# What is the period of f(t)=cos ( ( 9 t ) / 2 ) ?

$\frac{4}{9} \pi$
Using $\cos \left(k \left(t + \frac{2 \pi}{k}\right)\right) = \cos \left(k t + 2 \pi\right) = \cos \left(k t\right)$, we find that the period of cos ( kt ) is $\frac{2 \pi}{k}$.
Here, k = $\frac{9}{2}$, and so, $\frac{2 \pi}{k} = \frac{4}{9} \pi$