Prove the following trigonometric identity using the double-angle identity.

Divide the entire equation by 576:

$$\frac{16x^2}{576} - \frac{36y^2}{576} = \frac{576}{576}$$ $$\frac{x^2}{36} - \frac{y^2}{16} = 1$$

This is the standard form of a hyperbola centered at the origin $(0,0)$ with a horizontal transverse axis.
Here, $a^2 = 36 \implies a = 6$ and $b^2 = 16 \implies b = 4$.
The vertices are at $(\pm a, 0) = (\pm 6, 0)$.
The asymptotes are $y = \pm \frac{b}{a}x = \pm \frac{4}{6}x = \pm \frac{2}{3}x$.

1.1.1 Sketch the graph of the equation given above.
To sketch the graph, plot the vertices at $(\pm 6, 0)$. Draw the asymptotes $y = \frac{2}{3}x$ and $y = -\frac{2}{3}x$. Then, draw the two branches of the hyperbola opening left and right from the vertices, approaching the asymptotes.
(Since I cannot draw an image, I will describe the key features for a sketch).

• Center: $(0,0)$
• Vertices: $(\pm 6, 0)$
• Asymptotes: $y = \pm \frac{2}{3}x$
• The graph consists of two separate curves, one extending to the left from $x=-6$ and one extending to the right from $x=6$.

1.1.2 Is the graph a function or a relation?
The graph fails the vertical line test (for most $x$ values in its domain, there are two corresponding $y$ values).
It is a relation.

1.1.3 State the domain of the graph.
From $\frac{x^2}{36} - \frac{y^2}{16} = 1$, we have $\frac{x^2}{36} = 1 + \frac{y^2}{16}$. Since $\frac{y^2}{16} \ge 0$, then $\frac{x^2}{36} \ge 1$, which means $x^2 \ge 36$.
So, $x \le -6$ or $x \ge 6$.
The domain is $\boxed{(-\infty, -6] \cup [6, \infty)}$.

1.1.4 State the range of the graph.
From $\frac{y^2}{16} = \frac{x^2}{36} - 1$, for $x^2 \ge 36$, the term $\frac{x^2}{36} - 1$ can take any non-negative value, meaning $y^2$ can take any non-negative value.
Therefore, $y$ can be any real number.
The range is $\boxed{(-\infty, \infty)}$.

1.1.5 Is the graph $16x^2 - 36y^2 = 576$ continuous or discontinuous?
The graph of a hyperbola is a smooth curve without any breaks or jumps within its defined domain.
It is continuous.

1.1.6 Does the graph have an inverse function?
The graph fails the horizontal line test (for most $y$ values in its range, there are two corresponding $x$ values).
Therefore, it does not have an inverse function (without restricting its domain).