The Cosecant Function

Definition For a real number $x$, measured in radians, with $\sin x \ne 0$, the cosecant of $x$ is

Cosecant has the same sign as sine, positive where the unit circle point lies above the horizontal axis and negative where it lies below.

Definition Cosecant is defined exactly when $\sin x \ne 0$, that is, when $x \ne k\pi$ for every integer $k$. Its domain is

These are not the inputs excluded for secant. Secant loses $\frac{\pi}{2} + k\pi$, where cosine is zero, and cosecant loses $k\pi$, where sine is zero.

Definition The range of cosecant is

For every $x$ in the domain, $0 < \lvert \sin x \rvert \le 1$, so $\lvert \csc x \rvert = \frac{1}{\lvert \sin x \rvert} \ge 1$. Conversely, if $\lvert y \rvert \ge 1$ then $\frac{1}{y}$ is a nonzero number in $[-1, 1]$, so some $x$ has $\sin x = \frac{1}{y}$ and then $\csc x = y$. Cosecant has no zeros, because its numerator is $1$.

Definition The line $x = a$ is a vertical asymptote of the graph of $f$ if $f(x)$ grows without bound, toward $\infty$ or $-\infty$, as $x$ approaches $a$ from at least one side. Cosecant has a vertical asymptote at every excluded input $x = k\pi$, because sine approaches $0$ there. Sine changes sign at each multiple of $\pi$, so cosecant goes to $\infty$ on the side where sine is positive and to $-\infty$ on the side where sine is negative. The branches on the two sides are separate and must not be joined.

Definition For every $x$ in the domain, $-x$ is also in the domain and

So cosecant is an odd function, and its graph is symmetric about the origin. Secant, by contrast, is even.

Definition For every $x$ in the domain, $x + 2\pi$ is in the domain and $\csc(x + 2\pi) = \csc x$, because $\sin(x + 2\pi) = \sin x$. The least positive period is $2\pi$. If some $p$ with $0 < p < 2\pi$ were a period, then $\frac{\pi}{2} + p$ would be in the domain because $\frac{\pi}{2}$ is, and $\csc\left(\frac{\pi}{2} + p\right) = \csc \frac{\pi}{2} = 1$, which needs $\sin\left(\frac{\pi}{2} + p\right) = 1$. The only inputs with sine $1$ are $\frac{\pi}{2} + 2\pi n$ for integers $n$, and $\frac{\pi}{2} + p$ lies strictly between $\frac{\pi}{2}$ and $\frac{5\pi}{2}$, so no such $p$ exists.

Definition The graph of $y = \csc x$ is made of separate branches on the open intervals $(k\pi, (k+1)\pi)$ between consecutive asymptotes. On $(0, \pi)$ the sine satisfies $0 < \sin x \le 1$, with $\sin x = 1$ only at $x = \frac{\pi}{2}$, so $\csc x \ge 1$ there, with lowest point $\left(\frac{\pi}{2}, 1\right)$. On $(\pi, 2\pi)$ the sine satisfies $-1 \le \sin x < 0$, so $\csc x \le -1$, with highest point $\left(\frac{3\pi}{2}, -1\right)$. These two branches make one full cycle.

For a positive constant $A$, the graph of $y = A\csc x$ has the same asymptotes, every height is multiplied by $A$, and the range becomes

The band $-A < y < A$ is never reached.