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If “a” and “b” are non-zero real number and functions g(x) and h(x) are periodic functions having periods, “ ${T}_{1}$ ” and “ ${T}_{2}$ ”, then function $f\left(x\right)=ag\left(x\right)\pm bh\left(x\right)$ is also a periodic function. The period of f(x) is LCM of " ${T}_{1}$ " and " ${T}_{2}$ ".
LCM of integral numbers is obtained easily. There is, however, difficulty in finding LCM when numbers are fractions (like 3/4, 1/3 etc.) or irrational numbers (like π, 2√2 etc.).
For rational fraction, we can find LCM using following formula :
$$\mathrm{LCM}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$$
Consider fractions 3/5 and 2/3. The LCM of numerators 3 and 2 is 6. The HCF of denominators is 1. Hence, LCM of two fractions is 6/1 i.e. 6.
This rule also works for irrational numbers of similar type like 2√2/3 , 3√2/5 etc or π/2, 3π/2 etc. However, we can not find LCM of irrational numbers of different kind like 2√2 and π. Similarly, there is no LCM for combination of rational and irrational numbers.
For example, LCM of π/3 and 3π/2 is :
$$\mathrm{LCM}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}=\frac{\text{LCM of (\pi ,3\pi )}}{\text{HCF of (2,3)}}=\frac{\mathrm{3\pi}}{1}=\mathrm{3\pi}$$
Problem : Find period of
$$f\left(x\right)=\mathrm{sin}\left(2\pi x+\frac{\pi}{8}\right)+2\mathrm{sin}\left(3\pi x+\frac{\pi}{3}\right)$$
Solution : Period of $\mathrm{sin}\left(2\pi x+\pi /8\right)$ is :
$$\Rightarrow {T}_{1}=\frac{2\pi}{2\pi}=1$$
Period of $2\mathrm{sin}\left(3\pi x+\pi /3\right)$ is :
$$\Rightarrow {T}_{2}=\frac{2\pi}{3\pi}=\frac{2}{3}$$
LCM of numbers involving fraction is equal to the ratio of LCM of numerators and HCF of denominators. Hence,
$$\mathrm{LCM}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}=\frac{\text{LCM of (1,2)}}{\text{HCF of (1,3)}}=\frac{2}{1}=2$$
LCM rule is not always true. There are exceptions to this rule. We do not apply this rule, when functions are co-functions of each other or when functions are even functions. Further, if individual periods are rational and irrational numbers respectively, then LCM is not defined. As such, this rule can not be applied in such situation as well.
Two functions f(x) and g(y) are cofunctions if x and y are complimentary angles. The functions sinx and consx are cofunctions as :
$$\mathrm{sin}x=\mathrm{cos}(\frac{\pi}{2}-x)$$
Similarly, |cosx| and |sinx| are cofunctions as :
$$\left|\mathrm{cos}x\right|=\left|\mathrm{sin}\right(\frac{\pi}{2}-x\left)\right|$$
In such cases where LCM rule is not applicable, we proceed to apply definition of periodic function to determine period.
Problem : Find period of
$$f\left(x\right)=\left|\mathrm{cos}x\right|+\left|\mathrm{sin}x\right|$$
Solution : We know that |cosx| and |sinx| are co-functions. Recall that a function “f” is co-function of a function “g” if f(x) = g(y) where x and y are complementary angles. Hence, we can not apply LCM rule. But, we know that sin(x + π/2) = cosx. This suggests that the function may have the period "π/2". We check this as :
$$f\left(x+\frac{\pi}{2}\right)=\left|\mathrm{cos}\left(x+\frac{\pi}{2}\right)\right|+\left|\mathrm{sin}\left(x+\frac{\pi}{2}\right)\right|$$
$$\Rightarrow f\left(x+\frac{\pi}{2}\right)=|-\mathrm{cos}x|+\left|\mathrm{sin}x\right|=\left|\mathrm{cos}x\right|+\left|\mathrm{sin}x\right|=f\left(x\right)$$
Hence, period is “ $\pi /2$ ”.
It is intuitive here to work with this problem using LCM rule and compare the result. The period of modulus of all six trigonometric functions is π. The periods of |cosx| and |sinx| are π. Now, applying LCM rule, the period of given function is LCM of π and π, which is π.
Problem : Find period of
$$f\left(x\right)={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{4}x$$
Solution : Here, we see that f(-x) = - f(x).
$$f\left(\mathrm{-x}\right)={\mathrm{sin}}^{2}\left(\mathrm{-x}\right)+{\mathrm{cos}}^{4}\left(\mathrm{-x}\right)={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{4}x=f\left(x\right)$$
This means that given function is even function. As such, we can apply LCM rule. We, therefore, proceed to reduce the given function in terms of one trigonometric function type.
$$\Rightarrow f\left(x\right)=\mathrm{sin}{}^{2}x+\mathrm{cos}{}^{2}x\left(1-\mathrm{sin}{}^{2}x\right)=\mathrm{sin}{}^{2}x+\mathrm{cos}{}^{2}x-\mathrm{sin}{}^{2}X\mathrm{cos}{}^{2}x$$
$$\Rightarrow f\left(x\right)=1-\frac{1}{4}\mathrm{sin}{}^{2}2x=1-\frac{1}{4}X\frac{\left(1-\mathrm{cos}4x\right)}{2}$$
$$\Rightarrow f\left(x\right)=1-\frac{1}{8}+\frac{\left(\mathrm{cos}4x\right)}{8}$$
$$\Rightarrow T=\frac{2\pi}{4}=\pi /2$$
Note that if we apply LCM rule, then period evaluates to " $\pi $ ".
The results obtained in earlier sections are summarized here for ready reference.
1: All trigonometric functions are periodic on “R”. The functions $\mathrm{sin}x,\mathrm{cos}x,\mathrm{sec}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cosec}x$ have periodicity of " $2\pi $ ". On the other hand, periodicity of $\mathrm{tan}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cot}x$ is $\pi $ .
2: Functions ${\mathrm{sin}}^{n}x,{\mathrm{cos}}^{n}x,{\mathrm{cosec}}^{n}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{sec}}^{n}x$ are periodic on “R” with period “ $\pi $ ” when “n” is even and “ $2\pi $ ” when “n” is fraction or odd. On the other hand, Functions ${\mathrm{tan}}^{n}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{cot}}^{n}x$ are periodic on “R” with period “ $\pi $ ” whether n is odd or even.
3: Functions $\left|\mathrm{sin}x\right|,\left|\mathrm{cos}x\right|,\left|\mathrm{tan}x\right|,\left|\mathrm{cosec}x\right|,\left|\mathrm{sec}x\right|\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left|\mathrm{cot}x\right|$ are periodic on “R” with period “ $\pi $ ”.
4: A constant function is a periodic function without any fundamental period. For example,
$$f\left(x\right)=c$$
$$f\left(x\right)={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x$$
5: If “T” is the period of f(x), then period of function of the form given below is “T/|b|” :
$$af(bx+c)+d;\phantom{\rule{1em}{0ex}}\mathrm{a,b,c,d}\in Z$$
6: If f(x) is a periodic function with period “T” and g(x) is one one function (bijection), then "gof" is also periodic with period “T”.
7: If f(x) is a periodic function with a period T and its domain is a proper subset of domain of g(x), then gof(x) is a periodic function with a period T.
Problem : Find period of function :
$$f\left(x\right)=\mathrm{sin}\left(x\right)$$
where {} denotes fraction part function.
Solution : The fraction part function {x} is a periodic function with a period “1”. Its domain is R. On the other hand, sinx is a function having domain R. Therefore, domain of {x} is a proper subset of the domain of sinx. Hence, period of sin{x} is 1.
Problem : Find period of function tan⁻¹tanx
Solution : Inverse trigonometric function tan⁻¹x is one one function in [1,1] and tanx is a periodic function with period π in R. Hence, function tan⁻¹tanx function is a periodic function with period π.
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