Question

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

Answer 1

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

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

(iii) We have,
y= 3 cos (2x-1)
$\Rightarrow (y-0)=3cos2(x-\frac{1}{2})$
Shifting the origin at $(\frac{1}{2},0)$, We have
$x=X=\frac{1}{3}andy=Y+0$
Substituting these values in (i), we get
Y=3 cos 2X
Thus we draw the graph of Y= 3 cos 2x and shift it by $\frac{1}{2}$ to the right to get the required graph.

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