Sketch the graphs of the following trigonometric functions:
(i) $f(x)=\cos \left(x-\frac{\pi}{4}\right)$
(ii) $g(x)=\cos \left(x+\frac{\pi}{4}\right)$
(iii) $h(x)=\cos ^{2} 2 x$
(iv) $\phi(x)=2 \cos \left(x-\frac{\pi}{6}\right)$
(v) $\Psi(x)=\cos 3 x$
(vi) $u(x)=\cos ^{2} \frac{x}{2}$
(vii) $f(x)=\cos \pi x$
(viii) $g(x)=\cos 2 \pi x$
(i) $y=\cos \left(\mathrm{x}-\frac{\pi}{4}\right)$
$\Rightarrow y-0=\cos \left(x-\frac{\pi}{4}\right)$ ...(i)
On shifting the origin at $\left(\frac{\pi}{4}, 0\right)$, we get :
$\mathrm{x}=\mathrm{X}+\frac{\pi}{4}$ and $\mathrm{y}=\mathrm{Y}+0$
On subsititut $i n g$ the values in (i) we get:
$\mathrm{Y}=\cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{4}$ to the right.
Then, we obtain the following graph:.png)
(ii) $y=\cos \left(\mathrm{x}+\frac{\pi}{4}\right)$
$\Rightarrow y-0=\cos \left(x+\frac{\pi}{4}\right)$ ...(i)
On shifting the origin at $\left(-\frac{\pi}{4}, 0\right)$, we get:
$\mathrm{x}=\mathrm{X}-\frac{\pi}{4}$ and $\mathrm{y}=\mathrm{Y}+0$
On subsitituting the values in $(\mathrm{i})$, we get:
$\mathrm{Y}=\cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{4}$ to the left.
Then, we obtain the following graph:.png)
(iii) $y=\cos ^{2} 2 x$
The following graph is:
.png)
(iv) $y=2 \cos \left(x-\frac{\pi}{6}\right)$
$\Rightarrow y-0=2 \cos \left(x-\frac{\pi}{6}\right)$ ..(i)
On shifting the origin at $\left(\frac{\pi}{6}, 0\right)$, we get:
$x=X+\frac{\pi}{6}$ and $y=Y+0$
On subsitituting the values in $(\mathrm{i})$, we get:
$Y=2 \cos X$
Then, we draw the graph of $Y=\cos X$ and shift it by $\frac{\pi}{6}$ to the right.
Then, we obtain the following graph:
.png)
(v) $y=\cos 3 x$
The following graph is:
.png)
(vi) $y=\cos ^{2} \frac{x}{2}$
The following graph is:
.png)
(vii) $y=\cos \pi x$
The following graph is:
.png)
(viii)$y=\cos 2 \pi x$
The following graph is:
.png)