# What is the frequency of f(theta)= sin 3 t - cos 6 t ?

Feb 16, 2016

Frequency is $\frac{3}{2 \pi}$

#### Explanation:

A function in$\theta$ must have $\theta$ in RHS. It is assumed that function is $f \left(t\right) = \sin \left(3 t\right) - \cos \left(6 t\right)$

To find period (or frequency, which is nothing but inverse of period) of the function, we first need to find whether the function is periodic. For this, the ratio of the two related frequencies should be a rational number, and as it is $\frac{3}{6}$, the function $f \left(t\right) = \sin \left(3 t\right) - \cos \left(6 t\right)$ is a periodic function.

The period of $\sin \left(3 t\right)$ is $2 \frac{\pi}{3}$ and that of $\cos \left(6 t\right)$ is $2 \frac{\pi}{6}$

Hence, period of function is $2 \frac{\pi}{3}$ (for this we have to take LCM of two fractions $\frac{2 \pi}{3}$ and $\frac{2 \pi}{6}$, which is given by LCM of numerator divided by GCD of denominator).

Frequency being inverse of period is $\frac{3}{2 \pi}$