Question

How do you sketch $y=\sin \left( \dfrac{x}{2} \right)$.?

Answer 1

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 $y=\sin \left( \dfrac{x}{2} \right)$ is given by \left\\{ x:x\in R \right\\}. It means that the $y=\sin \left( \dfrac{x}{2} \right)$ function exists for all x values, 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 $y=\sin \left( \dfrac{x}{2} \right)$, we get
$y=\sin \left( \dfrac{x}{2} \right)\Rightarrow y=\sin \left( -\dfrac{\pi }{2} \right)$
Simplifying the expression, we get

& y=-\sin \left( \dfrac{\pi }{2} \right) \\\ & y=-1 \\\ \end{aligned}$$ Substituting $$x=-\dfrac{3\pi }{4}$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$ , we get $$\Rightarrow y=\sin \left( \dfrac{-\dfrac{3\pi}{4}}{2} \right)$$ $$\Rightarrow y=\sin \left( \dfrac{-3\pi }{8} \right)$$ As sine function is an odd function, so we can write as $$y=-\sin \left( \dfrac{3\pi }{8} \right)$$ Substituting the value of the angle, we get $$y=-1$$ Substituting $$x=-\dfrac{\pi }{2}$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\Rightarrow y=\sin \left( \dfrac{-\dfrac{\pi }{2}}{2} \right)$$ $$\Rightarrow y=\sin \left( \dfrac{-\pi }{4} \right)$$ As sine function is an odd function, so we can write as $$\Rightarrow y=-\sin \left( \dfrac{\pi }{4} \right)$$ Substituting the value of the angle, we get $$y=-\dfrac{1}{\sqrt{2}}$$ Substituting $$x=-\dfrac{\pi }{4}$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\Rightarrow y=\sin \left( \dfrac{-\dfrac{\pi }{4}}{2} \right)$$ $$\Rightarrow y=\sin \left( -\dfrac{\pi }{8} \right)$$ Substituting the value of the angle, we get $$y=-0.38269$$ Substituting $$x=0$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\begin{aligned} & \Rightarrow y=\sin \left( \dfrac{0}{2} \right) \\\ & \Rightarrow y=\sin \left( 0 \right) \\\ \end{aligned}$$ Substituting the value of the angle, we get $$y=0$$ Substituting $$x=\dfrac{\pi }{4}$$in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\Rightarrow y=\sin \left(\dfrac{ \dfrac{\pi }{4}}{2} \right)$$ $$\Rightarrow y=\sin \dfrac{\pi }{8}$$ Substituting the value of the angle, we get $$y=0.38269$$ Substituting $$x=\dfrac{\pi }{2}$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\begin{aligned} & \Rightarrow y=\sin \left( \dfrac{\dfrac{\pi }{2}}{2} \right) \\\ & \Rightarrow y=\sin \left( \dfrac{\pi }{4} \right) \\\ \end{aligned}$$ Substituting the value of the angle, we get $$y=\dfrac{1}{\sqrt{2}}$$ Substituting $$x=\dfrac{3\pi }{4}$$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\begin{aligned} & \Rightarrow y=\sin \left( \dfrac{\dfrac{3\pi }{4}}{2} \right) \\\ & \Rightarrow y=\sin \left( \dfrac{3\pi }{8} \right) \\\ \end{aligned}$$ Substituting the value of the angle, we get $$y=1$$ Substituting $$x=\pi $$ in the function $$y=\sin \left( \dfrac{x}{2} \right)$$, we get $$\Rightarrow y=\sin \left( \dfrac{\pi }{2} \right)$$ Substituting the value of the angle, we get $$y=1$$ Arranging the values of x and y in a table and writing the coordinates, we get $$x$$| $$y$$| $$\left( x,y \right)$$ ---|---|--- $$-\pi $$| $$-1$$ | $$\left( -\pi ,-1 \right)$$ $$-\dfrac{3\pi }{4}$$| $$-1$$ | $$\left( -\dfrac{3\pi }{4},-1 \right)$$ $$-\dfrac{\pi }{2}$$| $$\dfrac{-1}{\sqrt{2}}$$ | $$\left( -\dfrac{\pi }{2},-\dfrac{1}{\sqrt{2}} \right)$$ $$-\dfrac{\pi }{4}$$| $$-0.38269$$ | $$\left( -\dfrac{\pi }{4},-0.38269 \right)$$ $$0$$| $$0$$ | $$\left( 0,0 \right)$$ $$\dfrac{\pi }{4}$$| $$0.38269$$ | $$\left( \dfrac{\pi }{4},0.38269 \right)$$ $$\dfrac{\pi }{2}$$| $$\dfrac{1}{\sqrt{2}}$$ | $$\left( \dfrac{\pi }{2},\dfrac{1}{\sqrt{2}} \right)$$ $$\dfrac{3\pi }{4}$$| $$1$$ | $$\left( \dfrac{3\pi }{4},1 \right)$$ $$\pi $$| $$1$$ | $$\left( \pi ,1 \right)$$ Now we will use these coordinates of the points to plot the required graph. Plotting the graphs and joining the curve, we get  **This is the required graph of the function $$y=\sin \left( \dfrac{x}{2} \right)$$.** **Note:** Whenever such types of questions appear, make sure you have drawn the graph correctly. The period of a function $$y=\sin kx$$ is given by $$\dfrac{2\pi }{k}$$. The period of the function $$y=\sin \left( \dfrac{x}{2} \right)$$ is $$\dfrac{2\pi }{\dfrac{1}{2}}=4\pi $$.this means that the graph of $$y=\sin \left( \dfrac{x}{2} \right)$$ will repeat for every distance on the x-axis. The range of the sine function is from -1 to 1.