We now know how to graph common transformations of sine and cosine. Let's look at the rest of the graphs.
Tangent and cotangent have period $\pi$:
\[\tan(t + \pi) = \tan t \qquad\qquad \tan(t + \pi) = \tan t\]
while cosecant and secant have period $2\pi$:
\[\csc(t + 2\pi) = \csc t \qquad\qquad \sec(t + 2\pi) = \sec t\]
Draw one period of $f(x) = \tan(x)$ by hand.
The graph of $\tan(x)$ is especially interesting. As we approach $t = \frac{\pi}{2}$ from the left, the ratio $\frac{y}{x}$ becomes very large because $x$ is getting close to 0 while $y$ is getting close to $1$. Dividing a number close to one by a number close to zero results in extremely large values! In fact, these values have a name:
The vertical line $x = a$ is a vertical asymptote of the function $y = f(x)$ if $y$ approaches $\pm\infty$ as $x$ approaches $a$ from the right or left.
Usually, dividing by zero gives a vertical asymptote. This shows when $x = 0$ for a terminal point, $\tan(t)$ will have an asymptote.
Here are all the trigonometric graphs. Note the vertical asymptotes:
For reference:
$\tan(x)$ completes one complete period on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cot(x)$ completes one complete period on $\left[0, \pi\right]$
$\csc(x)$ completes one complete period on $\left[0, 2\pi\right]$
$\sec(x)$ completes one complete period on $\left[0, 2\pi\right]$
Transformations of tangent and cotangent
Given $k > 0$, the functions
\[y = a\tan k(x - b) \qquad\qquad y = a\cot k(x - b)\]
have period $\frac{\pi}{k}$ and horizontal shift of $+b$ units.
In the language of function transformations, what type of transformation is $a$, $k$ and $b$?
Find the period, and sketch by hand.
\[f(x) = \tan(2x) \qquad\qquad g(x) = \tan2 \left(x - \frac{\pi}{4}\right) \qquad\qquad h(x) = 2 \cot\left(3x - \frac{\pi}{4}\right)\]
Transformations of secant and cosecant
Given $k > 0$, the functions
\[y = a\csc k(x - b) \qquad\qquad y = a\sec k(x - b)\]
have period $\frac{2\pi}{k}$ and horizontal shift of $+b$ units.
Graph each function by hand.
\[y = \frac{1}{2}\csc 2x \qquad\qquad y = \frac{1}{2}\csc\left(2x + \frac{\pi}{2}\right)\]