Question

How do you sketch one cycle of \[y = \sec \left( {2x} \right)?\]

Answer 1

Hint:We need to know the trigonometric table value for \[\cos \theta \]and\[\sec \theta \]. Also, we need to know the relation between \[\cos \theta \] and \[\sec \theta \]. Also, this question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know how to assume the \[x\] value, by using the assumed \[x\] value we can set \[y\] the value. We need to know how to sketch graphs by using the \[x\] and \[y\] values.

Complete step by step solution:
The given equation is shown below,
\[y = \sec \left( {2x} \right) \to \left( 1 \right)\]
We know that,
\[\cos \theta = \dfrac{1}{{\sec \theta }}\]
So, first, we would find the value for \[\cos \theta \], next, we can easily find the values for \[\sec \theta \].
We get,
\[\cos 2x = \dfrac{1}{{\sec 2x}}\]
\[\sec 2x = \dfrac{1}{{\cos 2x}} \to \left( 2 \right)\]
We assume,
\[x = ....\dfrac{{ - \pi }}{2},0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi ,.....\]
Let substitute \[x = \dfrac{{ - \pi }}{2}\] in the equation \[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times \dfrac{{ - \pi }}{2}} \right) = \dfrac{1}{{\cos \left( {2 \times \dfrac{{ - \pi }}{2}}
\right)}} = \dfrac{1}{{\cos \left( { - \pi } \right)}} = \dfrac{1}{1} \\
\sec \left( {2 \times \dfrac{{ - \pi }}{2}} \right) = 1 \\
\]
Let substitute \[x = 0\]in the equation \[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times 0} \right) = \dfrac{1}{{\cos \left( {2 \times 0} \right)}} = \dfrac{1}{{\cos 0}} =
\dfrac{1}{1} \\
\sec \left( {2 \times 0} \right) = 1 \\
\]
Let’s substitute\[x = \dfrac{\pi }{2}\]in the equation\[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times \dfrac{\pi }{2}} \right) = \dfrac{1}{{\cos \left( {2 \times \dfrac{\pi }{2}}
\right)}} = \dfrac{1}{{\cos \left( \pi \right)}} = \dfrac{1}{{ - 1}} \\
\sec \left( {2 \times \dfrac{\pi }{2}} \right) = - 1 \\
\]
Let’s substitute \[x = \pi \]in the equation\[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times \pi } \right) = \dfrac{1}{{\cos \left( {2 \times \pi } \right)}} = \dfrac{1}{{\cos
2\pi }} = \dfrac{1}{1} \\
\sec \left( {2 \times \pi } \right) = 1 \\
\]
Let’s substitute \[x = \dfrac{{3\pi }}{2}\] in the equation\[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times \dfrac{{3\pi }}{2}} \right) = \dfrac{1}{{\cos \left( {2 \times \dfrac{{3\pi }}{2}}
\right)}} = \dfrac{1}{{\cos \left( {3\pi } \right)}} = \dfrac{1}{{ - 1}} \\
\sec \left( {2 \times \dfrac{{3\pi }}{2}} \right) = - 1 \\
\]
Let’s substitute \[x = 2\pi \] in the equation \[\left( 2 \right)\], we get
\[\left( 2 \right) \to \sec 2x = \dfrac{1}{{\cos 2x}}\]
\[
\sec \left( {2 \times 2\pi } \right) = \dfrac{1}{{\cos \left( {2 \times 2\pi } \right)}} = \dfrac{1}{{\cos
4\pi }} = \dfrac{1}{1} \\
\sec \left( {2 \times 2\pi } \right) = 1 \\
\]
By using these values we can make the following tabular column,

\[x\]	\[\dfrac{{- \pi}}{2}\]	\[0\]	\[\dfrac{\pi}{2}\]	\[\pi \]	\[\dfrac{{3\pi}}{2}\]	\[2\pi \]
\[y =\sec2x\]	\[1\]	\[1\]	\[ - 1\]	\[1\]	\[ - 1\]	\[1\]

By using this tabular column we can easily make the following graph,

The above graph shows the one cycle of \[\sec 2x\]

Note: Note that when \[\theta \] is the end with \[\dfrac{\pi }{2}\] the term, all \[\cos \theta \] value becomes zero. Also, note that when \[\theta \] has an odd \[\pi \] term, all the \[\cos \theta \] values become \[ - 1\] and when \[\theta \] has an even \[\pi \] term, all the \[\cos \theta \] values become \[1\]. Note that when the denominator value becomes zero, then the answer becomes undefined. This question involves the operation of addition / subtraction/ multiplication/ division. Also, note that \[\cos \theta \] is the inverse value of \[\sec \theta \]. Remember the trigonometric table values to make easy calculations.