Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for $\pi$.

\begin{enumerate}[label=(\roman*)]
\item Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre O. AB is one of the sides of the polygon. C is the midpoint of AB. Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon.

\includegraphics{figure_1}

\begin{enumerate}[label=(\Alph*)]
\item Show that $\text{AB} = 2 \sin 15°$. [2]

\item Use a double angle formula to express $\cos 30°$ in terms of $\sin 15°$. Using the exact value of $\cos 30°$, show that $\sin 15° = \frac{1}{2}\sqrt{2 - \sqrt{3}}$. [4]

\item Use this result to find an exact expression for the perimeter of the polygon.

Hence show that $\pi > 6\sqrt{2 - \sqrt{3}}$. [2]
\end{enumerate}

\item In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon.

\includegraphics{figure_2}

\begin{enumerate}[label=(\Alph*)]
\item Show that $\text{DE} = 2 \tan 15°$. [2]

\item Let $t = \tan 15°$. Use a double angle formula to express $\tan 30°$ in terms of $t$.

Hence show that $t^2 + 2\sqrt{3}t - 1 = 0$. [3]

\item Solve this equation, and hence show that $\pi < 12(2 - \sqrt{3})$. [4]
\end{enumerate}

\item Use the results in parts (i)(C) and (ii)(C) to establish upper and lower bounds for the value of $\pi$, giving your answers in decimal form. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q2 [19]}}