33. Periodicity of Trigonometrical Functions

Definition of Periodic Functions: A function $f(x)$ is said to be a periodic function if there exists positive number $T$ independent of $x$ such that $f(x + T) = f(x),$ for every $x\in$ domain $f$.

The least positive value of $T$ for which $f(x + T) = f(x)$, for every $x\in$ domain $f$ is called the period of fundadmental period of $f(x)$.

Example 1: Examine whether $\sin x$ is a periodic function or not. If yes, then find the period.

Sol. Given, $f(x) = \sin x$. Let $f(x + T) = f(x) \Rightarrow \sin(x + T) = \sin x$

$\Rightarrow x + T = n\pi + (-1)^nx$, where $n = 0, \pm1, \pm2, \ldots$

The positive values of $T$ independent of $x$ are given by $T = n\pi$, where $n = 2, 4, 6, \ldots$

The least positive value of $T = 2\pi$.

Thus, $\sin x$ is a periodic function with period $2\pi$.

Some Results

1. $\sin x, \cos x, sec x$ and $\cosec x$ are periodic functions with period $2\pi$.

2. $\tan x$ and $\cot x$ are periodic function with period $\pi$.

3. $|\sin x|, |\cos x|, |\tan x|, |\cot x|, |\sec x|$ and $|\cosec x|$ are periodic functions with period $\pi$.

4. $\sin^nx, \cos^nx, \sec^nx$ and $\cosec^nx$ are periodic functions with period $2\pi$ and $\pi$ according as $n$ is odd or even.

5. $\tan^nx$ and $\cot^nx$ are periodic functions with period $\pi$ irrespective of $n$ being odd or even.

6. If a circular funciton $f(x)$ is periodic function with period $T$, then $kf(ax + b)$ is also a periodic function with period $\frac{T}{|a|}$.

7. If circular functions $f(x)$ and $g(x)$ are periodic functions with period $T_1$ and $T_2$ then $af(x) + bg(x)$ is a periodic function with period $T$, where $T =$ L.C.M. of $T_1$ and $T_2$.

33.1. Problems

1. Which of the following functions are periodic? Also, find the period if the function is periodic.

   1. $f(x) = 10\sin3x$ ii. $f(x) = a\sin\lambda x + b\cos\lambda x$ iii. $f(x) = \sin^3x$ iv. $f(x) = \cos x^2$ v. $f(x) = \sin\sqrt{x}$ vi. $f(x) = \sqrt{\tan x}$ vii. $f(x) = x - |x|$ where $|x|$ is integral part of $x$ viii. $f(x) = x\cos x$

2. Which of the following functions are periodic? Also, find the period if the function is periodic and has fundamental period.

   1. $f(x) = 4\sin\left(3x + \frac{\pi}{4}\right)$ ii. $f(x) = 3\cos\frac{x}{2} + 4\sin\frac{x}{2}$ iii. $f(x) = \cot\frac{x}{2}$ iv. $f(x) = \sin^2x$ v. $f(x) = \sin x^2$ vi. $f(x) = \sin\frac{1}{x}$ vii. $f(x) = 1 + \tan x$ viii. $f(x) = [x]$ ix. $f(x) = 5$ x. $f(x) = |\cos x|$ xi. $f(x) = \sin^4x + \cos^4x$ xii. $f(x) = x + \sin x$ xiii. $f(x) = \cos\sqrt{x}$ xiv. $f(x) = \tan^{-1}(\tan x)$ xv. $f(x) = |\sin x| + |\cos x|$ xvi. $f(x) = \sin\frac{\pi x}{3} + \sin\frac{\pi x}{4}$ xvii. $f(x) = \sin\left(2\pi x + \frac{\pi}{3}\right) + 2\sin\left(3\pi x + \frac{\pi}{4}\right) + 3\sin\pi x$ xviii. $f(x) = \sin x + \cos\sqrt{x}$

3. Show that the function $f(x) = 2\sin x + 3\cos 2x$ is a periodic function of period $2\pi$.

4. For each of the following functions, mention whether the function is periodic and if yes, mention the period:

   1. $f(x) = 2x - [2x]$, where $[ ]$ denotes the integral part, and b. $g(x) = 1 + \frac{3}{2 - \sin^2x}$

5. Find the period of the function $1 - \frac{1}{4}\sin^2\left(\frac{\pi}{3} - \frac{3x}{2}\right)$.

33.2. Solutions

1. The solutions are given below:

   1. $f(x) = 10\sin3x$. Let $f(x + T) = f(x) \Rightarrow 10\sin3(x + T) = 10\sin3x$

      $\Rightarrow \sin3(x + T) = \sin3x$

      $\Rightarrow 3x + 3T = n\pi + (-1)^n3x$, where $n = 0, \pm1, \pm2, \pm3, \ldots$

      The positive value of $T$ independent of $x$ are given by $3T = n\pi$, where $n = 2, 4, 6, \ldots$

      Least positive value of $T = \frac{2\pi}{3}$.

      Hence, $f(x)$ is a periodic function with a period of $\frac{2\pi}{3}$.

   2. $f(x) = a\sin\lambda x + b\cos\lambda x = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\sin\lambda x + \frac{b}{\sqrt{a^2 + b^2}}\cos\lambda x\right)$

      $= \sqrt{a^2+b^2}(\cos\alpha\sin\lambda x + \sin\alpha\cos\lambda x)$, where $\cos\alpha = \frac{a}{\sqrt{a^2 + b^2}}$

      $= \sqrt{a^2 + b^2}\sin(\lambda x + \alpha)$

      which is a periodic function with a period of $\frac{2\pi}{|\lambda|}$.

   3. $f(x) = \sin^3x = \frac{3\sin x - \sin3x}{4} = \frac{3}{4}\sin x - \frac{1}{4}\sin3x$

      $\sin x$ is a periodic function with period $2\pi$ and $\sin3x$ is a periodic function with period $2\pi$ so the period of the required function will be L.C.M. of these two periods which will be $2\pi$.

   4. $f(x) = \cos x^2$. Let $f(x + T) = f(x) \Rightarrow \cos(x + T)^2 = \cos x^2$

      $\Rightarrow (x + T)^2 = 2n\pi \pm x^2$

      In the above expression $x$ cannot be elimminated until $T = 0$ so the given function is non-periodic.

   5. $f(x) = \sin\sqrt{x}$. Let $f(x + T) = f(x) \Rightarrow \sin\sqrt{x + T} = \sin\sqrt{x}$

      $\Rightarrow \sqrt{x + T} = n\pi + (-1)^n\sqrt{x}$

      which will give no positive value of $T$ independent of $x$ because $\sqrt{x}$ can be cancelled out only if $T = 0$. Hence, $f(x)$ is a non-periodic function.

   6. $f(x) = \sqrt{\tan x}$. Let $f(x + T) = f(x) \Rightarrow \sqrt{\tan(x + T)} = \sqrt{\tan x} \Rightarrow \tan(x + T) = \tan x$

      $\Rightarrow x + T = n\pi + x, n = 0, \pm1, \pm2, \ldots$

      From this positive values of $T$ independent of $x$ are given by $T = n\pi, n = 1, 2, 3, \ldots$

      $\therefore$ Least positive value of $T$ independent of $x$ is $\pi$. Hence, $f(x)$ is a periodic function of period $\pi$.

   7. $f(x) = x - [x]$, where $[x]$ denotes the integral part of $x$. Let $f(x + T) = f(x) \Rightarrow (x + T) - [x + T] = x - [x] \Rightarrow T = [x + T] - [x] =$ an integer

      Hence least positive value of $T$ independent of $x$ is $1$. Hence, $f(x)$ is a periodic function having a period of $1$.

   8. $f(x) = x\cos x$. Let $f(x + T) = f(x) \Rightarrow (x + T)\cos(x + T) = x\cos x$

      $\Rightarrow T\cos(x + T) = x[\cos x - \cos(x + T)]$

      From this no value of $T$ independent of $x$ can be found because on R.H.S. one factor is $x$ which is an algebraid function and on L.H.S. there is no algebraic function and hance $x$ cannot be eliminated.

      Hence $f(x)$ is a non-periodic function.

2. The solutions are given below:

   1. $f(x) = 4\sin\left(3x + \frac{\pi}{4}\right)$. From the sixth result of section Some Results we know that this is a periodic function with period $\frac{2\pi}{3}$ because $\sin3x$ is a periodic function with period $2\pi$.

   2. $f(x) = 3\cos\frac{x}{2} + 4\sin\frac{x}{2}$. We know that both $\sin x$ and $\cos x$ are periodic functions with period $2\pi$. Therefore $\sin\frac{x}{2}$ and $\cos\frac{x}{2}$ will have a period of $4\pi$. Now the function $f(x)$ will have period equal to L.C.M. of periods of these two funcitons which is equal to $4\pi$.

   3. $f(x) = \cot\frac{x}{2}$. We know that $\cot x$ has a period of $\pi$ therefore $f(x)$ will have period equal to $2\pi$.

   4. $f(x) = \sin^2x = \frac{1 - \cos 2x}{2}$. We know that $\cos x$ is a periodic function with a period of $2\pi$ therefore $f(x)$ will be a periodic function with period of $\pi$.

   5. $f(x) = \sin x^2$. Let $f(x + T) = f(x) \Rightarrow \sin(x + T)^2 = \sin x^2 \Rightarrow (x + T)^2 = n\pi + (-1)^nx^2$ which will yield no value of $T$ independent of $x$ unless $T = 0$. Thus, the given function is non-periodic.

   6. $f(x) = \sin\frac{1}{x}$. Let $f(x + T) = f(x) \Rightarrow \sin\frac{1}{x + T} = \sin\frac{1}{x} \Rightarrow \frac{1}{x + T} = n\pi + (-1)^n\frac{1}{x}$ which will give no value of $T$ independent of $x$ unless $T = 0$. Thus, the given function is non-periodic.

   7. $f(x) = 1 + \tan x$. We know that $\tan x$ is a periodic function with a period $\pi$. Hence, $f(x)$ will also be a periodic function with a period of $\pi$.

   8. $f(x) = [x]$, where $[x]$ is integral value of $x$. Let $f(x + T) = f(x) \Rightarrow [x + T] = [x] \Rightarrow [x + T] - [x] = 0$ which is not true for any value of $T$ as for any value of $T$ it is possiblel that $[x + T] - [x] = 1$. Thus, $f(x)$ is non-periodic.

   9. $f(x) = 5$. Let $f(x + T) = f(x) \Rightarrow 5 = 5$ which is true but gives us no value for $T$. Thus, the given function is periodic but has no fundamental period.

   10. $f(x) = |\cos x| \Rightarrow f(x) = -\cos x$ if $\cos x < 0$ and $f(x) = \cos x$ if $\cos x > 0$. We know that $\cos x$ has a period of $2\pi$ therefore $f(x)$ will have period equal to half the period of that of $\cos x$ i.e. $\pi$.

   11. $f(x) = \sin^4x + \cos^4x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x = 1 - \frac{\sin^22x}{2} = 1 - \frac{1 - \cos4x}{4} = \frac{3}{4} + \frac{1}{4}\cos4x$. We know that $\cos x$ is a function having period $2\pi$ therefore $f(x)$ will be a periodic function with a period $2\pi/4$ i.e. $\frac{\pi}{2}$.

   12. $f(x) = x + \sin x$. Let $f(x + T) = f(x) \Rightarrow x + T + \sin(x + T) = x + \sin x$ $\Rightarrow T = \sin x - \sin(x + T)$ which will give no value of $T$ independent of $x$ as R.H.S. is a trigonometric function in $x$ but L.H.S. is not. So the function $f(x)$ is non-periodic.

   13. $f(x) = \cos\sqrt{x}$. Following the fifth problem of previous problem we can deduce that given function is non-periodic.

   14. $f(x) = \tan^{-1}(\tan x)$. Let $f(x + T) = f(x) \Rightarrow \tan^{-1}\tan(x + T) = \tan^{-1}(\tan x) \Rightarrow \tan(x + T) = \tan x$ which gives $T = \pi$ as the period.

   15. $f(x) = |\sin x| + |\cos x|$ which will yield four different equations depending on whether $\sin x$ and $\cos x$ are positive or negative. Also, the period of $\sin x$ and $\cos x$ is $2\pi$ for both of the functions. Thus, the given function will have a period of $2\pi/4 = \frac{\pi}{2}$.

   16. $f(x) = \sin\frac{\pi x}{3} + \sin\frac{\pi x}{4}$. We know that $\sin x$ has a period of $2\pi$ therefore $\sin\frac{\pi x}{3}$ will have a period of $6$ and $\sin \frac{\pi x}{4}$ will have a period of $8$. The given function will have period equal to L.C.M. of $6$ and $8$ i.e. 24.

   17. $f(x) = \sin\left(2\pi x + \frac{\pi}{3}\right) + 2\sin\left(3\pi x + \frac{\pi}{4}\right) + 3\sin\pi x$. We know that the period of $\sin x$ has a period of $2\pi$ so the three terms will have period of $1, 2/3$ and $2$ respectively. Thus, given function will have period equal to L.C.M. of these three periods i.e. $2$.

   18. $f(x) = \sin x + \cos\sqrt{x}$. Now we have proven that $\cos\sqrt{x}$ is a non-periodic function therefore $f(x)$ will also be non-periodic.

3. $f(x) = 2\sin x + 3\cos 2x$. We know that both $\sin x$ and $\cos x$ have a period of $2\pi$ therefore period of first term would be $2\pi$ and of the second term will be $\pi$. $f(x)$ will have period equal to L.C.M. of these two terms i.e. $2\pi$.

4. a. Given, $f(x) = 2x - [2x]$. Let $f(x + T) = f(x) \Rightarrow 2(x + T) - [2(x + T)] = 2x - [2x]$

   $\Rightarrow T = \frac{[2x + 2T] - [2x]}{2} = \frac{\mathrm{an integer}}{2}$.

   Therefore, positive value of $T$ independentt of $x$ can be found and least such value is $\frac{1}{2}$.

   2. Given, $g(x) = 1 + \frac{3}{2 - \sin^2x}$. Let $g(x + T) = g(x)$

      $\Rightarrow \frac{3}{2 - \sin^2(x + T)} = \frac{3}{2 - \sin^2x}$

      $\Rightarrow \sin^2(x + T) = \sin^2x \Rightarrow x + T = n\pi +(-1)^n(\pm x) = n\pi\pm x$

      which gives us a periodic function with $T = \pi$.

5. $1 - \frac{1}{4}\sin^2\left(\frac{\pi}{3} - \frac{3x}{2}\right) = \frac{7}{8} + \frac{1}{8}\cos\left(3x - \frac{2\pi}{3}\right)$ which is a periodic function with period $2\pi/3$.