \dfrac{\cos x}{\sin x}$. This function behaves differently from $\csc x$ and $\sec x$ — it does not have a restricted range and it is a strictly decreasing function within each period:
- Period: $\pi$ (half the period of $\csc x$ and $\sec x$)
- Vertical asymptotes where $\sin x = 0$: at $x = n\pi$, $n \in \mathbb{Z}$ (same positions as $\csc x$)
- Zeros where $\cos x = 0$: at $x = \dfrac{\pi}{2} + n\pi$ (where $\tan x$ is undefined)
- Decreasing on every branch from $+\infty$ to $-\infty$
- Range: $(-\infty, +\infty)$ — takes all real values (unlike $\csc x$ and $\sec x$)
- Sign: positive in Q1 ($0 < x < \dfrac{\pi}{2}$), negative in Q2 ($\dfrac{\pi}{2} < x < \pi$)
Why period $\pi$? $\cot x = \dfrac{\cos x}{\sin x}$ and both numerator and denominator repeat with period $2\pi$, but their ratio repeats with period $\pi$ because $\cot(\pi + x) = \dfrac{\cos(\pi+x)}{\sin(\pi+x)} = \dfrac{-\cos x}{-\sin x} = \cot x$.
x = x{ x}; period = ; range = R (all real values); Vertical asymptotes at x = n (where x = 0)
Pause — copy the key features of $y = \cot x$ into your book: period $\pi$, range $\mathbb{R}$, vertical asymptotes at $x = n\pi$, zeros at $x = \frac{\pi}{2}+n\pi$, strictly decreasing on each branch.
For $y = \cot x$ on the interval $[0, \pi]$: state the asymptotes, find the zeros, and describe the behaviour of the function.
State the range of $y = \csc x$ and explain why the range of $y = \cot x$ is different.
So range of $\csc x = (-\infty,-1] \cup [1, +\infty)$.
So range of $\cot x = \mathbb{R}$.
Fill the gap: $y = \cot x$ has period while $y = \csc x$ has period .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $y = \cot x$ and $y = \csc x$ have vertical asymptotes at exactly the same $x$-values.
Activities · practice with the graphs
State all vertical asymptotes of $y = \csc x$ in the interval $[0, 2\pi]$ and their equations.
For $y = \sec x$ on $[-\pi, \pi]$: list the asymptote equations, the local minimum, and the local maximum.
Evaluate $\cot\!\left(\dfrac{3\pi}{4}\right)$ exactly and state which quadrant the angle is in.
Explain in your own words why the range of $y = \csc x$ does not include values between $-1$ and $1$.
Write a complete sketch description for $y = \cot x$ on $(0, \pi)$: asymptotes, zero crossing, and how the function changes.
Odd one out: Three of these statements about reciprocal trig graphs are correct. Which one is WRONG?
Earlier you were asked about vertical asymptotes of $\csc x$ and the value of $\csc x$ at the extremes of $\sin x$.
The answer: vertical asymptotes occur at $x = n\pi$ (wherever $\sin x = 0$). When $\sin x = 1$, $\csc x = \dfrac{1}{1} = 1$ — this is a local minimum. When $\sin x = -1$, $\csc x = \dfrac{1}{-1} = -1$ — this is a local maximum. The function is never defined between $-1$ and $1$.
Why does $\cot x$ have period $\pi$ while $\csc x$ has period $2\pi$? Because $\cot(x + \pi) = \cot x$ identically (both numerator and denominator change sign, cancelling), while $\csc(x + \pi) = -\csc x$ (not the same), so $\csc x$ needs a full $2\pi$ to return to its original value.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the equations of all vertical asymptotes of $y = \csc x$ in $[0, 2\pi]$. (2 marks)
Q2. Sketch $y = \sec x$ for $-\pi \le x \le \pi$, showing asymptotes and key points. (3 marks)
Q3. What is the range of $y = \csc x$? Justify your answer. (1 mark)
Comprehensive answers (click to reveal)
Activity 1: 1. $x=0$ and $x=2\pi$ · 2. Asymptotes at $x=-\frac{\pi}{2}$, $x=\frac{\pi}{2}$; local min $(0,1)$; local max $(-\pi,-1)$ and $(\pi,-1)$ · 3. $\cot(3\pi/4) = \cos(3\pi/4)/\sin(3\pi/4) = (-\frac{\sqrt2}{2})/(\frac{\sqrt2}{2}) = -1$; Q2 · 4. Because $|\sin x| \le 1$, so $|\csc x| = 1/|\sin x| \ge 1$ — the values between $-1$ and $1$ require $|\sin x| > 1$ which is impossible · 5. Asymptotes at $x=0^+$ and $x=\pi^-$; zero at $x=\pi/2$; strictly decreasing from $+\infty$ to $-\infty$
Q1 (2 marks): $\sin x = 0$ at $x = 0, \pi, 2\pi$ [1 each, max 2]. Equations: $x = 0$, $x = \pi$, $x = 2\pi$.
Q2 (3 marks): Asymptotes at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ [1]. Key points: $\sec(-\pi) = -1$, $\sec(0) = 1$, $\sec(\pi) = -1$ [1]. Three branches correctly drawn with correct direction (up/down) [1].
Q3 (1 mark): Range $= (-\infty,-1] \cup [1,\infty)$. Because $-1 \le \sin x \le 1$ (and $\sin x \ne 0$), taking the reciprocal gives $|\csc x| \ge 1$, so values in $(-1,1)$ are excluded [1].
Five timed questions on reciprocal trig graphs. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering reciprocal trig questions. Lighter alternative to the boss.
Next lesson: Pythagorean Identities.
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