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

1 Answer
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:
tan0=0 thus the function intersect the y -axis at (0,0) 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(3*x)=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(x)=3*x, and an outer function g(u)=tan(u). So the original function, tan(3*x), is equivalent to g(f(x)). Thus values of u=f(x) at zeros of this functions shall ensure that g(u), the outer function gives a value of zero.

tan(u)=0,
u=0+k*pi, where k is an integer (k in ZZ). 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*x=k*pi
x=k/3*pi

Thus coordinates of x -intercepts of this function shall fit into the general expression
(k/3*pi,0)

Taking k=-1, k=0, and k=1 gives x -intersects
(-k/3*pi,0), (0,0), and (k/3*pi,0).

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(x) is increasing.

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

For the composite tangent function here, the general expression for all the asymptotes would be
3*x=u=pi/2+k*pi and
x=(1/6+k/3)*pi

Evaluating at the expression at k=-1, k=0 and k=1 gives
x=-1/6*pi, x=1/6*pi, and x=1/2*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.

See also:
https://www.mathsisfun.com/geometry/unit-circle.html
Value of the tangent function on a unit circle