$a=A\sin \omega t_0$...$\left(1\right)$

$b=A\sin 2\omega t_0$...$\left(2\right)$

$c=A\sin 3\omega t_0$...$\left(3\right)$

adding equation $\left(1\right)$ and $\left(3\right)$

$a+c=A\left[\sin \omega t_0+\sin 3\omega t_0\right]=2A\sin 2\omega t_0\cos \omega t_0$...$\left(4\right)$

$\because \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$

from equation $\left(2\right)$ and $\left(4\right)$

$\frac{a+c}{b}=2\cos \omega t_0$

$\omega = \frac{1}{t_0}{\cos }^{-1}\left(\frac{a+c}{2b}\right) \Rightarrow f=\frac{1}{2\pi t_0}{\cos }^{-1}\left(\frac{a+c}{2b}\right)$