Periodicity of trigonometric functions

The graphs of the trigono­met­ric func­tions repeat them­selves: the wave of $\sin x$ over $[0, 2\pi]$ is copied again over $[2\pi, 4\pi]$, and so on for­ever. A func­tion that repeats like this is called peri­odic, and the length of the short­est repeat­ing piece is its period. This final les­son defines peri­od­ic­ity care­fully and gives the rules for find­ing the period of com­bi­na­tions such as $2\sin(3x + 5) + 8$, $\displaystyle \sin\frac{x}{2} + \cos\frac{x}{3}$, $\sin^2 x$ and $|\sin x| + |\cos x|$.

Peri­od­ic­ity of $\sin x$#

Let $f(x) = \sin x$. Adding a full turn changes noth­ing:

$$f(2\pi + x) = \sin(2\pi + x) = \sin x = f(x)$$

and like­wise $f(4\pi + x) = f(x)$, $f(6\pi + x) = f(x)$, and so on. Among all the pos­i­tive num­bers $T$ with $f(x + T) = f(x)$ for every $x$, the small­est is $2\pi$. So the period of $\sin x$ is $2\pi$.

Def­i­n­i­tion#

A func­tion $f : A \to \mathbb{R}$ is called a peri­odic func­tion with period $T$ if

$$f(x + T) = f(x), \quad \forall x \in A$$

The least pos­i­tive real value of $T$ with this prop­erty is called the period of $f$.

Peri­ods of the six func­tions#

$\tan$ and $\cot$ repeat twice as often, because $\tan(\pi + x) = \tan x$ already (both $\sin$ and $\cos$ change sign, and the signs can­cel).

Rules for find­ing peri­ods#

Rule 1. If $f$ is peri­odic with period $T$, then $-f$ and $\displaystyle \frac{1}{f}$ are also peri­odic with period $T$. For exam­ple, if $g(x) = -\sin x$, then $g(x + 2\pi) = -\sin(x + 2\pi) = -\sin x = g(x)$, so the period of $-\sin x$ is $2\pi$. Like­wise $\displaystyle \sec x = \frac{1}{\cos x}$ has the same period as $\cos x$.

Rule 2. If the period of $f(x)$ is $p$, then the period of $f(ax + b) + k$ is $\displaystyle \frac{p}{|a|}$. Mul­ti­ply­ing the whole func­tion by a num­ber, adding a con­stant $k$, or shift­ing by $b$ does not change the period; only the coef­fi­cient $a$ of $x$ does.

Exam­ple 1#

Find the peri­ods of $2\sin(3x + 5) + 8$ and $\tan(5 - 3x)$.

For $2\sin(3x + 5) + 8$: $a = 3$, so the period is $\displaystyle \frac{2\pi}{3}$.

For $\tan(5 - 3x)$: $a = -3$, so the period is $\displaystyle \frac{\pi}{|-3|} = \frac{\pi}{3}$.

Rule 3. If $f(x)$ and $g(x)$ have peri­ods $p_1$ and $p_2$, then the period of $f(x) \pm g(x)$ (or of $\displaystyle \frac{f(x)}{g(x)}$) is the L.C.M. of $p_1$ and $p_2$, pro­vided there is no pos­i­tive $\alpha$ with $f(x + \alpha) = g(x)$ and $g(x + \alpha) = f(x)$. (That pro­viso is explained in Exam­ple 6.)

To find the L.C.M. of frac­tions, use

$$\displaystyle \text{L.C.M. of } \left\{\frac{a}{b}, \frac{c}{d}\right\} = \frac{\text{L.C.M.}\{a, c\}}{\text{G.C.D.}\{b, d\}}$$

Exam­ple 2#

Find the period of $\displaystyle \sin\frac{x}{2} + \cos\frac{x}{3}$.

The peri­ods are $\displaystyle \frac{2\pi}{1/2} = 4\pi$ and $\displaystyle \frac{2\pi}{1/3} = 6\pi$. Their L.C.M. is $12\pi$ (since the L.C.M. of $4$ and $6$ is $12$). So the period is $12\pi$.

Pow­ers of trigono­met­ric func­tions#

Exam­ple 3#

Find the period of $\sin^2 x$.

Method 1. Lower the power: $\displaystyle \sin^2 x = \frac{1 - \cos 2x}{2} = \frac{1}{2} - \frac{1}{2}\cos 2x$. By Rule 2 the period is $\displaystyle \frac{2\pi}{2} = \pi$.

Method 2. Test directly. Cer­tainly $f(x + 2\pi) = f(x)$. But also

$$f(x + \pi) = \sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x = f(x)$$

and no smaller pos­i­tive num­ber works, so the period is $\pi$.

Exam­ple 4#

Find the period of $\sin^3 x$.

From $\sin 3x = 3\sin x - 4\sin^3 x$, we get $\displaystyle \sin^3 x = \frac{3}{4}\sin x - \frac{1}{4}\sin 3x$. The peri­ods of the two terms are $2\pi$ and $\displaystyle \frac{2\pi}{3}$, so the period is

$$\displaystyle \text{L.C.M.}\left\{\frac{2\pi}{1}, \frac{2\pi}{3}\right\} = \frac{\text{L.C.M.}\{2\pi, 2\pi\}}{\text{G.C.D.}\{1, 3\}} = \frac{2\pi}{1} = 2\pi$$

This time adding $\pi$ does not work: $\sin^3(x + \pi) = -\sin^3 x$.

Exam­ple 5#

Find the period of $\sin^4 x$.

$f(x + \pi) = \sin^4(x + \pi) = (-\sin x)^4 = \sin^4 x = f(x)$, and $\pi$ is the least such pos­i­tive num­ber. So the period is $\pi$.

Sum­mary.

An even power removes the minus sign that $\sin(x + \pi) = -\sin x$ pro­duces; an odd power keeps it.

Mod­uli of trigono­met­ric func­tions#

Note. The period of each of $|\sin x|$, $|\cos x|$, $|\tan x|$, $|\cot x|$, $|\sec x|$, $|\csc x|$ is $\pi$.

Proof for $|\sin x|$. $f(x + \pi) = |\sin(x + \pi)| = |-\sin x| = |\sin x| = f(x)$, and $\pi$ is the least such num­ber. The mod­u­lus removes the sign, just as an even power does.

Note. The period of $a|\sin x| + b|\cos x|$ (and like­wise of $a|\tan x| + b|\cot x|$, and of $a|\csc x| + b|\sec x|$) is $\pi$ if $a \neq b$, and $\displaystyle \frac{\pi}{2}$ if $a = b$.

For exam­ple, let $f(x) = |\sin x| + |\cos x|$. Then

$$\displaystyle f\left(x + \frac{\pi}{2}\right) = \left|\sin\left(x + \frac{\pi}{2}\right)\right| + \left|\cos\left(x + \frac{\pi}{2}\right)\right| = |\cos x| + |-\sin x| = f(x)$$

and $\displaystyle \frac{\pi}{2}$ is the least such num­ber, so the period is $\displaystyle \frac{\pi}{2}$, not the $\pi$ that Rule 3 would sug­gest.

Exam­ple 6#

Find the period of $\sin^4 x + \cos^4 x$.

Each term has period $\pi$, so Rule 3 would sug­gest $\pi$. But here $f(x) = \sin^4 x$ and $g(x) = \cos^4 x$ sat­isfy $\displaystyle f\left(x + \frac{\pi}{2}\right) = g(x)$ and $\displaystyle g\left(x + \frac{\pi}{2}\right) = f(x)$, which is exactly the excep­tion in Rule 3. Sim­plify instead:

$$\displaystyle \sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x = 1 - \frac{1}{2}\sin^2 2x = 1 - \frac{1}{4}(1 - \cos 4x) = \frac{3}{4} + \frac{1}{4}\cos 4x$$

So the period is $\displaystyle \frac{2\pi}{4} = \frac{\pi}{2}$.

Prac­tice#

Find the period of each func­tion.

1. $3\cos(2x - 1) + 4$
2. $\tan(4x + 3)$
3. $\sec(5 - 2x)$
4. $\sin 3x + \cos 2x$
5. $\displaystyle \sin\frac{x}{3} + \tan\frac{x}{2}$
6. $\cos^2 x$
7. $\cos^3 x$
8. $|\cos 2x|$
9. $|\sin x| + |\cos x|$
10. $2|\sin x| + 3|\cos x|$
11. $\tan^3 x$

Answers#