How do you graph \[y=\tan \left( 2x \right)\]?
Answer
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Hint: For solving this question, we should know that if \[f\left( x \right)\] (that is called a function of x) has a period of T, then the period of \[f\left( kx \right)\] will be \[\dfrac{T}{k}\]. In solving this question, first we will calculate the period of \[\tan \left( 2x \right)\]. After that we will draw the graph of \[\tan \left( 2x \right)\].
Complete step by step answer:
Let us solve the question.
As we know that period of \[\tan \left( x \right)\] is \[\pi \].
The graph of \[y=\tan \left( x \right)\] is:
Here, in the above graph, the value of x is given along x-axis and the value of \[\tan x\] is given along y-axis.
As it is seen from the above graph that after every \[\pi \] units, the \[\tan x\] is repeating.
Let us find out the period of \[\tan 2x\].
As we know that if period of a function \[f\left( x \right)\] is T. Then, the period of \[f(kx)\] is \[\dfrac{T}{k}\] , where k is any real number.
Now, applying the above procedure in \[\tan x\].
If \[\tan x\] has a period of \[\pi \]
Then, we can say that
\[\tan \left( 2x \right)\] has a period of \[\dfrac{\pi }{2}\].
If \[\tan x\] has a period of \[\pi \], that means the graph of \[\tan x\] is repeating after every \[\pi \] units.
Then, \[\tan \left( 2x \right)\] has a period of \[\dfrac{\pi }{2}\], that means the graph of \[\tan \left( 2x \right)\] will be repeating after every \[\dfrac{\pi }{2}\] units.
Therefore, the graph of \[\tan \left( 2x \right)\] will be:
In the above graph, the value of x is given along the x-axis and the value of \[\tan \left( 2x \right)\] is given along the y-axis.
We can see from the above graph that the \[\tan \left( 2x \right)\] is repeating after every \[\dfrac{\pi }{2}\] units.
This graph is 2 times faster than first.
Note: Remember the period of trigonometric functions to solve this type of problems. And we should have a proper knowledge in periodic functions also. As stated above that the period of \[f(kx)\] is \[\dfrac{T}{k}\]. Here, k can be any real number. But, make sure that k should not be zero. Otherwise, the process will be wrong in that case.
Complete step by step answer:
Let us solve the question.
As we know that period of \[\tan \left( x \right)\] is \[\pi \].
The graph of \[y=\tan \left( x \right)\] is:
Here, in the above graph, the value of x is given along x-axis and the value of \[\tan x\] is given along y-axis.
As it is seen from the above graph that after every \[\pi \] units, the \[\tan x\] is repeating.
Let us find out the period of \[\tan 2x\].
As we know that if period of a function \[f\left( x \right)\] is T. Then, the period of \[f(kx)\] is \[\dfrac{T}{k}\] , where k is any real number.
Now, applying the above procedure in \[\tan x\].
If \[\tan x\] has a period of \[\pi \]
Then, we can say that
\[\tan \left( 2x \right)\] has a period of \[\dfrac{\pi }{2}\].
If \[\tan x\] has a period of \[\pi \], that means the graph of \[\tan x\] is repeating after every \[\pi \] units.
Then, \[\tan \left( 2x \right)\] has a period of \[\dfrac{\pi }{2}\], that means the graph of \[\tan \left( 2x \right)\] will be repeating after every \[\dfrac{\pi }{2}\] units.
Therefore, the graph of \[\tan \left( 2x \right)\] will be:
In the above graph, the value of x is given along the x-axis and the value of \[\tan \left( 2x \right)\] is given along the y-axis.
We can see from the above graph that the \[\tan \left( 2x \right)\] is repeating after every \[\dfrac{\pi }{2}\] units.
This graph is 2 times faster than first.
Note: Remember the period of trigonometric functions to solve this type of problems. And we should have a proper knowledge in periodic functions also. As stated above that the period of \[f(kx)\] is \[\dfrac{T}{k}\]. Here, k can be any real number. But, make sure that k should not be zero. Otherwise, the process will be wrong in that case.
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