The value of c In Lagrange’s theorem for the following for the function $f\left(x\right)=\log \sin x$  in the interval $\left[\pi /6,5\pi /6\right]$

$f\left(\frac{5\pi }{6}\right)=\log \sin \left(\frac{5\pi }{6}\right)=\log \sin \frac{\pi }{6}$

               $=\log \frac{1}{2}=\log 2,$

$f\left(\frac{\pi }{6}\right)=\log \sin \frac{\pi }{6}=-\log 2.$

$f^{\text{'}}\left(c\right)=\frac{1}{\sin x}\cos x=\cot x$

By Lagrange’s mean value theorem,

$\frac{f\left(5\pi /6\right)-f\left(\pi /6\right)}{\left(5\pi /6\right)\left(\pi /6\right)}=\cot c$

$\Rightarrow \cot c=0\Rightarrow c=\frac{\pi }{2}$

Thus, $c=\frac{\pi }{2}\in \left(\pi /6,5\pi /6\right).$