# How do you graph two cycles of y=2tan(3theta)?

Mar 15, 2018

Try sketching with references to certain properties of the graph, noticeably intersects, monotonies, and asymptotes.
graph{tan(3x) [-2.23, 2.23, -1.162, 1.162]}

#### Explanation:

Note that all $\theta$ has been replaced with $x$ in the following explanation.

1. Intersections
Evaluate the function at $x = 0$ to find the $y$ -intersect:
$\tan 0 = 0$ thus the function intersect the $y$ -axis at $\left(0 , 0\right)$ and therefore passes through the origin.

We can find $x$ -intersects, or "zeros," of the function by setting its value to zero and solving for $x$.
Let $\tan \left(3 \cdot x\right) = y = 0$

This explanation shows how to solve the equation by considering the composite nature of the function: it consists of two parts, an inner function $f \left(x\right) = 3 \cdot x$, and an outer function $g \left(u\right) = \tan \left(u\right)$. So the original function, $\tan \left(3 \cdot x\right)$, is equivalent to $g \left(f \left(x\right)\right)$. Thus values of $u = f \left(x\right)$ at zeros of this functions shall ensure that $g \left(u\right)$, the outer function gives a value of zero.

$\tan \left(u\right) = 0$,
$u = 0 + k \cdot \pi$, where $k$ is an integer ($k \in \mathbb{Z}$). Here $u$ has more than one possible value since the function $\tan u$, is cyclical with a period of $\pi$.

Substituting $u$ with an expression about $x$ gives
$3 \cdot x = k \cdot \pi$
$x = \frac{k}{3} \cdot \pi$

Thus coordinates of $x$ -intercepts of this function shall fit into the general expression
$\left(\frac{k}{3} \cdot \pi , 0\right)$

Taking $k = - 1$, $k = 0$, and $k = 1$ gives $x$ -intersects
$\left(- \frac{k}{3} \cdot \pi , 0\right)$, $\left(0 , 0\right)$, and $\left(\frac{k}{3} \cdot \pi , 0\right)$.

2. Monotonies
The tangent function always increases as the angle grows, as seen from a unit circle. Therefore the graph of $\tan x$ should slope upwards and extend to the upper-right corner of the Cartesian plane. This observation is also the case for a composite tangent function like $\tan u$ as long as the inner function $u = f \left(x\right)$ is increasing.

3.Asymptotes
The tangent function is not defined at the sum of $\frac{\pi}{2}$ any integer multiple of $\pi$ and shows asymptotic behavior at each of these values of $x$. That is
$x = \frac{\pi}{2} + k \cdot \pi$ where $k$ is an integer.

For the composite tangent function here, the general expression for all the asymptotes would be
$3 \cdot x = u = \frac{\pi}{2} + k \cdot \pi$ and
$x = \left(\frac{1}{6} + \frac{k}{3}\right) \cdot \pi$

Evaluating at the expression at $k = - 1$, $k = 0$ and $k = 1$ gives
$x = - \frac{1}{6} \cdot \pi$, $x = \frac{1}{6} \cdot \pi$, and $x = \frac{1}{2} \cdot \pi$.

Now plot all three of these features on the graph, and the curve you sketch should:
a. Slopes upwards;
b. Passes through all of the intersections, and
c. Approaches, but never touches each of the asymptotes.