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Question

Question: What is the period of \(y = 3\cos 5x\) ?...

What is the period of y=3cos5xy = 3\cos 5x ?

Explanation

Solution

Here we are going to find the period of trigonometric function using the definition of periodic function.
Definition used:
A function is said to be periodic if there exists a positive real number TT such that f(x+T)=f(x)f\left( {x + T} \right) = f(x) for all xDx \in D where DD is the domain of the functionf(x)f(x) .
Now for trigonometric function, graphs of trigonometric function clearly show that periods of cosx\cos x is 2π2\pi . Here we shall mathematically determine periods of few of these trigonometric functions using definition of period.

Complete step-by-step solution:
For cosx\cos x to be periodic function,
cos(x+T)=cosx\cos \left( {x + T} \right) = \cos x
x+T=2nπ±x\Rightarrow x + T = 2n\pi \pm x , nZn \in Z
Therefore x+T=2nπ+xx + T = 2n\pi + x or x+T=2nπxx + T = 2n\pi - x
First set of value is independent of xx hence,
T=2nπ,nZT = 2n\pi , n \in Z
The least positive value of TT that is the period of the function is T=2πT = 2\pi .
Now we have to find the periodic function y=3cos5xy = 3\cos 5x . We know that the period of the function cosx\cos x is 2π2\pi .
If we have a function f(x)=cos(ax)f(x) = \cos (ax) where a>0a > 0 is the coefficient of the xx term, then the period of the functions is T=2πaT = \dfrac{{2\pi }}{a} .
Here we have the periodic function is f(x)=3cos5xf(x) = 3\cos 5x The formula for the period of the function is T=2πaT = \dfrac{{2\pi }}{a}.
From given function, a=5a = 5
Hence the period of the function T=2π5T = \dfrac{{2\pi }}{5}

Note: The smallest positive αR\alpha \in \mathbb{R} for a periodic function ff is defined as the fundamental period of ff.
Ex:
The function f=sinxf = \sin x is periodic function with set of periods \left\\{ {2n\pi :n \in \mathbb{Z}} \right\\} fundamental period of f=sinxf = \sin x is 2π2\pi

Periodicity is the domain based property.
A periodic function may or may not have a fundamental period.
Ex:
f:RRf: R \to R
f(x)=c,xR,cRf(x) = c, \forall x \in \mathbb{R},\,c \in {\mathbb{R}}
Then for any αRf(x+α)=f(x)=c,xR\alpha \in \mathbb{R}\,\,\,\,\,\,f\left( {x + \alpha } \right) = f(x) = c, \forall x \in \mathbb{R}
Set of periods is R\mathbb{R}
But the fundamental period does not exist.
Sum or difference of periodic function may not be periodic. Sum of two non periodic functions may be a periodic function.