Chapter 5.4 Measurement of a Circle
In this section, we talk about an approximation scheme we can use to get the circumference and area of the unit circle. It is here where we combine sequences and polygons to get these results.
Length and Area
The principle of finite approximation can help us show that the length of an arc and the area of a sector arises from approximation procedures.
This is an arc, \(\mathcal{A}\), of a circle from point \(p\) to point \(q\):
This is a sector of the disk that is associated to \(\mathcal{A}\).
Take \(g_3(m)\) to be the regular inscribed polygon with \(3\cdot 2^m\) sides, where \(m\) is in \(\mathbb N_0\).
Specifically, \(g_3(m)\) is the polygon with ordered vertex set \(((1,0), p, p^2, \dots, p^{3\cdot 2^m-1})\), where \(p\) is in \(\mathcal{C}\) and \[\mathrm{Frac}(p) = \frac{1}{3\cdot 2^m}.\]
Take \(G_3(m)\) to be a regular circumscribed polygon with \(3\cdot2^m\) sides, where \(m\) is in \(\mathbb N_0\).
This means that all edge lengths of \(G_3(m)\) are equal and the midpoints of each edge of \(G_3(m)\) is a vertex of \(g_3(m)\).
Here are the inscribed polygons \(g_3(0)\), \(g_3(1)\), and \(g_3(2)\)
Here are the circumscribed polygons \(G_3(0)\), \(G_3(1)\), and \(G_3(2)\)
This is the region between \(g_3(0)\) and \(G_3(0)\)
These are the respective regions between \(g_3(1)\) and \(G_3(1)\), and \(g_3(2)\) and \(G_3(2)\):
Notice that the region that is between the inscribed and circumscribed polygons gets small very quickly with increasing \(m\).
Take \(\Lambda(g_3(m))\) and \(\Lambda(G_3(m))\) to be perimeters of \(g_3(m)\) and \(G_3(m)\).
Take \(\mathcal{A}(g_3(m))\), and \(\mathcal{A}(G_3(m))\) to be the areas of \(g_3(m)\) and \(G_3(m)\).
It can be shown that \(\displaystyle \lim_{m\to\infty}\Lambda(g_3(m))=\displaystyle \lim_{m\to\infty}\Lambda(G_3(m))=2\pi r\) and \(\displaystyle \lim_{m\to\infty}\mathcal{A}(g_3(m))=\displaystyle \lim_{m\to\infty}\mathcal{A}(G_3(m))=\pi r^2\), where \(r\) is the radius of the circle.
Limits Involving the Trigonometric Functions
Now we will learn about using the approximation scheme to compute two important limits involving sine and cosine.
Take \((\theta_n)\) to be a positive null sequence in \((0, \frac{\pi}{4})\), meaning that \(\displaystyle \lim_{n\to\infty}\theta_n=0\).
For each \(n\), since the arc from \((1,0)\) to the point \(p_n\) that has length \(\theta_n\) determines a sector with area equal to \(\frac{\theta_n}{2}\), the figure below shows that \[\tfrac{1}{2}\sin(\theta_n)\cos(\theta_n) < \tfrac{\theta_n}{2} < \tfrac{1}{2}\tan(\theta_n).\]
Rearrange terms to obtain the inequalities \[\cos(\theta_n) < \tfrac{\theta_n}{\sin(\theta_n)} < \tfrac{1}{\cos(\theta_n)}.\]
Since \((\theta_n)\) is a positive null sequence, \((\cos(\theta_n))\) tends to \(1\) as \(n\) tends to \(\infty\), therefore the squeeze theorem and the limit law for quotients together imply that \[\lim_{n\to\infty} \tfrac{\sin(\theta_n)}{\theta_n} = 1.\]
Note that \(\sin\) is an odd function, so for any null sequence \((\theta_n)\) that has only nonzero terms, \[\tfrac{\sin(\theta_n)}{\theta_n} = \tfrac{\sin(|\theta_n|)}{|\theta_n|},\] and so for any null sequence \((\theta_n)\) that has only nonzero terms, \[\lim_{n\to\infty} \tfrac{\sin(\theta_n)}{\theta_n} = 1.\]
Use this fact to compute the following limit.
Example 1
Suppose that \((\theta_n)\) is a null sequence with only nonzero terms. Calculate \[\lim_{n\to\infty} \tfrac{\sin(3\theta_n)}{\theta_n}.\]
Notice that \(\tfrac{\sin(3\theta_n)}{\theta_n}=\tfrac{\sin(3\theta_n)}{\theta_n}\cdot\tfrac{3}{3}=3\cdot \tfrac{\sin(3\theta_n)}{3\theta_n}.\)
Since \(\theta_n\) is a null sequence with only nonzero terms, \(3\theta_n\) is a still a null sequence.
Hence \(\displaystyle\lim_{n\to\infty}\tfrac{\sin(3\theta_n)}{3\theta_n}=1.\) Thus
\[\lim_{n\to\infty} \tfrac{\sin(3\theta_n)}{\theta_n}=\lim_{n\to\infty}3\cdot \tfrac{\sin(3\theta_n)}{3\theta_n}=\lim_{n\to\infty}3\cdot \lim_{n\to\infty}\tfrac{\sin(3\theta_n)}{3\theta_n}=3\cdot 1=3. \]
Here is another example of a well-known limit.
Example 2
Suppose that \((\theta_n)\) is a null sequence with only nonzero terms. Calculate \[\lim_{n\to\infty} \tfrac{1-\cos(\theta_n)}{\theta_n}.\]
Notice that \[ \begin{align*} \tfrac{1-\cos(\theta_n)}{\theta_n}&=\tfrac{1-\cos(\theta_n)}{\theta_n}\cdot \tfrac{1+\cos(\theta_n)}{1+\cos(\theta_n)}\\&=\tfrac{1-\cos^2(\theta_n)}{\theta_n}\cdot \tfrac{1}{1+\cos(\theta_n)}\\ &=\tfrac{\sin^2(\theta_n)}{\theta_n}\cdot \tfrac{1}{1+\cos(\theta_n)}\\ &=\tfrac{\sin(\theta_n)}{\theta_n}\cdot \tfrac{\sin(\theta_n)}{1+\cos(\theta_n)}.\\ \end{align*} \]
Hence \[ \begin{align*} \lim_{n\to\infty}\tfrac{1-\cos(\theta_n)}{\theta_n}&=\lim_{n\to\infty}\tfrac{\sin(\theta_n)}{\theta_n}\cdot \tfrac{\sin(\theta_n)}{1+\cos(\theta_n)}\\ &=\lim_{n\to\infty}\tfrac{\sin(\theta_n)}{\theta_n}\cdot \lim_{n\to\infty}\tfrac{\sin(\theta_n)}{1+\cos(\theta_n)}\\ &=1\cdot \tfrac{0}{1+1}\\ &=1\cdot 0\\ &=0. \end{align*} \]
It is certainly possible to define the cosine and sine functions with respect to different angle measures.
For instance, take \(\sin_d(\theta)\) to give the \(y\)-coordinate of the point on \(\mathcal{C}\) whose degree measure is \(\theta\).
In this case, if \((\theta_n)\) is a null sequence with nonzero terms, then it will no longer be the case that \(\Big(\frac{\sin_d(\theta_n)}{\theta_n}\Big)\) converges to \(1\).
We will soon see why it is nice to have angles given by radian measures.
Example 3
Suppose that \((\theta_n)\) is a null sequence with only nonzero terms. Calculate \[\lim_{n\to\infty} \tfrac{\sin_d(\theta_n)}{\theta_n}.\]
Note that \(\sin_d(\theta_n)=\sin\left(2\pi \cdot \frac{\theta_n}{360}\right)=\sin\left(\theta_n\cdot\frac{\pi}{180}\right)\).
Since \(\theta_n\) is a null sequence with only nonzero terms, \(\theta_n\cdot\frac{\pi}{180}\) is a still a null sequence.
Hence \[ \begin{align*}\lim_{n\to\infty} \tfrac{\sin_d(\theta_n)}{\theta_n}&=\lim_{n\to\infty} \tfrac{\sin\left(\theta_n\cdot\frac{\pi}{180}\right)}{\theta_n}\\ &=\lim_{n\to\infty}\frac{\pi}{180}\cdot \lim_{n\to\infty}\tfrac{\sin(\theta_n\cdot\frac{\pi}{180})}{\frac{\pi}{180}\theta_n}\\ &=\frac{\pi}{180}\cdot 1\\ &=\frac{\pi}{180} \end{align*} \]