Question

Sketch the following graphs :
(i) y=2 sin 2x
(ii) y= 3 sin x
(iii) $y=2sin(x-\frac{\pi }{4})$
(iv) y=2 sin (2x-1)
(v) y=3 sin(3x+1)
(vi) $y=3sin(2x-\frac{\pi }{4})$

Answer 1

(i)To obtain the graph of y = 2sin 2x we first draw the graph of y = sinx in the interval [0,2x] and then divide the x-coordinates of the points where it crosses x-axis by 2. The maximum and minimum values are 2 and -2 respectively.

(ii) To obtain the graph of y=3sin x we first draw the graph of y=sinx in the interval $[0,2\pi ]$. The maximum and minimum values are 3 and -3 respectively.

(iii) We have
$y=2sin(x-\frac{\pi }{4})$
$\Rightarrow (y-0)=2sin(x-\frac{\pi }{4})$
Shifting the origin at $(\frac{\pi }{4},0)$, we have
$x=X+\frac{\pi }{4}andy=Y+0$
Substituting these values in (i), we get
Y=2 sin X
Thus we draw the graph of Y=2 sin X and shift it by $\frac{\pi }{4}$to the right to get the required graph.

(iv) We have,
y=2 sin (2x-1)
$\Rightarrow (y-0)=2sin2(x-\frac{1}{2})$
Shifting the origin at $\left(\frac{1}{2}\right)$, we have
$x=X+\frac{1}{2}andy=Y+0$
Substituing these values in (i), we get
Y=2 sin 2 x
Thus we draw the graph of Y=2 sin 2x and shift it by $\frac{1}{2}$to the right to get the requuired graph.

(v) We have,
y=3 sin (3x+1)
$\Rightarrow (y-0)=3sin3(x+\frac{1}{3})$
Shifting the origin at $(-\frac{1}{3},0)$, we have x=X-$\frac{1}{3}$ and y = Y+0
Substituting these values in (i), we get
Y=3 sin 3X
Thus we draw the graph of Y=3 sin 3X and shift it by $\frac{1}{3}$to the left to get the required graph.

(vi) We have,
$y=3sin(2x-\frac{\pi }{4})$
$\Rightarrow (y-0)=3sin2(x-\frac{\pi }{8})$
Shifting the origin at $(\frac{\pi }{8},0)$, we have
x = X + $\frac{\pi }{8}$ and y=Y+0
Substituting these values in (i), we get Y=3 sin 2X
Thus we draw the graph of Y=3 sin 2X and shift it by $\frac{\pi }{8}$ to the right to get the required graph.