Why is the period of $\sec$ and $\csc 2\pi$ instead of $\pi$?

Question

Hi and thanks in advance.

When $\tan x = \frac{(2x + 1)\pi}{2}$, the function is undefined, resulting in a period of simply $\pi$. However, $\csc$ and $\sec$ also include undefined points. Taking this discontinuity into account, why are their periods $2\pi$ instead of $\pi$?

Answer 1

The period of $\csc x$ is $2\pi$ because the period of $\sin x$ is $2\pi$, and $\csc x = \frac{1}{\sin x}$. It doesn't have to do with the discontinuity. The same thing holds for $\sec x$.

The period of $\tan x$ is the outlier here. Note that $\sin (x + \pi) = -\sin x$, and similarly, $\cos (x + \pi) = -\cos x $. So, after a rotation of $\pi$, neither $\sin x$ nor $\cos x$ has returned to its starting point. But, here's what happens to tangent:

$$ \tan (x+\pi) = \frac{\sin (x + \pi)}{\cos (x + \pi)} = \frac{-\sin x}{-\cos x} = \frac{\sin x}{\cos x} = \tan x . $$

The same thing happens with $\cot x$. So, while most trig functions have a period of $2\pi$, $\tan x$ and $\cot x$ have a period of $\pi$.

Answer 2

The basic trigonometric functions are $\sin$ and $\cos$. $\cos\theta$ and $\sin\theta$ are, respectively, the $x$- and $y$-coordinates that you arrive at after tracing an arc of $\theta$ units counterclockwise around the unit circle, starting from the point $(1,0)$.

From this, the other four functions can be defined in the following way: \begin{align} \sec\theta &= \frac{1}{\cos\theta} & \csc\theta &= \frac{1}{\sin\theta} \\[6pt] \tan\theta &= \frac{\sin\theta}{\cos\theta} & \cot\theta &= \frac{\cos\theta}{\sin\theta} \end{align} Because $\tan\theta$ is defined as the ratio of $\sin\theta$ to $\cos\theta$, it has a nice geometric interpretation as the slope of the line segment joining the origin with the point $(x,y)=(\cos\theta,\sin\theta)$. From this it follows that $$ \tan(\theta + \pi) = \tan\theta \, , $$ since a rotation of $\pi$ units simply maps the point $(x,y)$ to $(-x,-y)$. By contrast, $\sec$ and $\csc$ are simply the reciprocal functions of $\cos$ and $\sin$, meaning that their periodicity must be $2\pi$. If you want more of a justification, then consider that $\cos(\theta + \pi) = -\cos\theta$, and so $$\sec(\theta + \pi) = \frac{1}{\cos(\theta + \pi)} = \frac{1}{-\cos\theta} = -\sec\theta \, .$$ A similar argument can be made for $\csc$. Hence, as Amaan has already mentioned, $\tan$ is the outlier.