Cosecant Function

Approach:

The cosecant is the reciprocal of the sine, so it is undefined exactly at the zeros of the sine, and its absolute value can never drop below $1$.

Steps:

1. By the unit-circle definition $\csc x = \frac{1}{\sin x}$, so the expression is undefined when $\sin x = 0$, that is when $x = k\pi$.
2. Since $|\sin x| \leq 1$ with equality exactly at $x = \frac{\pi}{2} + k\pi$, the value of $\frac{1}{\sin x}$ always satisfies $\left|\frac{1}{\sin x}\right| \geq 1$.
3. Both bounds are attained, so the range is $(-\infty, -1] \cup [1, +\infty)$.

Answer:

Domain: $x \neq k\pi$; Range: $(-\infty, -1] \cup [1, +\infty)$.

Approach:

Read the period off the addition formula for the sine and check its parity by comparing $\csc(-x)$ with $\csc x$.

Steps:

1. The addition formula gives $\sin(x + 2\pi) = \sin x$, so $\csc(x + 2\pi) = \csc x$ and $2\pi$ is a period.
2. Computing $f(-x) = \csc(-x) = \frac{1}{\sin(-x)} = -\frac{1}{\sin x} = -f(x)$, the function is odd.
3. Hence the period is $2\pi$ and the graph is symmetric about the origin.

Answer:

Period is $2\pi$, it is an odd function.