Draw a (fairly large) circle on a blank sheet of paper. We’ll think of this as a unit circle.
Divide the unit circle into \(2^2 = 4\) equal wedges each with its vertex at the center of the circle \(O\). On each wedge, call the
two corners of the wedge that lie on the circle \(A\) and \(B_2\). Let \(\mathcal {A}_2\) denote the area of the triangle \(\triangle OAB_2\) and let \(\theta _2\) denote the
measure of the angle at \(O\). Explain how to estimate the area of the circle with triangle \(\triangle OAB_2\). What is your estimate?
Divide the unit circle into \(2^3 = 8\) equal wedges each with its vertex at the center of the circle \(O\). On each wedge, call the
two corners of the wedge that lie on the circle \(A\) and \(B_3\). Let \(\mathcal {A}_3\) denote the area of the triangle \(\triangle OAB_3\) and let \(\theta _3\) denote the
measure of the angle at \(O\). Explain how to estimate the area of the circle with triangle \(\triangle OAB_3\). What information do you
need to know to actually do this computation?
Given an angle \(\theta \), explain the relation of \(\sin (\theta )\) and \(\cos (\theta )\) to the unit circle. How could these values help with the calculation
described above?
Divide the unit circle into \(2^n\) equal wedges each with its vertex at the center of the circle \(O\). On each wedge, call
the two corners of the wedge that lie on the circle \(A\) and \(B_n\). Let \(\mathcal {A}_n\) denote the area of the triangle \(\triangle OAB_n\) and
let \(\theta _n\) denote the measure of the angle at \(O\). Explain why someone would be interested in the value of:
\[ \sin \left (\frac {\theta _n}{2}\right ) \]
Recalling that:
\[ \sin \left (\frac {\theta }{2}\right ) = \sqrt {\frac {1-\cos (\theta )}{2}} \qquad \text {and}\qquad \cos (\theta )^2 + \sin (\theta )^2 = 1 \]
Explain why: \[ 2 \mathcal {A}_{n+1} = \sqrt {\frac {1 - \sqrt {1 - (2\mathcal {A}_n)^2}}{2}} \]
Let’s fill out the following table (a calculator will help!):
2025-01-06 15:56:25 \[ \begin{array}{| c || c | c | c | c | c |} \hline n & \mathcal {A}_n & \text {Approx. Area} & \sqrt {1-(2\mathcal {A}_n)^2} & \frac {1 - \sqrt {1-(2\mathcal {A}_n)^2} }{2} &\rule [0mm]{0mm}{6mm}2\mathcal {A}_{n+1} =\sqrt {\frac {1 - \sqrt {1-(2\mathcal {A}_n)^2} }{2}} \\ \hline \hline 2 &\rule [7mm]{20mm}{0mm}\hspace {20mm} &\hspace {20mm} &\hspace {20mm} & \hspace {20mm} & \hspace {20mm}\\ \hline 3 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline 4 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline 5 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline 6 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline 7 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline 8 &\rule [0mm]{0mm}{7mm} & & & & \\ \hline \end{array} \]
What do you notice?