Exercises — Graphs of cosec x, sec x, cot x

WHAT we do: find where each bottom is zero, because $\frac{1}{0}$ makes the wall.

• $\csc x=\frac{1}{\sin x}$ — bottom is $\sin x$. It is $0$ at $x=n\pi$. So asymptotes at $x=n\pi$.
• $\sec x=\frac{1}{\cos x}$ — bottom is $\cos x$. It is $0$ at $x=\frac{\pi }{2}+n\pi$. So asymptotes at $x=\frac{\pi }{2}+n\pi$.
• $\cot x=\frac{\cos x}{\sin x}$ — bottom is $\sin x$ again. Asymptotes at $x=n\pi$.

Answer: $\csc$: $x=n\pi$; $\sec$: $x=\frac{\pi }{2}+n\pi$; $\cot$: $x=n\pi$.

WHAT: flip the cosine. WHY: $\sec$ is $1/\cos$. $\cos \frac{\pi }{3}=\frac{1}{2}\Rightarrow \sec \frac{\pi }{3}=\frac{1}{1/2}=2.$ Answer: $\sec \left(\frac{\pi }{3}\right)=2$. Notice $2\ge 1$ — it obeys the forbidden band (nothing lands strictly between $-1$ and $1$).

WHAT/WHY: a fraction is zero only when its top is zero.

• $\csc x=\frac{1}{\sin x}$: top is $1$, never $0$ → no zeros.
• $\sec x=\frac{1}{\cos x}$: top is $1$, never $0$ → no zeros.
• $\cot x=\frac{\cos x}{\sin x}$: top is $\cos x$, which is $0$ at $x=\frac{\pi }{2}+n\pi$ → has zeros.

Answer: $\csc$ and $\sec$ have no zeros; $\cot$ has zeros at $x=\frac{\pi }{2}+n\pi$.

Step 1 — flip both sides. WHY: turn the unfamiliar $\csc$ into the familiar $\sin$. $\csc x=2\Rightarrow \frac{1}{\sin x}=2\Rightarrow \sin x=\frac{1}{2}.$ Step 2 — solve the sine. $\sin x=\frac{1}{2}$ at $x=\frac{\pi }{6}$ (Quadrant I) and $x=\frac{5\pi }{6}$ (Quadrant II — sine is still positive there). See Solving Trigonometric Equations. Answer: $x=\frac{\pi }{6},\frac{5\pi }{6}$. Geometrically these are where the $\csc$ curve crosses the line $y=2$ on its two upper branches.

Step 1 — flip. $\sec x=-\sqrt{2}\Rightarrow \cos x=\frac{1}{-\sqrt{2}}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}.$ Step 2 — negative cosine lives in Quadrants II and III. The reference angle with $\cos =\frac{\sqrt{2}}{2}$ is $\frac{\pi }{4}$.

• Quadrant II: $x=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$.
• Quadrant III: $x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$.

Answer: $x=\frac{3\pi }{4},\frac{5\pi }{4}$.

Step 1 — turn into tan. $\cot x=\frac{1}{\tan x}=1\Rightarrow \tan x=1$. WHY: $\tan$ is the graph we know best. Step 2 — $\tan x=1$ where sine and cosine are equal and same-signed: Quadrant I at $x=\frac{\pi }{4}$, Quadrant III at $x=\frac{\pi }{4}+\pi =\frac{5\pi }{4}$ (tangent has period $\pi$). Answer: $x=\frac{\pi }{4},\frac{5\pi }{4}$.

WHAT: watch the sign of $\sin x$ as it slides through $0$ at $x=\pi$.

• Just below $\pi$ (say $x=3.0$), $\sin x$ is a small positive number. So $\csc x=\frac{1}{\text{small }+}=+\infty$.
• Just above $\pi$ (say $x=3.3$), $\sin x$ is a small negative number. So $\csc x=\frac{1}{\text{small }-}=-\infty$.

Answer: $x\to {\pi }^{-}$: $\csc x\to +\infty$; $x\to {\pi }^{+}$: $\csc x\to -\infty$. The branch on the left of the wall shoots up; the branch on the right dives down. Look at the master figure across $x=\pi$.

Step 1 — bound the bottom. For every $x$, $-1\le \cos x\le 1$, so $|\cos x|\le 1$. Step 2 — flip the inequality on magnitudes. For nonzero $u$, if $|u|\le 1$ then $\left|\frac{1}{u}\right|\ge 1$. (Dividing $1$ by something no bigger than $1$ gives something no smaller than $1$.) Step 3 — apply. Hence $|\sec x|=\frac{1}{|\cos x|}\ge 1$, i.e. $\sec x\ge 1$ or $\sec x\le -1$. Answer: No value lands in the open band $(-1,1)$. The band is forbidden. Same argument gives the range of $\csc$; see Domain and Range.

Endpoints: as $x\to 0^{+}$, $\sin x\to 0^{+}$ and $\cos x\to 1$, so $\cot x=\frac{+1}{\text{small }+}\to +\infty$. As $x\to {\pi }^{-}$, $\cos x\to -1$ and $\sin x\to 0^{+}$, so $\cot x=\frac{-1}{\text{small }+}\to -\infty$. Midpoint: at $x=\frac{\pi }{2}$, $\cos x=0$ so $\cot x=0$. WHAT this tells us: the value marches $+\infty \to 0\to -\infty$ as $x$ goes $0\to \frac{\pi }{2}\to \pi$ — steadily downhill. Answer: $\cot x$ is decreasing on $(0,\pi )$ (and on every $(n\pi ,(n+1)\pi )$). This is opposite to $\tan x$, which climbs.

Step 1 — choose the right identity. WHY this one: $1+{\cot }^{2}x={\csc }^{2}x$ directly links the quantity we have to the one we want. See Trigonometric Identities. ${\csc }^{2}x=1+{\cot }^{2}x=1+{\left(\frac{3}{4}\right)}^2=1+\frac{9}{16}=\frac{25}{16}.$ Step 2 — take the root, choose the sign. In Quadrant I, $\sin x>0$ so $\csc x>0$. $\csc x=+\sqrt{\frac{25}{16}}=\frac{5}{4}.$ Answer: $\csc x=\frac{5}{4}$.

WHAT: replace $x$ by $-x$ inside each definition and use $\cos (-x)=\cos x$, $\sin (-x)=-\sin x$.

• $\sec (-x)=\frac{1}{\cos (-x)}=\frac{1}{\cos x}=\sec x$ (even).
• $\csc (-x)=\frac{1}{\sin (-x)}=\frac{1}{-\sin x}=-\csc x$ (odd).

Answer: $\sec (-x)+\csc (-x)=\sec x-\csc x$.

Step 1. $\csc \left(x+\frac{\pi }{2}\right)=\frac{1}{\sin \left(x+\frac{\pi }{2}\right)}$. Step 2 — use the co-function shift. $\sin \left(x+\frac{\pi }{2}\right)=\cos x$. $\csc \left(x+\frac{\pi }{2}\right)=\frac{1}{\cos x}=\sec x.✓$ Meaning: the $\sec$ graph is the $\csc$ graph shifted left by $\frac{\pi }{2}$ — same shape, walls just relocated from $x=n\pi$ to $x=\frac{\pi }{2}+n\pi$.

Step 1 — flip. $\csc x=\frac{1}{2}\Rightarrow \sin x=2$. Step 2 — check feasibility. But $-1\le \sin x\le 1$, and $2>1$. Impossible. Why this had to fail: $\frac{1}{2}$ lies inside the forbidden band $(-1,1)$, and $\csc$ never enters that band (proved in L3.2, mirror argument). Answer: No solutions. A value strictly between $-1$ and $1$ can never be a $\csc$ (or $\sec$) output.

Step 1 — flip. $\sec x=-1\Rightarrow \cos x=-1$. Step 2 — solve. $\cos x=-1$ only at $x=\pi$ in $[0,2\pi )$. Step 3 — interpret geometrically. $y=-1$ is the top of the lower branches; the $\sec$ curve touches it (a tangent kiss, a maximum of the lower branch) rather than crossing. Answer: exactly one contact, at $x=\pi$. This is a touch point $(\pi ,-1)$, not a crossing.

Walls (asymptotes): where $\sin x=0$ inside $(0,2\pi )$ → only $x=\pi$ (the ends $0,2\pi$ are also walls at the border). Zeros: where $\cos x=0$ → $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$. Trend: decreasing on $(0,\pi )$ from $+\infty$ to $-\infty$, then again decreasing on $(\pi ,2\pi )$ from $+\infty$ to $-\infty$. Period: $\pi$, so the two branches are identical copies. Point check: $\cot \left(\frac{7\pi }{6}\right)$. Here $\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2}$, $\sin \frac{7\pi }{6}=-\frac{1}{2}$, so $\cot \frac{7\pi }{6}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}.$ Fits? $\frac{7\pi }{6}$ sits just past the wall at $\pi$, on the upper part of the second branch, where $\cot$ is large and positive — and $\sqrt{3}\approx 1.73>0$. ✓ Answer: $\cot \left(\frac{7\pi }{6}\right)=\sqrt{3}$.

At $x=0$: $\cot 0=\frac{\cos 0}{\sin 0}=\frac{1}{0}$ — dividing by zero, undefined; that is the wall. Just off $0$: for any $x$ with $0<x<\pi$, $\sin x>0$ (strictly), so the denominator is a genuine nonzero number and $\frac{\cos x}{\sin x}$ is a finite real value. As $x\to 0^{+}$ that finite value grows without bound ($\to +\infty$), confirming the wall is a limit, never an attained point. Answer: undefined only where the denominator is exactly $0$ ($x=0,\pi ,\dots$); finite everywhere strictly inside.

Every problem here was the same move wearing different clothes: flip the bottom, watch the sign, respect the range. Flip to turn a scary $\csc /\sec /\cot$ into a friendly $\sin /\cos /\tan$. Watch the sign to know which quadrants (and which way to $\pm \infty$ near a wall). Respect the range so you never chase a solution inside the forbidden band. Master those three habits and no exam version of this topic can surprise you.