Chapter 5.3 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.
Fractions of a Circle
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.
To begin, we need to define some things.
This is an arc, \(\mathcal{A}\), of a circle from point \(p\) to point \(q\):
We learned about this before in Chapter 2: it is the angle \(\angle pOq\).
This is a sector of the disk that is associated to \(\mathcal{A}\).
The goal of this first subsection is to
- Show that circles are simple closed curves. Simple closed curves, by the Jordan Curve Theorem, have a well-defined area.
- Establish an algebraic notation of a fraction of a circle in order to define an arc for all real numbers in \([0,1)\).
To begin, \(\mathcal{C}\), the unit circle, is a simple closed curve because it has the following parameterization which is a simple, closed path:
\[c(t)=\begin{cases}(1-t,\sqrt{1-(t-1)^2})&\text{if }0\leq t\leq 2\\ (t-3,-\sqrt{1-(t-3)^2})&\text{if }2<t\leq 4.\end{cases}\]
This is known as the square root parameterization of \(\mathcal{C}.\)
This comes from using the equation \[x(t)^2+y(t)^2=1\] and solving for \(y\) to get
\[y(t)=\pm\sqrt{1-x(t)^2}.\]
For the above half of the circle, take \(x(t)=1-t\) to get \[c(t)=(1-t,\sqrt{1-(1-t)^2})\quad \text{for }0\leq t\leq 2.\]
For the bottom half of the circle. take \(x(t)=t-3\) to get \[c(t)=(t-3,\sqrt{1-(t-3)^2})\quad \text{for }2< t\leq 4.\]
What about any circle \(\mathcal{C}_r(a,b)\), with radius \(r\) and center \((a,b)\)? A parameterization of \(\mathcal{C}_r(a,b)\) can be written like this: \[\gamma=\langle a,b\rangle +\sigma_r\circ c.\] This is a simple, closed path so, any circle is a simple closed curve.
In this example, write out the parameterization.
Example 1
Take \((a,b)\) to be a point in \(\mathbb{R}^2\) and take \(r\) to be a positive real number. Simulate the position of a particle whose position at time \(t\) is \(\gamma\), where \(\gamma\) is the function given above that parameterizes the circle \(\mathcal{C}_r(a,b)\). How does the position of the particle change with respect to time near the coordinate axes?
The parameterization looks like this: \[ \begin{align*} \gamma(t)&=\langle a,b\rangle+\sigma_r(c(t))\\ &=\begin{cases} (r(1-t)+a,r\sqrt{1-(1-t)^2}+b)&\text{if }0\leq t<2\\ (r(t-3)+a,-r\sqrt{1-(t-3)^2}+b)&\text{if }2< t\leq 4. \end{cases} \end{align*} \]
This is a simulation of the parameterization.
Near the \(y\)-axis, the position of the particle tends to slow down. However, near the \(x\)-axis, the position of the particle tends to speed up.
Now that we know area is defined for a circle, we next want to be able define an arc relative to a fraction of a circle.
Specifically, to say that an arc \(\mathcal{A}\) of the unit circle \(\mathcal{C}\) is a certain fraction \(s\) of \(\mathcal{C}\) should mean that
- the ratio of the length of \(\mathcal{A}\) to the circumference of \(\mathcal{C}\) is equal to \(s\);
- the ratio of the area of the sector defined by \(\mathcal{A}\) to the area that is bounded by \(\mathcal{C}\) is equal to \(s.\)
To properly define these conditions, we will algebraically define what we mean a fraction of a circle.
For any point \(p\) in \(\mathcal{C}\), Take \(\mathcal{A}_p\) to be the arc from \((1,0)\) to \(p\) and recall that \(p^n\) is defined as
\[ p^{n}=\underbrace{p\star\cdots\star p}_{n\text{-times}}, \] where \(\star\) is the rotation operation we defined back in Chapter 2.
For each \(i\) in the set \(\{0,1,2\dots,n\}\) there is a \(t_i\) in the interval \([0,4]\) so that
\[c(t_i)=p^i,\quad \text{where}\quad p^0=(1,0).\]
If \((t_0,\dots,t_n)\) is an epoch in \([0,4]\) and \(p^n\) is equal to \(p^n\) is equal to \((1,0)\), then the fraction of \(\mathcal{C}\) that is given by \(\mathcal{A}_p\), denoted by \(\mathrm{Frac}(p)\), is the quantity
\[\mathrm{Fac}(p)=\frac{1}{n}.\]
Essentially, the fraction of \(p\) is the smallest natural number \(n\) so that \(p^n\) equals \((1,0)\).
Practice with this example.
Example 2
Show that \[ \mathrm{Frac}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{1}{3} \]
and use this to determine the unique point \(p\) on \(\mathcal{C}\) so that \(\mathrm{Frac}(p)\) is equal to \(\frac{1}{6}.\) Describe the polygon that has \(p\) as its first vertex that is counterclockwise from \((1,0).\)
To show that the \(\mathrm{Frac}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) is equal to \(\frac{1}{3}\), calculate \(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)^3\):
\[\begin{align*} \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)&=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\\ &=(1,0). \end{align*}\]
Therefore, \(\mathrm{Frac}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{1}{3}.\)
To determine a point \(p\) with fraction equal to \(\frac{1}{6}\), identify the midpoint \(q\) of the line segment \(\overline{(1,0)\left(-\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right)}:\)
\[q=\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right).\]
Now project \(q\) onto the unit circle to get
\[p=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right).\]
Notice that \[p^2=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right),\]
and so
\[p^6=\left(-\frac{1}{2},\sqrt{3}{2}\right)^3=(1,0)\]
Thus \(\mathrm{Frac}\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\frac{1}{6}.\)
\[\begin{align*} \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)&=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\\ &=(1,0). \end{align*}\]
The polygon that has \(p\) as its first vertex that is counterclockwise from \((1,0)\) is a regular hexagon.
In Example 2, we obtained a regular polygon. Does the above process always produce a regular polygon? The answer is yes.
That is, for any natural numbers \(m\) and \(n\) greater than or equal to \(3\), \(n>m\), and \(p\) in \(\mathcal{C}\) with \(\mathrm{Frac}(p)=\frac{1}{n}\), the length of the line segment \(\overline{p^mp^{m+1}}\) is equal to the length of the line segment \(\overline{(1,0)p}.\)
Thus, the oridered vertex set \(((1,0),p,p^2,\dots,p^{n-1})\) is the positively oriented vertex set of a polygon \(g(n).\) In fact, \(g(n)\) is a regular inscribed polygon. That is, all the vertices of \(g(n)\) are in \(\mathcal{C}\) and all the edges of \(g(n)\) have the same length.
If \(q\) is the \(m^{\text{th}}\) vertex of \(g(n)\), then define \(\mathrm{Frac}(q)\) by \[\mathrm{Frac}(q)=\frac{m}{n}.\]
Example 3
The polygons given below are regular inscribed polygons.
Label each vertex by the fraction of the circle that is associated to it.
Describe each regular inscribed polygon \(g(n)\) for which the given vertex is the first vertex of \(g(n)\) that is counterclockwise from \((1,0)\).
- Here are the vertices for each polygon
- The regular polygons are a equilaterial triangle, square, regular hexagon, and a regular dodecagon, respectively.
The next thing we discuss is how \(\mathrm{Frac}\) interacts with rotation. First, for any \(q_1\) and \(q_2\) in \(\mathcal{C}\) with
\[\mathrm{Frac}(q_1)=\frac{m1}{n1}\quad\text{and}\quad\mathrm{Frac}(q_2)=\frac{m_2}{n_2},\] \(q_1\) and \(q_2\) are, respectively, the \(m_1n2\) and \(m_2n_1\) vertex of \(g(n_1n_2)\) from \((1,0).\)
Here are some statements that follow from the fact that \(g(n_1n_2)\) has a first vertex \(p\) that is counterclockwise from \((1,0)\) and \(((1,0),p,\dots,p^{n1n_2-1})\) is the positively oriented vertex set of \(g(n_1n_2)\):
- The ordered vertex set \(((1,0),q_1,q_2)\) is positively oriented if and only if \[\mathrm{Frac}(q_1)<\mathrm{Frac}(q_2),\]
- The function \(\mathrm{Frac}\) is additive in the sense that \[\mathrm{Frac}(q_1\star q_2)=\begin{cases} \mathrm{Frac}(q_1)+\mathrm{Frac}(q_2)&\text{if }\mathrm{Frac}(q_1)+\mathrm{Frac}(q_2)\in [0,1)\\ \mathrm{Frac}(q_1)+\mathrm{Frac}(q_2)-1&\text{if }\mathrm{Frac}(q_1)+\mathrm{Frac}(q_2)\in [1,2). \end{cases}\]
Here is an example to practice using the formula.
Example 4
Take \(p\) and \(q\) to be points in \(\mathcal{C}\) with \[\mathrm{Frac}(p)=\frac{5}{6}\quad\text{and}\quad\mathrm{Frac}(q)=\frac{1}{3}.\] Evaluate \(\mathrm{Frac}(p\star q)\) by using the additivity of \(\mathrm{Frac}\) as well as by finding \(p\) and \(q\), and computing \(p\star q\) directly.
First, use the additivity of \(\mathrm{Frac}\) to evaluate \(\mathrm{Frac}(p\star q)\). Because \(\mathrm{Frac}(p)+\mathrm{Frac}(q)>1\), we have that \[\mathrm{Frac}(p\star q)=\frac{5}{6}+\frac{1}{3}-1=\frac{1}{6}.\]
Now, evaluate \(\mathrm{Frac}(p\star q)\) by finding \(q\) and \(p\) directly. Because \(q\) is \(\frac{1}{3}\) of \(\mathcal{C}\) and \(p\) is \(\frac{5}{6}\) of \(\mathcal{C}\), their radian angle measure are \(\frac{2\pi}{3}\) and \(\frac{5\pi}{3}\) respectively. Therefore, \(q=\left(-\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right)\) and \(p=\left(\tfrac{1}{2},-\tfrac{\sqrt{3}}{2}\right)\), and so \[p\star q=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right).\] To determine its \(\mathrm{Frac}\), notice that \((p\star q)^2=q\) and since \(\mathrm{Frac}(q)=\frac{1}{3}\), we have that \[\mathrm{Frac}(p\star q)=\frac{1}{2}\mathrm{Frac}(q)=\frac{1}{6}.\]
Now that we have defined fraction of a circle algebraically, we now want to define an approximation scheme for the area and circumference of the unit circle.
In order to create an approximation scheme, we will work with regular inscribed polygons.
For any natural number \(n\), that is at least \(3\), a regular refinement of a regular inscribed \(n\)-gon is a regular inscribed \(2n\)-gon.
For any \(p\) in the first quadrant that is not on a coordinate axis, take \(q_m(p)\) and \(Q_m(p)\) to be the adjacent vertices of \(g(n2^m)\) so that \((q_m(p),p,Q_m(p))\) is positively oriented and \(m\) is large enough so that both \(q_m(p)\) and \(Q_m(p)\) are also in the first quadrant.
The sequences \((q_m(p))\) and \((Q_m(p))\) are, respectively, increasing and decreasing sequences (with respect to the ordering on \(\mathcal{C}\)).
To see how this works, look at the following example.
Example 5
Take \((a,b)\) to be the first vertex of \(g(n2^m)\) that is counterclockwise from \((1,0)\) and take \((c,d)\) to be the first vertex of \(g(n2^{m+1})\) that is counterclockwise from \((1,0).\) Write \(c\) and \(d\) in terms of \(a\) and \(b\) and compare \(\mathrm{Frac}(a,b)\) and \(\mathrm{Frac}(c,d).\)
The midpoint of \(\overline{(1,0)(a,b)}\) is \(p\), where \(p=\left(\tfrac{a+1}{2},\tfrac{b}{2}\right).\)
Take \(V=p-(0,0)=\langle \tfrac{a+1}{2},\tfrac{b}{2}\rangle\) so that we can project \(p\) onto the unit circle to get \((c,d)\):
\[(c,d)=\hat{V}+(0,0)=\frac{1}{\sqrt{\left(\frac{a+1}{2}\right)^2+\left(\frac{b}{2}\right)^2}}\langle \tfrac{a+1}{2},\tfrac{b}{2}\rangle+(0,0).\]
To simplify, notice that \((a,b)\) in \(\mathcal{C}\) so \(a^2+b^2=1\). Thus, \[(1+a)^2+b^2=1+a^2+b^2+2a=2(a+1)\] which means
\[\begin{align*} \left(\frac{a+1}{2}\right)^2+\left(\frac{b}{2}\right)^2&=\frac{(1+a)^2+b^2}{4}\\ &=\frac{2(a+1)}{4}\\ &=\frac{a+1}{2}. \end{align*}\]
So
\[\begin{align*} (c,d)&=\frac{1}{\sqrt{\left(\frac{a+1}{2}\right)^2+\left(\frac{b}{2}\right)^2}}\left(\tfrac{a+1}{2},\tfrac{b}{2}\right)\\ &=\frac{1}{\sqrt{\frac{a+1}{2}}}\left(\tfrac{a+1}{2},\tfrac{b}{2}\right)\\ &=\tfrac{\sqrt{2}}{\sqrt{a+1}}\left(\tfrac{a+1}{2},\tfrac{b}{2}\right)\\ &=\left(\tfrac{\sqrt{1+a}}{\sqrt{2}},\tfrac{b}{\sqrt{2}\sqrt{a+1}}\right)\\ &=\left(\tfrac{\sqrt{1+a}}{\sqrt{2}},\tfrac{\sqrt{1-a^2}}{\sqrt{2}\sqrt{a+1}}\right) &&\text{since }b=\sqrt{1-a^2}\\ &=\left(\tfrac{\sqrt{1+a}}{\sqrt{2}},\tfrac{\sqrt{1-a}}{\sqrt{2}}\right). \end{align*}\]
Notice that \((c,d)\star(c,d)=(a,b)\) so \(\mathrm{Frac}(c,d)=\tfrac{1}{2}\mathrm{Frac}(a,b).\)
Length and Area
Now we are ready to describe an approximation scheme.
Take \(g_3(m)=g(3\cdot 2^m)\) 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
What we learned from the approximation scheme is that for any point \(p\) on \(\mathcal{C}\), the length of the arc from \((1,0)\) to \(p\) is equal to \(2\pi \mathrm{Frac}(p)\) and the area of the sector that is given by that arc will be \(\pi\mathrm{Frac}(p).\)
Thefore, the trigonometric functions really are just functions of a length of arc on \(\mathcal{C}\).
Example 6
Take \(c\) to be the function that is given for each \(t\) in \(\mathbb{R}\) by \[c(t)=(\cos(t),\sin(t)).\] Simulate the position of a particle that has position \(c(t)\) at time \(t\). Discuss the ways in which the behavior of this particle differs from the one that is given by the square root parameterization of \(\mathcal{C}\).
This is a link to the simulation.
The speed is constant for this path. In time \(t\), the particle moves a distance \(t\).
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 \(p_n=(\cos(\theta_n),\sin(\theta_n))\) that has length \(\theta_n\) determines a sector with area equal to \(\frac{\theta_n}{2}\), the figure below shows that \[\tfrac{1}{2}\sin(\theta_n)\cos(\theta_n) < \tfrac{\theta_n}{2} < \tfrac{1}{2}\tan(\theta_n).\]
Note: \(q_n\) is the point on the line \(y=\tan(\theta_n)x\) with \(x\)-coordinate equal to \(1\), so \(q_n=(1,\theta_n).\)
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 7
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, \((\phi_n)=(3\theta_n)\) is a still a null sequence with only nonzero terms. Hence \(\lim\limits_{n\to\infty}\tfrac{\sin(\phi_n)}{\phi_n}=1.\)
Thus
\[\lim_{n\to\infty} \tfrac{\sin(3\theta_n)}{\theta_n}=\lim_{n\to\infty}3\cdot \tfrac{\sin(3\theta_n)}{3\theta_n}=\lim_{n\to\infty}3\cdot \lim_{n\to\infty}\tfrac{\sin(\phi_n)}{\phi_n}=3\cdot 1=3. \]
Here is another example of a well-known limit.
Example 8
Suppose that \((\theta_n)\) is a null sequence with only nonzero terms. Calculate \[\lim_{n\to\infty} \tfrac{1-\cos(\theta_n)}{\theta_n}.\]
Rewrite the expression like this: \[\begin{align*} \tfrac{1-\cos(\theta_n)}{\theta_n}&=\tfrac{1-\cos(\theta_n)}{\theta_n}\cdot \tfrac{1+\cos(\theta_n)}{1+\cos(\theta_n)}\\&=\tfrac{1-\cos^2(\theta_n)}{\theta_n}\cdot \tfrac{1}{1+\cos(\theta_n)}&&\text{since }\sin^2(\theta)+\cos^2(\theta)=1\text{ implies }\cos^2(\theta)=1-\sin^2(\theta)\\ &=\tfrac{\sin^2(\theta_n)}{\theta_n}\cdot \tfrac{1}{1+\cos(\theta_n)}\\ &=\tfrac{\sin(\theta_n)}{\theta_n}\cdot \tfrac{\sin(\theta_n)}{1+\cos(\theta_n)}.\\ \end{align*}\]
This rewriting together with the equalities
\[\lim_{n\to\infty}(1+\cos(\theta_n))=1+\lim_{n\to\infty}\cos(\theta_n)=1\]
and
\[\lim_{n\to\infty}\sin(\theta_n)=0\]
imply that \[ \begin{align*} \lim_{n\to\infty}\tfrac{1-\cos(\theta_n)}{\theta_n}&=\lim_{n\to\infty}\tfrac{\sin(\theta_n)}{\theta_n}\cdot \tfrac{\sin(\theta_n)}{1+\cos(\theta_n)}\\ &=\lim_{n\to\infty}\tfrac{\sin(\theta_n)}{\theta_n}\cdot \tfrac{\lim\limits_{n\to\infty}\sin(\theta_n)}{\lim\limits_{n\to\infty}(1+\cos(\theta_n))}\\ &=1\cdot \tfrac{0}{1+1}\\ &=0. \end{align*} \]
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 9
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, \((\phi_n)=\left(\theta_n\cdot\frac{\pi}{180}\right)\) is a still a null sequence with only nonzero terms.
Hence \[ \begin{align*}\lim_{n\to\infty} \tfrac{\sin_d(\theta_n)}{\theta_n}&=\lim_{n\to\infty} \tfrac{\sin\left(\theta_n\cdot\frac{\pi}{180}\right)}{\theta_n}\\ &=\lim_{n\to\infty}\frac{\pi}{180}\cdot \lim_{n\to\infty}\tfrac{\sin(\theta_n\cdot\frac{\pi}{180})}{\theta_n\cdot\frac{\pi}{180}}\\ &=\lim_{n\to\infty}\frac{\pi}{180}\cdot \lim_{n\to\infty}\tfrac{\sin(\phi_n)}{\phi_n}\\ &=\frac{\pi}{180}\cdot 1\\ &=\frac{\pi}{180} \end{align*} \]