5.3 Periodicity and periods  (Page 2/2)

Problem : Find period of function :

$$f\left(x\right)=3\sin 2\left\{3x\right\}+5\cos 3\left\{2x\right\}$$

where {} denotes fraction part function.

Solution : The period of {x} is 1. Therefore, period of function {3x} is 1/3. Domains of sin2x and FPF functions are R. Thus, domain of FPF is proper subset of domain of sin2x. Hence, period of 3sin2{3x} is 1/3. On the other hand, period of function {2x} is 1/2. Domains of cos3x and FPF functions are R. Hence, period of 5cos3{2x} is 1/2. LCM of 1/2 and 1/3 is :

$$\Rightarrow LCM=\frac{\text{LCM of}\left(\mathrm{1,1}\right)}{\text{HCF of}\left(\mathrm{2,3}\right)}=\frac{1}{1}=1$$

Problem : Find period of function :

$$f\left(x\right)=3\sin 2\sqrt{3}x+2\cos 5\sqrt{3}x$$

Solution : Period of $3\sin 2\sqrt{3}x$ is :

$$\Rightarrow T_1=\frac{2\pi }{2\sqrt{3}}$$

Period of $2\cos 5\sqrt{3}x$ is :

$$\Rightarrow T_2=\frac{2\pi }{5\sqrt{3}}$$

Two irrational periods are of same kind. Hence, period of given function is :

$$\Rightarrow LCM=\frac{\text{LCM of}\left(2\pi ,2\pi \right)}{\text{HCF of}\left(2\sqrt{\mathrm{3,}}5\sqrt{3}\right)}=\frac{2\pi }{\sqrt{3}}$$

Problem : Find period of function :

$$f\left(x\right)=\sin \sqrt{2}x+\cos \sqrt{3}x$$

Solution : Period of $\sin \sqrt{2}x$ is :

$$\Rightarrow T_1=\frac{2\pi }{\sqrt{2}}$$

Period of $\cos \sqrt{3}x$ is :

$$\Rightarrow T_2=\frac{2\pi }{\sqrt{3}}$$

Two irrational periods are not of same kind. Hence, function is not periodic.

Problem : Determine whether the function given below is periodic?

$$f\left(x\right)=\cos \sqrt{x}$$

If the function is periodic, then find its period.

Solution : We need to check periodicity applying definition of periodic function. According to definition of periodic function,

$$f\left(x+T\right)=f\left(x\right)$$

$$\Rightarrow \cos \sqrt{\left(x+T\right)}=\cos \sqrt{x}$$

$$\Rightarrow \sqrt{\left(x+T\right)}=2n\pi \pm \sqrt{x},n\in Z$$

Squaring both sides and solving for “T”, we have :

$$\Rightarrow \left(x+T\right)={\left(2n\pi \pm \sqrt{x}\right)}^2$$

$$\Rightarrow T={\left(2n\pi \pm \sqrt{x}\right)}^2-x$$

If we expand the square term, then we find that the expression of “T” is not independent of “x”. Hence, given function is not a periodic function.

Problem : Find period of function :

$$f\left(x\right)=\sin x+\tan \frac{x}{2}$$

Solution : The function is addition of two trigonometric functions. Each of the functions is periodic. We see that neither sine nor tangent function is even function. Therefore, we apply LCM rule to find the period of the function.

The period of “sin x” is “2π”. Hence,

$$\Rightarrow T_1=2\pi$$

We also know that period of g(ax+b) is equal to the period of g(x), divided by “|a|”. The period of second term of “f(x)”, therefore, is equal to the period of “tanx”, divided by “1/2”. Thus, period of “tanx” is “π”.

$$\Rightarrow T_2=\frac{\pi }{\frac{1}{2}}=2\pi$$

Now, LCM of “2π” and “2π” is “2π”. Hence,

$$\Rightarrow T=2\pi$$

Problem : Determine whether the function given below is periodic?

$$f\left(x\right)=x\cos x$$

If the function is periodic, then find its period.

Solution : The function is product of algebraic function “x” and trigonometric function “cosx”. We need to check periodicity applying definition of a periodic function.

Let the function be periodic. Then,

$$f\left(x+T\right)=f\left(x\right)$$

$$\Rightarrow \left(x+T\right)\cos \left(x+T\right)=x\cos x$$

$$\Rightarrow T\cos \left(x+T\right)=x\cos x-x\cos \left(x+T\right)=x\{\cos x-\cos \left(x+T\right)\}$$

We see that right hand side is product of algebraic function “x” and trigonometric function. On the left hand side, there is only trigonometric function (apart from “T”, which is a constant). There is no algebraic function in “x” on the left which can cancel "x" on the right. Thus, we conclude that “T” is not independent of “x” and as such give function is not periodic.

Problem : If the function $f\left(x\right)=\sin x+\cos ax$ be periodic, then prove that “a” is rational.

Solution :

Statement of the problem : The function is sum of two trigonometric functions. It is given that the function is a periodic function.

Let “T” be the period, then according to definition :

$$\sin \left(x+T\right)+\cos a\left(x+T\right)=\sin x+\cos ax$$

Putting, x = 0, we have :

$$\Rightarrow \sin T+\cos aT=1$$

Putting, x = -T, we have :

$$\Rightarrow \sin 0+\cos 0=-\sin T+\cos T$$

$$\Rightarrow -\sin T+\cos aT=1$$

Subtracting one equation from another,

$$\sin T=0$$

$$\Rightarrow T=n\pi ,n\in Z$$

and two equations :

$$\cos aT=1$$

$$\Rightarrow aT=2m\pi ,n\in Z$$

Combining two results :

$$\Rightarrow a=\frac{\mathrm{2m}}{n}$$

Hence, “a” is a rational number.

Problem : Find value of “n”, if period of given function is π/4.

$$f\left(x\right)=\frac{4\sin 2nx}{1+{\cos }^{2}nx};n\in N$$

Solution : The numerator and denominator repeat simultaneously. Hence, period of given function is LCM of the periods of numerator and denominator. Now, period of 4sin2nx is :

$$\Rightarrow T_1=\frac{2\pi }{2n}=\frac{\pi }{n}$$

Using trigonometric identity, the denominator of the given function is :

$$\Rightarrow 1+\cos {}^{2}nx=1+\frac{1+\cos 2nx}{2}$$

The period of denominator is :

$$\Rightarrow T_2=\frac{2\pi }{2n}=\frac{\pi }{n}$$

Hence, period of given function is LCM of π/n and π/n, which is π/n. According to question,

$$\Rightarrow \frac{\pi }{n}=\frac{\pi }{4}$$

$$\Rightarrow n=4$$

Problem : Find period of function :

$$f\left(x\right)={\sin }^{3}x+{\sin }^2\left(x+\frac{\pi }{3}\right)-\cos x\cos \left(x+\frac{\pi }{3}\right)$$

Solution : The given function is not an even function as ${\sin }^3\left(-x\right)=-{\sin }^{3}x$ . We can use LCM rule to determine period of given function. The periods of ${\sin }^{3}x$ and $\sin {}^2\left(x+\frac{\pi }{3}\right)$ are 2π and π respectively. Using trigonometric identity,

$$\cos x\cos \left(x+\frac{\pi }{3}\right)=\frac{1}{2}\cos \left(2x+\frac{\pi }{3}\right)+\cos \left(\frac{\pi }{3}\right)$$ $$\Rightarrow \cos x\cos \left(x+\frac{\pi }{3}\right)=\frac{1}{2}\cos \left(2x+\frac{\pi }{3}\right)+\frac{1}{2}$$

The period of cos (2x+π/3) is 2π/2 = π. Now, periods of three terms of given function are 2π,π and π. Hence, period of given function is LCM of three terms i.e. 2π.