$\blacktriangleright$ 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 $0$.

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.

$$G(t) = \frac {\sin (t)}{1 + \cos (t)} = \frac {\sin (t)}{1 + \cos (t)} \cdot 1 = \frac {\sin (t)}{1 + \cos (t)} \cdot \frac {1-\cos (t)}{1-\cos (t)} = \frac {\sin (t) (1-\cos (t))}{1-\cos ^2(t)}$$

$$G(t) = \frac {1-\cos (t)}{\sin (t)}$$

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.

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.