What is the period of the function y = 3 csc x?

Step-by-step solution

1. 1

   Recall what the period measures. The period is the smallest positive $p$ with $f(x+p) = f(x)$ for every $x$ in the domain — the horizontal length of one full cycle before the graph repeats.

2. 2

   Use the reciprocal definition. $\csc x = \dfrac{1}{\sin x}$. If $\sin(x + p) = \sin x$ for all $x$, then taking reciprocals gives $\csc(x+p) = \csc x$ wherever both are defined. So cosecant repeats exactly when sine repeats, and its period is the same as sine's: $2\pi$.

3. 3

   Confirm nothing smaller works. Could the period be $\pi$? Test $x = \tfrac{\pi}{2}$: $\csc\tfrac{\pi}{2} = 1$, while $\csc\bigl(\tfrac{\pi}{2}+\pi\bigr) = \csc\tfrac{3\pi}{2} = -1$. Since $1 \neq -1$, $\pi$ is not a period. The same test rules out $\tfrac{\pi}{2}$, so $2\pi$ really is the smallest.

4. 4

   See why the coefficient 3 is irrelevant. Multiplying by $3$ stretches the graph vertically, so $3\csc(x+2\pi) = 3\csc x$ holds for exactly the same shifts. Only a coefficient on $x$ — as in $\csc(bx)$, period $2\pi/|b|$ — changes the period. Here $b = 1$, so

   $\text{period} = \frac{2\pi}{|1|} = 2\pi$

5. 5

   State the answer. The period of $y = 3\csc x$ is $2\pi$. As a final sanity check, $\csc$ has vertical asymptotes wherever $\sin x = 0$, i.e. at $x = 0, \pi, 2\pi, \ldots$ — spaced $\pi$ apart — but the pattern of upward and downward branches only repeats every $2\pi$.