Reciprocal Curves

You do not memorise the shape of cosec x. You construct it from sin x. Same for sec x from cos x, and cot x from tan x.

Throughout this page, n denotes any integer — that is, $n \in \mathbb{Z}$, meaning $n$ can be $\ldots, -2, -1, 0, 1, 2, \ldots$

The construction rule

At every point where $\sin x = 0$, the value $\frac{1}{\sin x}$ is undefined. That gives a vertical asymptote.

At every point where $\sin x = 1$ (a peak), $\cosec x = \frac{1}{1} = 1$ — the curves touch.

At every point where $\sin x = -1$ (a trough), $\cosec x = -1$ — the curves touch.

In between, as $\sin x$ rises from $0$ toward $1$, $\cosec x$ falls from $+\infty$ toward $1$. As $\sin x$ falls from $1$ back to $0$, $\cosec x$ rises from $1$ back to $+\infty$.

The result: $\cosec x$ forms U-shaped branches opening upward between $x = 0$ and $x = \pi$ (where $\sin x$ is positive), and inverted U-shaped branches between $x = \pi$ and $x = 2\pi$ (where $\sin x$ is negative).

Drawing procedure

1. Draw $\sin x$ lightly in pencil
2. Mark every zero of $\sin x$ — draw vertical dashed asymptote lines there
3. At each peak of $\sin x$, mark the corresponding touch-point for $\cosec x$
4. At each trough, mark the touch-point below the x-axis
5. Draw the U-curves between each pair of asymptotes, touching at the marked points

The four reciprocal pairs

The zeros of $\sin x$ occur at $x = n\pi$ (for any integer $n$). The zeros of $\cos x$ occur at $x = \frac{\pi}{2} + n\pi$.

Function	Asymptotes at	Touches original at
$\cosec x = \frac{1}{\sin x}$	$x = n\pi$	peaks and troughs of $\sin x$
$\sec x = \frac{1}{\cos x}$	$x = \frac{\pi}{2} + n\pi$	peaks and troughs of $\cos x$

Note: $\cot x = \frac{\cos x}{\sin x}$ is not drawn the same way as a simple reciprocal — its asymptotes are at $x = n\pi$ (where $\tan x = 0$), not at $x = \frac{\pi}{2} + n\pi$.

Domain and range

$\cosec x : \text{domain} = \mathbb{R} \setminus \{n\pi\}, \quad \text{range} = (-\infty, -1] \cup [1, +\infty)$

$\sec x : \text{domain} = \mathbb{R} \setminus \left\{\frac{\pi}{2} + n\pi\right\}, \quad \text{range} = (-\infty, -1] \cup [1, +\infty)$

The range says: $\cosec x$ and $\sec x$ are never between $-1$ and $1$. This is the algebraic echo of the geometric fact that $\frac{1}{\sin x}$ blows up near the zeros of $\sin x$.

Exam application

Given the graph of $y = 2\sin\!\left(\frac{x}{2}\right)$, sketch $y = 2\cosec\!\left(\frac{x}{2}\right)$.

Step 1. The zeros of $2\sin\!\left(\frac{x}{2}\right)$ occur when $\frac{x}{2} = n\pi$, i.e., at $x = 2n\pi$. Draw vertical asymptotes there.

Step 2. The peaks of $2\sin\!\left(\frac{x}{2}\right)$ are at $y = 2$. The $\cosec$ curve touches these peaks.

Step 3. The troughs at $y = -2$ are touch-points for the lower branches.

The amplitude of the original scales the touch-points of the reciprocal. The period of the reciprocal equals the period of the original.