Completely analyze \(G(t) = \frac {\sin (t)}{1 + \cos (t)}\)

\(\blacktriangleright \) Domain:

First, \(G\) has been described with a formula. Therefore, we are seeking a natural or implied domain. \(G\) is a quotient function (not a rational function), therefore our focus is on real numbers that make the denominator equal to \(0\).

\begin{align*} 1 + \cos (t) & = 0 \\ \cos (t) & = -1 \end{align*}

We need to exclude all odd multiples of \(\pi \). The natural domain is \(\{ r \in \mathbb {R} \, | \, r \ne (2k+1)\pi \text { where } k \in \mathbb {Z} \}\)

Therefore, we only need to analyze one period. Let’s investigate \([0, 2\pi ) - \{ \pi \}\).

• \(G(0) = 0\) Closed point on the graph
• \(G(2\pi ) = 0\) Open point on the graph.
• No point for \(\theta = \pi \).

Therefore, we will analyze \(G\) on the interval \([0, 2\pi )\), except \(\pi \), and then extend that information periodically to all real numbers.

\(\blacktriangleright \) Zeros:

On \([0, 2\pi )\), the numerator has a zero at \(0\) and \(\pi \). The denominator is not \(0\) at \(0\), therefore, the whole fraction is \(0\) at \(0\). \(G\) has a zero at \(0\).

Remember, we are only examining one period. Extending this information tells us that \(G\) has an infinite number of zeros at all even-\(\pi \).

\(\{ 2k \pi \, | \, k \in \mathbb {Z} \}\)

\(\pi \) is a different story. First, \(\pi \) is not in the domain. So, it will not be a zero of \(G\).

We are looking at a singularity here. However, we still need to figure out what \(G\) is doing around \(\pi \). Both the numerator and denominator are \(0\) at \(\pi \), which signals that anything can happen. We’ll need to think in more detail around \(\pi \).

\(\blacktriangleright \) Continuity:

\(G\) is a quotient function of continuous functions. Therefore, \(G\) is continuous on its domain and has no discontinuities.

What about singularities?

What happens near \(\pi \)? Numbers near \(\pi \) make the denominator near \(0\), but they also make the numerator near \(0\), since \(\sin (\pi )=0\).

It might be heplful to view the function with an equivalent formula.

This has changed the description around \(\pi \).

Near \(\pi \), \(1-\cos (t)\) is near \(2\) and \(\sin (t)\) is near \(0\). Therefore, \(G(t)\) is unbounded near \(\pi \). The graph will have a vertical asymptote at \(\pi \).

• \(1-\cos (t)\) is near \(2\), therefore, the numerator is positive when \(t\) is near \(\pi \).
• the denominator is \(\sin (t)\). It is near \(0\), when \(t\) is near \(\pi \). But, it is positive on one side and negative on the other side.
• therefore, the whole fraction is unbounded as \(t\) approaches \(\pi \). It is positive on the left and negative on the right.

On the left side of \(\pi \), \(\sin (t) > 0\), which makes \(G(t) > 0\). \(\lim \limits _{t \to \pi ^{-}}G(t) = \infty \)

On the right side of \(\pi \), \(\sin (t) < 0\), which makes \(G(t) < 0\). \(\lim \limits _{t \to \pi ^{+}}G(t) = -\infty \)

Every odd-\(\pi \) is a singularity of \(G\).

The graph contains the vertical asymptote \(t=\pi \) and then vertical asymptotes at all odd multiples of \(\pi \)’s.

\(\{ \theta = (2k+1) \pi \, | \, k \in \mathbb {Z} \}\)

Our graph of \(y=G(t)\) is piecing together.

\(\blacktriangleright \) Behavior:

We now know that \(G(0) = 0\) and \(\lim \limits _{t \to \pi ^{-}}G(t) = \infty \). So, let’s think about the interval \([0, \pi )\).

As \(\theta \) slowly moves away from \(0\), both the numerator and denominator will be positive, which means \(G\) is positive.

We also know that \(G(2\pi ) = 0\) and \(\lim \limits _{t \to \pi ^{+}}G(t) = -\infty \). So, let’s think about the interval \((\pi , 2\pi )\).

As \(\theta \) slowly moves away from \(2\pi \) (in our interval), both the numerator will be negative and the denominator will be positive, which means \(G\) is negative.

Let’s think of the interval \([0, \pi )\) in two pieces \(\left [ 0, \frac {\pi }{2} \right ) \cup \left ( \frac {\pi }{2}, \pi \right )\).

On \(\left [ 0, \frac {\pi }{2} \right )\), \(\sin (\theta )\) is increasing and \(1 + \cos (\theta )\) is decreasing. This makes the fraction increase.

On \(\left ( \frac {\pi }{2}, \pi \right )\), we know that both the numerator and denominator are approaching \(0\). This makes it difficult to conclude if \(G\) is still increasing or not. However, we know that \(\lim \limits _{t \to \pi ^{-}}G(t) = \infty \) and that is a pretty good signal that \(G\) is increasing.

On the interval \((\pi , 2\pi )\), similar reasoning will tell us that \(G\) is increasing.

From this we can sketch in a graph.

And, this is extended periodically.

Actually, once we extend this periodically, we might have a better idea.

Hmmm, maybe we need a different period to examine. Let’s take a look at \((-\pi , \pi )\).

That gives a better idea of what is going on. Remember, this is just one period.

0.1 Summary

\(\blacktriangleright \) \(G\) is a continuous function and has no discontinuities.

\(\blacktriangleright \) All of the odd-\(\pi \)’s are asymptotic singularities. The graph has vertical asymptotes.

• \(\lim \limits _{t \to \pi ^{-}}G(t) = \infty \)
• \(\lim \limits _{t \to \pi ^{+}}G(t) = -\infty \)

\(\blacktriangleright \) This tells us that \(G\) has no global maximums or minimums.

\(\blacktriangleright \) Since \(G\) is continuous, this also tells us that the range is \((-\infty , \infty )\).

\(\blacktriangleright \) \(G\) is an increasing function on any interval in the domain.

\(\blacktriangleright \) \(G\) is not an increasing function on its domain. Pick any pair in one interval, there is a “lower” pair in the next interval.

\(\blacktriangleright \) \(G\) is a strictly increasing function on any interval in the domain.

\(\blacktriangleright \) Since \(G\) is strictly increasing, there are no local maximums or minimums.