Answer
$\kappa \left( x \right) = \frac{{\left| { - \sin x} \right|}}{{{{\left( {1 + {{\cos }^2}x} \right)}^{3/2}}}}$
Work Step by Step
We have $y = \sin x$.
Write $y = f\left( x \right) = \sin x$. So,
$f'\left( x \right) = \cos x$ and $f{\rm{''}}\left( x \right) = - \sin x$.
By Eq. (5) of Theorem 2, the curvature is given by
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( x \right) = \frac{{\left| { - \sin x} \right|}}{{{{\left( {1 + {{\cos }^2}x} \right)}^{3/2}}}}$
Using a computer algebra system we plot the curvature and the graph is shown in the figure attached. From the graph we see that the maximum of the curvature is at $x = \frac{\pi }{2}$ and $\frac{{3\pi }}{2}$.
We notice that the maximum of the numerator of $\kappa \left( x \right)$ occurs at $x = \frac{\pi }{2}$ and $\frac{{3\pi }}{2}$. While at $x = \frac{\pi }{2}$ and $\frac{{3\pi }}{2}$, the denominator of $\kappa \left( x \right)$ has its minimum. Therefore, we conclude that the maximum of the curvature occurs at $x = \frac{\pi }{2}$ and $\frac{{3\pi }}{2}$.
