Answer
At $x = \frac{\pi }{2}$, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{\pi }{2},0} \right) + \left( {\cos t,\sin t} \right)$
Work Step by Step
We have $y = \sin x$. Write $f\left( x \right) = y = \sin x$. So, $f'\left( x \right) = \cos x$ and $f{\rm{''}}\left( x \right) = - \sin x$.
Step 1. Find the radius of the osculating circle
Applying Eq. (5), the curvature is
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( x \right) = \frac{{\left| { - \sin x} \right|}}{{{{\left( {1 + {{\cos }^2}x} \right)}^{3/2}}}} = \frac{{\sin x}}{{{{\left( {1 + {{\cos }^2}x} \right)}^{3/2}}}}$
Since $R = \frac{1}{\kappa }$, at $x = \frac{\pi }{2}$ the osculating circle has radius $R=1$.
Step 2. Find the normal vector
The curve can be parametrized by setting $x=t$. So, ${\bf{r}}\left( t \right) = \left( {t,\sin t} \right)$. The derivative is ${\bf{r}}'\left( t \right) = \left( {1,\cos t} \right)$.
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( {1,\cos t} \right)}}{{\sqrt {\left( {1,\cos t} \right)\cdot\left( {1,\cos t} \right)} }} = \left( {\frac{1}{{\sqrt {1 + {{\cos }^2}t} }},\frac{{\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}} \right)$
To find the derivative of ${\bf{T}}\left( t \right)$:
Evaluate $\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + {{\cos }^2}t} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + {{\cos }^2}t} }}} \right) = \frac{d}{{dt}}\left( {{{\left( {1 + {{\cos }^2}t} \right)}^{ - 1/2}}} \right)$
$\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {1 + {{\cos }^2}t} }}} \right) = - \frac{{ - 2\cos t\sin t}}{{2{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}} = \frac{{\cos t\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}$
Evaluate $\frac{d}{{dt}}\left( {\frac{{\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}} \right)$
$\frac{d}{{dt}}\left( {\frac{{\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}} \right) = \frac{{\left( { - \sin t} \right)\sqrt {1 + {{\cos }^2}t} - \left( {\cos t} \right)\left( {\frac{{ - 2\cos t\sin t}}{{2\sqrt {1 + {{\cos }^2}t} }}} \right)}}{{1 + {{\cos }^2}t}}$
$\frac{d}{{dt}}\left( {\frac{{\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}} \right) = \frac{{\left( { - \sin t} \right)\left( {1 + {{\cos }^2}t} \right) + {{\cos }^2}t\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}} = - \frac{{\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}$
Thus,
${\bf{T}}'\left( t \right) = \left( {\frac{{\cos t\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}, - \frac{{\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}} \right)$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\cos }^2}t{{\sin }^2}t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^3}}} + \frac{{{{\sin }^2}t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^3}}}$
$||{\bf{T}}'\left( t \right)|{|^2} = \frac{{{{\sin }^2}t\left( {1 + {{\cos }^2}t} \right)}}{{{{\left( {1 + {{\cos }^2}t} \right)}^3}}} = \frac{{{{\sin }^2}t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^2}}}$
$||{\bf{T}}'\left( t \right)|| = \frac{{\sin t}}{{1 + {{\cos }^2}t}}$
The normal vector at $t$ is
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \left( {\frac{{1 + {{\cos }^2}t}}{{\sin t}}} \right)\left( {\frac{{\cos t\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}, - \frac{{\sin t}}{{{{\left( {1 + {{\cos }^2}t} \right)}^{3/2}}}}} \right)$
${\bf{N}}\left( t \right) = \left( {\frac{{\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}, - \frac{1}{{\sqrt {1 + {{\cos }^2}t} }}} \right)$
Step 3. Find the the center of the osculating circle
By Eq. (9), the center of the osculating circle is given by
$\overrightarrow {OQ} = {\bf{r}}\left( {{t_0}} \right) + \frac{1}{{{\kappa _P}}}{\bf{N}}$
At $x = t = \frac{\pi }{2}$, the center of curvature is
$\overrightarrow {OQ} = {\bf{r}}\left( {\frac{\pi }{2}} \right) + {\bf{N}}\left( {\frac{\pi }{2}} \right)$
$\overrightarrow {OQ} = \left( {\frac{\pi }{2},1} \right) + \left( {0, - 1} \right)$
$\overrightarrow {OQ} = \left( {\frac{\pi }{2},0} \right)$
Step 4. Parametrize the osculating circle
At $x = t = \frac{\pi }{2}$, the osculating circle has radius $R=1$ and center $\left( {\frac{\pi }{2},0} \right)$. So, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{\pi }{2},0} \right) + \left( {\cos t,\sin t} \right)$
