Reciprocal Trig Functions

You already know the reciprocal trig ratios cosecant, secant, and cotangent. Treating them as functions of an angle gives three new graphs, and each one can be sketched straight from the graph of its “partner”: $\sin x$, $\cos x$, or $\tan x$. All angles on this page are in radians.

Key ideas

Sketching a reciprocal from its partner

$$\csc x = \frac{1}{\sin x} \qquad\qquad \sec x = \frac{1}{\cos x} \qquad\qquad \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}$$

The rules are the same as for any reciprocal function $y = \dfrac{1}{f(x)}$:

• Where $f(x) = 0$, the reciprocal has a vertical asymptote.
• Where $f(x) = 1$ or $-1$, the reciprocal has the same value (since $\tfrac{1}{1} = 1$ and $\tfrac{1}{-1} = -1$). The graphs touch there.
• Where $f(x)$ is small (close to $0$), the reciprocal is large, and the other way around.
• The reciprocal has the same sign as $f(x)$.

y = csc x

Because $-1 \le \sin x \le 1$, its reciprocal is always $1$ or more, or $-1$ or less. Where $\sin x$ has a maximum of $1$ (at $\tfrac{\pi}{2}$), $\csc x$ has a local minimum of $1$; where $\sin x$ has a minimum of $-1$, $\csc x$ has a local maximum of $-1$. The graph is a row of U-shapes, opening up and down alternately.

y = sec x

The same idea with $\cos x$. The U-shapes touch the cosine curve at its maximums and minimums, $(0, 1)$, $(\pi, -1)$, $(2\pi, 1)$, and the asymptotes are where $\cos x = 0$. The graph of $y = \sec x$ is the graph of $y = \csc x$ shifted $\tfrac{\pi}{2}$ to the left, just as cosine is sine shifted $\tfrac{\pi}{2}$ to the left.

$y = \csc x$ and $y = \sec x$ (solid) with their partners $y = \sin x$ and $y = \cos x$ (dashed).

y = cot x

$\cot x = \dfrac{\cos x}{\sin x}$ has asymptotes where $\sin x = 0$, at $x = n\pi$, and zeros where $\cos x = 0$, at $x = \tfrac{\pi}{2} + n\pi$. Those zeros are exactly where $\tan x$ has its asymptotes: when $\tan x$ is huge, its reciprocal is close to $0$. Since $\tan x$ increases on each branch, $\cot x$ decreases on each branch.

$y = \cot x$ (solid) and $y = \tan x$ (dashed). The zeros of one are the asymptotes of the other.

Summary of properties

For each, $n$ is any integer ($n \in \mathbb{Z}$).

	$y = \csc x$	$y = \sec x$	$y = \cot x$
Domain	$\{x \in \mathbb{R} \mid x \ne n\pi\}$	$\left\{x \in \mathbb{R} \mid x \ne \tfrac{\pi}{2} + n\pi\right\}$	$\{x \in \mathbb{R} \mid x \ne n\pi\}$
Range	$\{y \in \mathbb{R} \mid y \le -1 \text{ or } y \ge 1\}$	$\{y \in \mathbb{R} \mid y \le -1 \text{ or } y \ge 1\}$	$\{y \in \mathbb{R}\}$
Period	$2\pi$	$2\pi$	$\pi$
Asymptotes	$x = n\pi$	$x = \tfrac{\pi}{2} + n\pi$	$x = n\pi$
Zeros	none	none	$x = \tfrac{\pi}{2} + n\pi$

Each reciprocal has the same period as its partner, because if $f$ repeats, so does $\tfrac{1}{f}$.

Notation: reciprocal is not inverse

The reciprocal of $\sin x$ can be written as

$$\csc x \qquad\text{or}\qquad \frac{1}{\sin x} \qquad\text{or}\qquad (\sin x)^{-1}$$

but not as $\sin^{-1} x$. The notation $\sin^{-1} x$ (like $f^{-1}(x)$ for inverse functions) means the inverse sine: the angle whose sine is $x$. That’s a completely different thing. For example:

$$\sin^{-1}(0.5) = \frac{\pi}{6} \approx 0.524 \qquad\text{but}\qquad \csc(0.5) = \frac{1}{\sin 0.5} \approx 2.086$$

This is an awkward exception: $\sin^2 x$ does mean $(\sin x)^2$, but $\sin^{-1} x$ does not mean $(\sin x)^{-1}$.

Worked examples

Example 1: Sketching y = sec x from y = cos x

Describe how to sketch $y = \sec x$ for $0 \le x \le 2\pi$, starting from the graph of $y = \cos x$.

Solution. Work through the features of $y = \cos x$ one at a time (see the bottom graph above).

1. Zeros of cos x become asymptotes. $\cos x = 0$ at $x = \tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$, so draw dashed vertical asymptotes there.
2. Points where cos x = ±1 stay put. Mark $(0, 1)$, $(\pi, -1)$, and $(2\pi, 1)$. They’re on both graphs.
3. Same sign, opposite size. From $0$ to $\tfrac{\pi}{2}$, $\cos x$ falls from $1$ toward $0$, so $\sec x$ rises from $1$ toward the asymptote. From $\tfrac{\pi}{2}$ to $\tfrac{3\pi}{2}$, $\cos x$ is negative, so $\sec x$ is a downward U with its highest point at $(\pi, -1)$. From $\tfrac{3\pi}{2}$ to $2\pi$, $\sec x$ comes down from the asymptote to $(2\pi, 1)$.

Check a point: $\sec\tfrac{\pi}{3} = \dfrac{1}{\cos(\pi/3)} = \dfrac{1}{1/2} = 2$, on the first rising piece. ✓

Example 2: Properties of y = csc x on an interval

For $-2\pi \le x \le 2\pi$, give the equations of the vertical asymptotes of $y = \csc x$, and all its local maximum and minimum points.

Solution. Asymptotes are where $\sin x = 0$:

$$x = -2\pi,\quad x = -\pi,\quad x = 0,\quad x = \pi,\quad x = 2\pi$$

Local minimums (value $1$) are where $\sin x = 1$: $\left(-\tfrac{3\pi}{2}, 1\right)$ and $\left(\tfrac{\pi}{2}, 1\right)$.

Local maximums (value $-1$) are where $\sin x = -1$: $\left(-\tfrac{\pi}{2}, -1\right)$ and $\left(\tfrac{3\pi}{2}, -1\right)$.

Example 3: Reciprocal or inverse?

Evaluate, if possible, to $4$ decimal places: (a) $\csc 1.2$ (b) $\sin^{-1} 1.2$ (c) $\cot 2$

Solution. In radian mode:

(a) $\csc 1.2 = \dfrac{1}{\sin 1.2} \approx \dfrac{1}{0.9320} \approx 1.0729$

(b) $\sin^{-1} 1.2$ asks for an angle whose sine is $1.2$. There isn’t one, because sine is never bigger than $1$, so the calculator gives an error. This is a good reminder that $\sin^{-1}$ is not cosecant.

(c) $\cot 2 = \dfrac{1}{\tan 2} \approx \dfrac{1}{-2.1850} \approx -0.4577$

Check (c) with CAST: $2$ is between $\tfrac{\pi}{2}$ and $\pi$, in quadrant II, where tangent and cotangent are negative. ✓

Example 4: Explaining the asymptotes and zeros of cot x

For $0 \le x \le 2\pi$, where is $y = \cot x$ undefined, and where is it zero? Explain each using $\cot x = \dfrac{\cos x}{\sin x}$.

Solution.

• Undefined where the denominator $\sin x = 0$: $x = 0$, $\pi$, and $2\pi$. Near these values, $\sin x$ is tiny while $\cos x$ is close to $\pm 1$, so $\cot x$ is huge: these are vertical asymptotes.
• Zero where the numerator $\cos x = 0$ (and $\sin x \ne 0$): $x = \tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$.

Check: $\cot\tfrac{\pi}{2} = \dfrac{\cos(\pi/2)}{\sin(\pi/2)} = \dfrac{0}{1} = 0$. ✓ (You can’t get this from $\dfrac{1}{\tan(\pi/2)}$, since $\tan\tfrac{\pi}{2}$ is undefined, which is why the form $\dfrac{\cos x}{\sin x}$ is useful.)

Common mistakes

Writing sin⁻¹ x for csc x. $\sin^{-1} x$ is the inverse sine (an angle). The reciprocal is $\csc x$, $\dfrac{1}{\sin x}$, or $(\sin x)^{-1}$.

Putting the asymptotes in the wrong places. The reciprocal’s asymptotes are at the zeros of its partner. For $\sec x$, that’s where $\cos x = 0$, at $\tfrac{\pi}{2} + n\pi$, not where $\sin x = 0$.

Drawing csc x or sec x between −1 and 1. Their range is $y \le -1$ or $y \ge 1$. If a branch dips below $1$ (or rises above $-1$), it’s wrong: the U-shapes only touch $y = \pm 1$ at their turning points.

Mixing up maximums and minimums. A maximum of $\sin x$ becomes a local minimum of $\csc x$, because the U opens away from the axis.

Giving cot x a period of 2π. Like tangent, cotangent repeats every $\pi$.

Practice

1. (Warm-up) Find the exact value: (a) $\csc\dfrac{\pi}{6}$ (b) $\sec\dfrac{\pi}{3}$ (c) $\cot\dfrac{\pi}{4}$

Solution

(a) $\dfrac{1}{\sin(\pi/6)} = \dfrac{1}{1/2} = 2$

(b) $\dfrac{1}{\cos(\pi/3)} = \dfrac{1}{1/2} = 2$

(c) $\dfrac{1}{\tan(\pi/4)} = \dfrac{1}{1} = 1$

2. (Warm-up) Evaluate to $3$ decimal places: (a) $\sec 2.5$ (b) $\csc 0.3$ (c) $\cot 4$

Solution

(a) $\sec 2.5 = \dfrac{1}{\cos 2.5} \approx -1.248$

(b) $\csc 0.3 = \dfrac{1}{\sin 0.3} \approx 3.384$

(c) $\cot 4 = \dfrac{1}{\tan 4} \approx 0.864$

3. (Warm-up) Which of these mean the reciprocal of $\sin x$? (a) $\sin^{-1} x$ (b) $\csc x$ (c) $\dfrac{1}{\sin x}$ (d) $(\sin x)^{-1}$

Solution

(b), (c), and (d). Choice (a), $\sin^{-1} x$, is the inverse sine function.

4. (Core) State the domain, range, and period of $y = \sec x$, and the equations of its asymptotes for $-\pi \le x \le \pi$.

Solution

Domain $\left\{x \in \mathbb{R} \mid x \ne \tfrac{\pi}{2} + n\pi,\ n \in \mathbb{Z}\right\}$. Range $\{y \in \mathbb{R} \mid y \le -1 \text{ or } y \ge 1\}$. Period $2\pi$.

Asymptotes for $-\pi \le x \le \pi$: $x = -\dfrac{\pi}{2}$ and $x = \dfrac{\pi}{2}$.

5. (Core) For $0 \le x \le 2\pi$: (a) where does $\csc x = 1$? (b) where does $\csc x = -1$? (c) where does $\sec x = -1$?

Solution

(a) $\csc x = 1$ when $\sin x = 1$: $x = \dfrac{\pi}{2}$.

(b) $\csc x = -1$ when $\sin x = -1$: $x = \dfrac{3\pi}{2}$.

(c) $\sec x = -1$ when $\cos x = -1$: $x = \pi$.

6. (Core) $y = \sin x$ has a maximum at $\left(\tfrac{\pi}{2}, 1\right)$, but $y = \csc x$ has a local minimum there. Explain why.

Solution

Near $x = \tfrac{\pi}{2}$, $\sin x$ is at most $1$, and slightly less than $1$ on either side. Dividing $1$ by a number slightly less than $1$ gives a number slightly more than $1$. So $\csc x$ equals $1$ at $\tfrac{\pi}{2}$ and is bigger than $1$ on either side, which makes $\left(\tfrac{\pi}{2}, 1\right)$ a local minimum. The biggest value of $\sin x$ gives the smallest value of its reciprocal.

7. (Core) For $-2\pi \le x \le 2\pi$, give the equations of the asymptotes and the zeros of $y = \cot x$.

Solution

Asymptotes where $\sin x = 0$: $x = -2\pi$, $x = -\pi$, $x = 0$, $x = \pi$, $x = 2\pi$.

Zeros where $\cos x = 0$: $x = -\dfrac{3\pi}{2}, -\dfrac{\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}$.

8. (Challenge) Find all $x$ with $0 \le x \le 2\pi$ such that (a) $\sec x = 2$ (b) $\csc x = -\sqrt{2}$.

Solution

(a) $\sec x = 2$ means $\cos x = \tfrac{1}{2}$. The related angle is $\tfrac{\pi}{3}$, and cosine is positive in quadrants I and IV:

$$x = \frac{\pi}{3} \qquad\text{or}\qquad x = \frac{5\pi}{3}$$

(b) $\csc x = -\sqrt{2}$ means $\sin x = -\tfrac{1}{\sqrt{2}} = -\tfrac{\sqrt{2}}{2}$. The related angle is $\tfrac{\pi}{4}$, and sine is negative in quadrants III and IV:

$$x = \frac{5\pi}{4} \qquad\text{or}\qquad x = \frac{7\pi}{4}$$

9. (Challenge) State the period and range of $y = \csc(2x)$, and the equations of its asymptotes for $0 \le x \le 2\pi$.

Solution

$y = \sin(2x)$ has period $\tfrac{2\pi}{2} = \pi$, so its reciprocal $y = \csc(2x)$ also has period $\pi$. A horizontal compression doesn’t change the $y$-values, so the range is still $\{y \in \mathbb{R} \mid y \le -1 \text{ or } y \ge 1\}$.

Asymptotes where $\sin(2x) = 0$: $2x = n\pi$, so $x = \tfrac{n\pi}{2}$:

$$x = 0,\quad x = \frac{\pi}{2},\quad x = \pi,\quad x = \frac{3\pi}{2},\quad x = 2\pi$$