The value of c In Lagrange’s theorem for the following for the function f(x)=log sinx  in the interval [π/6,5π/6]

f(5π6)=log sin(5π6)=log sinπ6                =log12=log2, f(π6)=logsinπ6=−log2.  f'(c)=1sinxcosx=cotx By Lagrange’s mean value theorem,f(5π/6)−f(π/6)(5π/6)(π/6)=cotc ⇒cotc=0⇒c=π2 Thus, c=π2∈(π/6,5π/6).