Question

Sketch the following graphs:

(i) y = 2 sin 2x
(ii) y = 3 sin x
(iii) $y=2\sin \left(x-\frac{\mathrm{\pi }}{4}\right)$
(iv) y = 2 sin (2x − 1)
(v) y = 3 sin (3x + 1)
(vi) $y=3\sin \left(2x-\frac{\mathrm{\pi }}{4}\right)$

Answer 1

(i)
Step I- We find the value of c and a by comparing y = 2 sin 2x with y = c sin ax, i.e. c = 2 and a = 2.
Step II- Then, we draw the graph of y = sin x and mark the point where it crosses the x-axis.
Step III- Divide the x-coordinates of the points where y = sin x crosses x-axis by 2 (i.e. a = 2) and mark the maximum value (i.e. c = 2) and minimum value (i.e.$-$c = $-$2).
Then, we obtain the following graph:

(ii)
Step I- We find the value of c and a by comparing y = 3 sin x with y = c sin ax, i.e. c = 3 and a = 1.
Step II- Then, we draw the graph of y = sin x and mark the point where it crosses the x-axis.
Step III- Divide the x-coordinates of the points where y = sin x crosses x-axis by 1 (i.e. a = 1) and mark the maximum value (i.e. c = 3) and minimum value (i.e.$-$c = $-$3).
Then, we obtain the following graph:


(iii)
$$\begin{gathered} y=2\sin \left(\mathrm{x}-{\displaystyle \frac{\mathrm{\pi }}{4}}\right) \\ \Rightarrow y-0=2\sin \left(x-\frac{\mathrm{\pi }}{4}\right)...\left(\mathrm{i}\right) \\ On\mathrm{shifting}\mathrm{the}\mathrm{origin}\mathrm{at}\left(\frac{\mathrm{\pi }}{4},0\right),weget: \\ \mathrm{x}=\mathrm{X}+\frac{\mathrm{\pi }}{4}\mathrm{and}\mathrm{y}=\mathrm{Y}+0 \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Y}=2\mathrm{sinX} \\ \mathrm{Then},\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}Y=2\sin X\mathrm{and}\mathrm{shift}\mathrm{it}\mathrm{by}\frac{\mathrm{\pi }}{4}\mathrm{to}\mathrm{the}\mathrm{right}. \end{gathered}$$
Then, we obtain the following graph:


(iv)
$$\begin{gathered} y=2\sin \left(2\mathrm{x}-1\right) \\ \Rightarrow y-0=2\sin 2\left(x-\frac{1}{2}\right)...\left(\mathrm{i}\right) \\ On\mathrm{shifting}\mathrm{the}\mathrm{origin}\mathrm{at}\left(\frac{1}{2},0\right),\mathrm{we}get: \\ \mathrm{x}=\mathrm{X}+\frac{1}{2}\mathrm{and}\mathrm{y}=\mathrm{Y}+0 \\ \mathrm{On}\mathrm{subsitituting}\mathrm{the}\mathrm{values}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Y}=2\sin 2X \\ \mathrm{Then},\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}Y=2\sin 1X\mathrm{and}\mathrm{shift}\mathrm{it}\mathrm{by}\frac{1}{2}\mathrm{to}\mathrm{the}\mathrm{right}. \end{gathered}$$
Then, we obtain the following graph:


(v)

$$\begin{gathered} y=3\sin \left(3\mathrm{x}+1\right) \\ \Rightarrow y-0=2\sin 3\left(\mathrm{x}+\frac{1}{3}\right)...\left(\mathrm{i}\right) \\ On\mathrm{shifting}\mathrm{the}\mathrm{origin}\mathrm{at}\left(-\frac{1}{3},0\right),\mathrm{we}get: \\ \mathrm{x}=\mathrm{X}-\frac{1}{3}\mathrm{and}\mathrm{y}=\mathrm{Y}+0 \\ \mathrm{On}\mathrm{subsitituting}\mathrm{the}\mathrm{values}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Y}=3\sin 3X \\ \mathrm{Then},\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}Y=3\sin 3X\mathrm{and}\mathrm{shift}\mathrm{it}\mathrm{by}\frac{1}{3}\mathrm{to}\mathrm{the}\mathrm{left}. \end{gathered}$$
Then, we obtain the following graph:



(vi)
$$\begin{gathered} y=3\sin \left(\mathrm{x}-{\displaystyle \frac{\mathrm{\pi }}{4}}\right) \\ \Rightarrow y-0=3\sin 2\left(x-\frac{\mathrm{\pi }}{8}\right)...\left(\mathrm{i}\right) \\ \mathrm{On}\mathrm{shifting}\mathrm{the}\mathrm{origin}\mathrm{at}\left(\frac{\mathrm{\pi }}{8},0\right),\mathrm{we}\mathrm{get}: \\ \mathrm{x}=\mathrm{X}+\frac{\mathrm{\pi }}{8}\mathrm{and}\mathrm{y}=\mathrm{Y}+0 \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Y}=3\sin 2X \\ \mathrm{Then}.\mathrm{we}\mathrm{draw}\mathrm{the}\mathrm{graph}\mathrm{of}Y=3\sin 2X\mathrm{and}\mathrm{shift}\mathrm{it}\mathrm{by}\frac{\mathrm{\pi }}{8}\mathrm{to}\mathrm{the}\mathrm{right}. \end{gathered}$$
Then, we obtain the following graph: