Question

Arrange the following function in ascending order of their fundamental period.

• A. $\cos \left(3x-\frac{\pi }{2}\right)-2$
• B. $2\cos x$
• C. $\cos 4(x+\pi )$
• D. $6-\cos 2\left(\frac{x}{4}+\pi \right)$

Answer 1

We know $\cos x$ has a period $2\pi .$
Now, we know if a function $f\left(x\right)$ has a fundamental period $T$ and it is transformed to $af(bx+c)+d,$ then the period of this function is $\frac{T}{|b|}.$

Using this if $f\left(x\right)=\cos x,$ then $2\cos x=2f\left(x\right)$
Hence the period will remain the same i.e. $2\pi .$

Now, for $\cos 4(x+\pi )$ it can be expressed as $f(bx+c),$ where $b=4\&c=\pi$
Hence, it's period will be $\frac{2\pi }{4}=\frac{\pi }{2}$

Similarly, for $\cos (3x-\frac{\pi }{2})-2$ it can be expressed as: $f(bx+c)+d,$ where $b=3,c=-\frac{\pi }{2}\&d=-2$

Hence, it's period will be $\frac{2\pi }{3}.$

Now for the last transformation as $6-\cos 2(\frac{x}{4}+\pi ),$ it can be written as $d+af(bx+c),$ where $d=6,a=-1,b=\frac{1}{2},c=\pi$
Thus, it's period will be $\frac{2\pi }{1/2}=4\pi$
Thus, in ascending order of fundamental period of trigonometric functions, we can arrange them as:
$\cos 4(x+\pi ),\cos \left(3x-\frac{\pi }{2}\right)-2,2\cos x,6-\cos 2\left(\frac{x}{4}+\pi \right)$