Question

What is the Amplitude and Period of \[y = \cos 9x\]?

Answer 1

Hint: We use some of the concepts of geometry and trigonometry to solve this problem. We will get to know about the terms like period, amplitude and other in this sum. We will also check whether our answer is correct or not, using some examples.
Modulus of a number always gives a positive number as a result.

Complete step by step solution:
In a graph of functions like sine or cosine, there is a term namely Amplitude which means the maximum value or the minimum value the function gains. It can also be defined as half the distance between maximum and minimum values of the function.
And if a function repeats its values at regular intervals, then it is called a periodic function and the interval is called the period of that function.
For a function of kind \[y = a\cos (bx)\] or \[y = a\sin (bx)\] ,
The amplitude is equal to modulus of \[a\] i.e., amplitude \[A = |a|\]
And, the period is equal to \[P = \dfrac{{2\pi }}{{|b|}}\]
So, for example, take \[y = - 2\sin \left( {\dfrac{{3x}}{7}} \right)\]
So, here amplitude is \[A = | - 2| = 2\] and period is \[P = \dfrac{{2\pi }}{{\left| {\dfrac{3}{7}} \right|}}\]
And now, in the question, it is given as \[y = \cos 9x\] and we need to find the amplitude and period.
On comparing this equation with \[y = a\cos (bx)\], we get, \[a = 1\] and \[b = 9\]
So, amplitude is \[A = |a| = 1\] and period is \[P = \dfrac{{2\pi }}{{|b|}} = \dfrac{{2\pi }}{9}\]. So, the period is \[P = \dfrac{{2\pi }}{9}\]. So, for every \[\dfrac{{2\pi }}{9}\] interval, this function gains the same values.
At \[x = 0\] and we get \[y = \cos (0) = 1\]
At \[x = \dfrac{{2\pi }}{9}\] we get \[y = \cos \left( {9.\dfrac{{2\pi }}{9}} \right) = \cos \left( {2\pi } \right) = 1\]
At \[x = \dfrac{{4\pi }}{9}\] we get \[y = \cos \left( {9.\dfrac{{4\pi }}{9}} \right) = \cos \left( {4\pi } \right) = 1\]
Therefore, we can conclude that, for every \[\dfrac{{2\pi }}{9}\] interval, we are getting the same values.

Note:
For sine and cosine functions, the period is \[\dfrac{{2\pi }}{{|b|}}\], but whereas for the function of kind \[y = a\tan (bx)\] the amplitude is \[|a|\] and period is equal to \[\dfrac{\pi }{{|b|}}\]. We can also find the amplitude and period by drawing the graph of the function. Also be careful while taking the values of \[a,b\], because they have to be taken with modulus.