2 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).
- Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) 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.
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\caption{Fig. 8.1}
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(A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
(B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
(C) Use this result to find an exact expression for the perimeter of the polygon.
Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\). - 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.
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\includegraphics[alt={},max width=\textwidth]{54a69773-651f-4e2f-9a3c-06ea7c07098b-2_450_420_1562_938}
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\caption{Fig. 8.2}
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(A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
(B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\).
Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
(C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\). - 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.