Question

How do you graph \[y = \sin 2x\]?

Answer 1

Hint: We need to graph the given function. We will use the domain and some values of \[x\] lying between \[ - \pi \] and \[\pi \] to find some values of \[y\]. Then, we will use the values of \[y\] to find the coordinates of points lying on the required graph, and use the coordinates obtained to graph the function.

Complete step-by-step solution:
The domain of all sine functions is the set of all real numbers.
Thus, the domain of the function \[y = \sin 2x\] is given by \[\left\{ {x:x \in R} \right\}\]. This means that the function \[y = \sin 2x\] exists for all values of \[x\], and is a continuous function.
Now, we will find some values of \[y\] for some values of \[x\] lying between \[ - \pi \] and \[\pi \].
Substituting \[x = - \pi \] in the function \[y = \sin 2x\], we get
\[\begin{array}{l}y = \sin \left( {2\left( { - \pi } \right)} \right)\\ \Rightarrow y = \sin \left( { - 2\pi } \right)\end{array}\]
The sine of an angle \[ - x\] can be written as \[\sin \left( { - x} \right) = - \sin x\].
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow y = - \sin \left( {2\pi } \right)\\ \Rightarrow y = 0\end{array}\]
Substituting \[x = - \dfrac{{3\pi }}{4}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( { - \dfrac{{3\pi }}{4}} \right)} \right)\\ \Rightarrow y = \sin \left( { - \dfrac{{3\pi }}{2}} \right)\end{array}\]
As sine function is an odd function, so we can writ above equation as
\[ \Rightarrow y = - \sin \left( {\dfrac{{3\pi }}{2}} \right)\]
Substituting the value of the angle, we get
\[ \Rightarrow y = 1\]
Substituting \[x = - \dfrac{\pi }{2}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( { - \dfrac{\pi }{2}} \right)} \right)\\ \Rightarrow y = \sin \left( { - \pi } \right)\end{array}\]
As sine function is an odd function, so we can writ above equation as
\[\begin{array}{l} \Rightarrow y = - \sin \left( \pi \right)\\ \Rightarrow y = 0\end{array}\]
Substituting \[x = - \dfrac{\pi }{4}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( { - \dfrac{\pi }{4}} \right)} \right)\\ \Rightarrow y = \sin \left( { - \dfrac{\pi }{2}} \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = - 1\]
Substituting \[x = 0\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( 0 \right)} \right)\\ \Rightarrow y = \sin \left( 0 \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = 0\]
Substituting \[x = \dfrac{\pi }{4}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( {\dfrac{\pi }{4}} \right)} \right)\\ \Rightarrow y = \sin \left( {\dfrac{\pi }{2}} \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = 1\]
Substituting \[x = \dfrac{\pi }{2}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( {\dfrac{\pi }{2}} \right)} \right)\\ \Rightarrow y = \sin \left( \pi \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = 0\]
Substituting \[x = \dfrac{{3\pi }}{4}\] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( {\dfrac{{3\pi }}{4}} \right)} \right)\\ \Rightarrow y = \sin \left( {\dfrac{{3\pi }}{2}} \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = - 1\]
Substituting \[x = \pi \] in the function \[y = \sin 2x\], we get
\[\begin{array}{l} \Rightarrow y = \sin \left( {2\left( \pi \right)} \right)\\ \Rightarrow y = \sin \left( {2\pi } \right)\end{array}\]
Substituting the value of the angle, we get
\[ \Rightarrow y = 0\]
Arranging the values of \[x\] and \[y\] in a table and writing the coordinates, we get

\[x\]	\[y\]	\[\left( {x,y} \right)\]
\[ - \pi \]	\[0\]	\[\left( { - \pi ,0} \right)\]
\[ - \dfrac{{3\pi }}{4}\]	\[1\]	\[\left( { - \dfrac{{3\pi }}{4},1} \right)\]
\[ - \dfrac{\pi }{2}\]	\[0\]	\[\left( { - \dfrac{\pi }{2},0} \right)\]
\[ - \dfrac{\pi }{4}\]	\[ - 1\]	\[\left( { - \dfrac{\pi }{4}, - 1} \right)\]
\[0\]	\[0\]	\[\left( {0,0} \right)\]
\[\dfrac{\pi }{4}\]	\[1\]	\[\left( {\dfrac{\pi }{4},1} \right)\]
\[\dfrac{\pi }{2}\]	\[0\]	\[\left( {\dfrac{\pi }{2},0} \right)\]
\[\dfrac{{3\pi }}{4}\]	\[ - 1\]	\[\left( {\dfrac{{3\pi }}{4}, - 1} \right)\]
\[\pi \]	\[0\]	\[\left( {\pi ,0} \right)\]

Now, we will use the coordinates of the points to plot the required graph.
Plotting the graphs and joining the curve, we get

This is the required graph of the function \[y = \sin 2x\].

Note:
The period of a function \[y = \sin kx\] is given by \[\dfrac{{2\pi }}{k}\]. The period of the function \[y = \sin 2x\] is \[\dfrac{{2\pi }}{2} = \pi \]. This means that the graph of \[y = \sin 2x\] will repeat for every \[\pi \] distance on the \[x\]-axis. It can be observed that the pattern and shape of the graph of \[y = \sin 2x\] is the same from \[ - \pi \] to 0, and from 0 to \[\pi \]. The range of the sine function is from \[ - 1\] to 1.