Visual walkthrough — Graphs of cosec x, sec x, cot x

This is the visual companion to the parent topic. If any word below feels new, the prerequisite lives in the Graphs of sin x, cos x, tan x and Vertical Asymptotes and Limits notes.

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Step 1 — What "upside-down twin" even means

WHAT. Pick any number $u$ that is not zero. Its reciprocal is $\frac{1}{u}$ — read aloud as "one divided by $u$". We will call it the twin of $u$.

WHY. All three functions $\csc ,\sec ,\cot$ are nothing but reciprocals: $\csc x=\underset{\text{twin of }\sin x}{\underbrace{\frac{1}{\sin x}}},\sec x=\underset{\text{twin of }\cos x}{\underbrace{\frac{1}{\cos x}}},\cot x=\underset{\text{twin of }\tan x}{\underbrace{\frac{1}{\tan x}}}.$ So before touching any wave, we must understand exactly how the twin $y=\frac{1}{u}$ behaves as $u$ slides along the number line. Here every symbol: $u$ is the input (the value of $\sin$, $\cos$ or $\tan$ later); $y$ is the output (the value of $\csc$, $\sec$ or $\cot$).

PICTURE.

That last line is the seed of the forbidden band, planted here and harvested in Step 5.

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Step 2 — Lay the sine wave underneath

WHAT. Draw $\sin x$ lightly across $(-\frac{\pi }{2},2\pi )$. Mark its three special heights: where it equals $0$, where it equals $+1$, where it equals $-1$.

WHY. By Step 1 the twin only cares about three things in the input: is it $0$, is it $\pm 1$, or is it somewhere between? The sine wave is our supply of $u$-values, so we must locate exactly those three heights before flipping.

Why this exact window? $\sin x$ repeats every $2\pi$ — one full up-and-down happens over any stretch of length $2\pi$. So if we understand what the twin does over a single copy, we understand it everywhere: just clone the pattern left and right forever. The interval $(-\frac{\pi }{2},2\pi )$ is chosen only because it comfortably shows more than one full copy, so you can literally see the repetition begin. Everything we derive here therefore extends to all real $x$ by copying with period $2\pi$.

$\sin x=0\text{ at }x=0,\pi ,2\pi ;\sin x=1\text{ at }x=\frac{\pi }{2};\sin x=-1\text{ at }x=\frac{3\pi }{2}.$ Here $x$ is the angle (horizontal axis); $\sin x$ is the height fed in as $u$.

PICTURE.

The red vertical dashes sit at the zeros — memorise their locations, because in the next step they become walls.

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Step 3 — Turn every zero into a wall (asymptotes of $\csc x$)

WHAT. At each $x$ where $\sin x=0$, draw a vertical asymptote — a wall the graph races toward but never touches.

WHY. Feed $u=\sin x=0$ into the twin $\frac{1}{u}$. Division by zero has no answer, and from Step 1 we saw $\frac{1}{u}$ shoots to $\pm \infty$ as $u$ nears $0$. So the graph explodes precisely at $x=n\pi$. Because $\sin$ has zeros at every multiple of $\pi$, these walls march off in both directions forever, spaced by $\pi$. (The idea of a wall from $1/0$ is developed fully in Vertical Asymptotes and Limits.)

Which way does it explode? Read the sign of $\sin x$ on each side: $\underset{u>0}{\underbrace{\sin x>0}}\Rightarrow \csc x=\frac{1}{u}>0(\text{up}),\underset{u<0}{\underbrace{\sin x<0}}\Rightarrow \csc x<0(\text{down}).$

PICTURE.

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Step 4 — Pin the touch points and draw the U-branches

WHAT. Where $\sin x=1$, plant a dot at $y=1$; where $\sin x=-1$, plant a dot at $y=-1$. Connect each dot smoothly up to the neighbouring walls.

WHY. From the four facts: $u=1\Rightarrow y=1$ exactly (twin unchanged), and around it $|\sin x|$ drops below $1$, so the twin grows — meaning $(\frac{\pi }{2},1)$ is the lowest point of an upward U. Symmetrically $(\frac{3\pi }{2},-1)$ is the highest point of a downward U.

$\underset{\text{output}}{\underbrace{\csc \frac{\pi }{2}}}=\frac{1}{\underset{=1}{\underbrace{\sin \frac{\pi }{2}}}}=1(\text{a minimum}),\csc \frac{3\pi }{2}=\frac{1}{-1}=-1(\text{a maximum}).$

PICTURE.

Now the full $\csc x$ curve exists: U's opening upward sitting on $y=1$, U's opening downward hanging from $y=-1$, one between each pair of walls. And since the sine wave underneath repeats every $2\pi$, this whole up-U-then-down-U pattern repeats with period $2\pi$ across the entire real line.

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Step 5 — The forbidden band (a degenerate-value case, not an $x$ case)

WHAT. Shade the horizontal strip $-1<y<1$. The $\csc$ graph never enters it.

WHY. This is a value edge case, not an $x$ edge case. Chase the logic: $\sin x$ always satisfies $|\sin x|\le 1$. Flipping an inequality of positive numbers reverses it: $0<|\sin x|\le 1⟹\left|\frac{1}{\sin x}\right|\ge 1⟹|\csc x|\ge 1.$ Term by term: the left says "sine's size is at most $1$"; taking reciprocals (which flips $\le$ into $\ge$ for positives) says "cosecant's size is at least $1$". So $\csc x\ge 1$ or $\csc x\le -1$ — nothing in between.

PICTURE.

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Step 6 — Shift the recipe to get $\sec x$

WHAT. Repeat Steps 2–5 with $\cos x$ instead of $\sin x$. Walls now land where $\cos x=0$, i.e. $x=\frac{\pi }{2}+n\pi$; touch points at $(0,1)$ and $(\pi ,-1)$.

WHY the shift. Because $\cos x=\sin \left(x+\frac{\pi }{2}\right)$, the cosine wave is the sine wave slid left by $\frac{\pi }{2}$. Reciprocating a shifted wave just shifts the reciprocal the same way, so the entire $\sec$ graph is the $\csc$ graph shifted left by $\frac{\pi }{2}$ — same forbidden band, same U-branches, walls in new spots. Like $\cos$, it also repeats with period $2\pi$.

WHY each branch opens the way it does (the Step-3 analogue for secant). Don't take the shift on faith — run the sign analysis directly on $\cos x$, exactly as we did for $\sin x$:

• Around $x=0$: $\cos x=1$ (its maximum), so $\sec x=\frac{1}{1}=1$, and nearby $|\cos x|<1$ makes the twin grow — an upward U floored at $(0,1)$.
• Crossing $x=\frac{\pi }{2}$: just left $\cos x\to 0^{+}$ so $\sec x\to +\infty$; just right $\cos x\to 0^{-}$ so $\sec x\to -\infty$ — the sign flip of $\cos$ across its zero flips the branch, just like $\sin$ did.
• Around $x=\pi$: $\cos x=-1$ (its minimum), so $\sec x=\frac{1}{-1}=-1$, floored from above — a downward U capped at $(\pi ,-1)$.

So each secant U opens upward where $\cos$ is positive and downward where $\cos$ is negative — the identical rule as $\csc$, just read off the cosine wave.

PICTURE.

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Step 7 — $\cot x$ has no forbidden band (why it's different)

WHAT. $\cot x=\frac{\cos x}{\sin x}$. Walls where $\sin x=0$ ($x=n\pi$); zeros where $\cos x=0$ ($x=\frac{\pi }{2}+n\pi$). Each branch falls from $+\infty$ to $-\infty$.

WHY. Unlike $\csc$ and $\sec$, the numerator here is $\cos x$, which itself sweeps between $-1$ and $1$ — so the output is not a plain reciprocal of a bounded thing; it can take any real value. Read one interval $(0,\pi )$:

• Near $x=0^{+}$: $\cos x\to 1$ (positive), $\sin x\to 0^{+}$ (tiny positive) $\Rightarrow \cot x\to +\infty$.
• At $x=\frac{\pi }{2}$: $\cos x=0\Rightarrow \cot x=0$ (a clean zero-crossing, midway between walls).
• Near $x={\pi }^{-}$: $\cos x\to -1$ (negative), $\sin x\to 0^{+}$ (tiny positive) $\Rightarrow \cot x\to -\infty$.

So it slides downward the whole way — the reciprocal flips $\tan$'s upward trend. Crucially, this falling picture over $(0,\pi )$ is complete on its own: the very next interval $(\pi ,2\pi )$ is an identical copy, so $\cot x$ repeats with period $\pi$ (half the period of $\csc$ and $\sec$).

PICTURE.

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The one-picture summary

All three at once: $\csc$ (coral) and $\sec$ (lavender) hug the shaded forbidden band $(-1,1)$ with U-branches; $\cot$ (mint) marches through zero on every fall, band-free. The walls sit exactly at the zeros of the wave each function came from, and every pattern tiles the whole line — period $2\pi$ for $\csc$ and $\sec$, period $\pi$ for $\cot$. The relevant identities $1+{\cot }^{2}x={\csc }^{2}x$ live in Trigonometric Identities; the symmetry of each curve in Odd and Even Functions.

Recall Feynman: the whole walkthrough in plain words

Start with one number and its upside-down twin: near zero the twin is a giant, at $\pm 1$ the twin high-fives the original. Now lay down the sine wave — it hands us zeros and $\pm 1$'s. Every zero becomes a wall (you can't flip zero), and which side the branch climbs depends on whether sine was a tiny-plus or tiny-minus just there. Every $\pm 1$ becomes a gentle touch point that anchors a U. Because sine never exceeds $1$, its twin never shrinks below $1$ — that carves out the empty middle stripe. The whole pattern repeats every $2\pi$, so you draw one copy and clone it. Slide the copy left by a quarter-turn and you've got secant for free. Cotangent is the odd one out: its top isn't a $1$, it's the whole cosine wave, so it has no stripe — it just tips downhill from sky to floor between every pair of walls, repeating every $\pi$. Nothing memorised; everything watched.

Recall

Why does $\csc x$ have no zeros? ::: Because $\frac{1}{\sin x}$ can never equal $0$ — a fraction with numerator $1$ is never zero. Which side does a $\csc$ branch climb just left of $x=\pi$? ::: Upward, because $\sin$ is a tiny positive there, so its twin is huge positive. Why is the band $(-1,1)$ forbidden for $\csc$? ::: Since $0<|\sin x|\le 1$, reciprocating flips the inequality to $|\csc x|\ge 1$. How is $\sec x$ obtained from $\csc x$? ::: Shift left by $\frac{\pi }{2}$, since $\cos x=\sin (x+\frac{\pi }{2})$. Why does $\cot x$ decrease while $\tan x$ increases? ::: Taking the reciprocal flips the direction of the run. What are the periods of $\csc$, $\sec$, $\cot$? ::: $2\pi$, $2\pi$ and $\pi$.

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Connections

• Parent topic
• Graphs of sin x, cos x, tan x — the waves we flipped.
• Vertical Asymptotes and Limits — the wall from $1/0$.
• Domain and Range — the forbidden band.
• Trigonometric Identities · Odd and Even Functions · Solving Trigonometric Equations