Tangent

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quotient

Tangent is the quotient of sine and cosine.

\[ \tan (\theta ) = \frac {\sin (\theta )}{\cos (\theta )} \]

zeros: $\tan (\theta )$ has a zero everywhere that $\sin (\theta )$ has a zero: all of the whole-$\pi $’s.
singluaries: $\tan (\theta )$ has a singularity everywhere that $\cos (\theta )$ has a zero: all of the half-$\pi $’s.

This means intercepts at the whole-$\pi $’s and vertical asymptotes at the half-$\pi $’s.

Tangent is always increasing.
Tangent has no maximums or minimum, since it is unbounded near the half-$\pi $’s.
The period for tangent is $\pi $.

Analyze

The usual interval people look at tangent is $\left ( \frac {-\pi }{2}, \frac {\pi }{2} \right )$. This interval has singularities on both ends, which signal vertical asymptotes on the graph. $\frac {-\pi }{2}$ and $\frac {\pi }{2}$ are what we need the inside of our function to equal.

\begin{align*} \frac {t}{2} + \pi & = \frac {-\pi }{2} \\ \frac {t}{2} & = \frac {-3\pi }{2} \\ t & = \answer {-3 \pi } \end{align*}

\begin{align*} \frac {t}{2} + \pi & = \frac {\pi }{2} \\ \frac {t}{2} & = \frac {-\pi }{2} \\ t & = \answer {-\pi } \end{align*}

The period is $\answer {2 \pi }$. One of the arms of this tangent function sits on the interval $[ -3 \pi , - \pi ]$.

Then the $-1$ coefficient flips the graph upside-down.

There are no maximums or minimums.

$K(t)$ is always decreasing.

The zeros are $\{ 2 k \pi \, | \, k \in \mathbb {Z} \}$

With these ideas, we can write an algebraic analysis.

We are going to view this as a composition of three functions.

\[ K = Out \circ \tan \circ In \]

where

$Out(x) = -x$

$In(y) = \frac {y}{2} + \pi $

Along with $\tan (\theta )$, these three component functions, we get

\[ K(t) = (Out \circ \tan \circ In)(t) = Out(tan(In(t))) = -\tan \left ( \frac {t}{2} + \pi \right ) \]

$K$ is a periodic function, since $\tan (\theta )$ is a periodic function.

The principal interval of $\tan (\theta )$ is $\left ( -\frac {\pi }{2}, \frac {\pi }{2} \right )$. The principal interval of $K$ will be the values of $t$ that make the values of $In(y)$ run from $-\frac {\pi }{2}$ to $\frac {\pi }{2}$.

$\frac {y}{2} + \pi = -\frac {\pi }{2}$ at $y = -3\pi $

$\frac {y}{2} + \pi = \frac {\pi }{2}$ at $y = -\pi $

The principle interval of $T$ is $\left ( -3\pi , \pi \right )$.

The length of this interval is $2\pi $, which is the period of $K$.

$\blacktriangleright $ desmos graph

This browser does not support embedded elements.

$K$ is a periodic function with period $2\pi $.

Therefore, our analysis will focus on the principal interval of $\left ( -3\pi , -\pi \right )$.

All of the features and characteristics we discover will repeat with a period of $2\pi $.

Domain

The domain of $Out(x) = -x$ is $(-\infty , \infty )$, because $Out$ is a linear function.

This includes any value of $\tan (\theta )$. That means we can use the entire domain of $\tan (\theta )$, which excludes $\frac {\pi }{2} + n \pi $ where $n \in \mathbb {Z}$.

This would be $-\pi $ plus or minus any whole numbers of $2\pi $.

We need to exclude all $\pi + 2n\pi $ where $n \in \mathbb {Z}$.

The domain of $T$ is $(-\infty , \infty )$ except $\left \{ \pi + 2n\pi \, | \, n \in \mathbb {Z} \right \}$.

We will be examining $K(t)$ on $(-3\pi , -\pi )$.

Zeros

Since the values of $K$ come from the values of $Out$, we are first looking for zeros of $Out$. $Out(x) = -x$ is a linear function and has only one zero, $0$.

We are looking for where $x = \tan (\theta ) = 0$.

The only number in our principal interval where this happens is $\theta = 0$.

We need the values of $y$ where $\theta = In(y) = \frac {y}{2} + \pi = 0$.

$y = -2\pi $

[ These agree with the graph. ]

And, these repeat every $2\pi $ for $K$.

Continuity

The component functions of the composition are linear and tangent, all continuous.

The composition of continuous function is continuous.

$K$ is continuous. It has no disontinuities.

However, $K$ does have singularities. In our principal interval, $K$ has singularities at $-3\pi $ and $\pi $. These repeat witha period of $2\pi $.

The singularities of $K$ are $\{ (2n+1)\pi \, | \, n \in \mathbb {Z} \}$.

We also need the behvaior of $K$ at these singularities.

\[ \lim \limits _{y to -3\pi ^+} In(y) = \lim \limits _{y to -3\pi ^+} \frac {y}{2} + \pi = \left ( \frac {5\pi }{2} \right )^+ \]

\[ \lim \limits _{\theta to \left ( \tfrac {5\pi }{2} \right )^+} \tan (\theta ) = -\infty \]

\[ \lim \limits _{x to -\infty } Out(x) = \lim \limits _{x to -\infty } (-x) = \infty \]

\[ \lim \limits _{x to -3\pi ^+} K(x) = \infty \]

Tangent has the opposite behavior on the other side of the principal interval.

\[ \lim \limits _{x to -\pi ^-} K(x) = -\infty \]

Notes:

$\blacktriangleright $ Comparing $K(t)$ to $\tan (t)$, the inside has been multiplied by $\frac {1}{2}$. This would indicate the perioid is doubled from $\pi $ to $2\pi $, which has happened.

$\blacktriangleright $ Additionally, an $\pi $ hs been added to the inside. This would indicate a horizontal shift left, which has happened.

End-Behavior

$K$ is a periodic function with period $2\pi $.

It either has no end-behavior or the end-behavior is that it is periodic.

Behavior

We need the behavior of each of the three component funcitons and then we will compose them together.

$Out(x) = -x$ is a linear function with a negative leading coefficient, so it is an decreasing function.

$In(y) = \frac {y}{2} + \pi $ is a linear function with a positive leading coefficient, so it is an increasing function.

$\tan (\theta )$ is an increasing function.

Each of the component functions is monomtonic function, which means $K$ will not change behavior.

\[ K(t) = negative \circ increasin \circ increasing = decreasing \]

[ This agrees with the graph. ]

Global Maximum and Minimum

The singularity behavior has already shown us that $K$ is unbounded. There is no global maximum or minimum.

Local Maximum and Minimum

$K$ is monotonic (its bahavior doesn’t change) and it is unbounded. There are no local minimums or minimums.

Range

Since $T$ is continuous and the singularity behavior has already shown us that $K$ is unbounded, then the range is $(-\infty , \infty )$.

[ This agrees with the graph. ]

Geometry

Sine and cosine come from measurements of the unit circle. What about tangent?

Draw our usual right triangle for sine and cosine on the unit circle. Draw a tangent line at the point on the circle. Extend this line to the horizontal axis and use this to draw a second right triangle. The two right triangles have an angle $\theta $, which means all three angles are equal, which means the triangles are similar.

These two right triangles are similar. There is also a third big right triangle.

Let’s label some sides.

In the diagram above, we know that $a = \cos (\theta )$ and $b = \sin (\theta )$.

We also know the unit circle right triangles are both similar to the big right triangle, since they all include the angle $\theta $.

Since, they are similar, we know that

\[ \frac {opp}{adj} = \frac {b}{a} = \frac {\sin (\theta )}{\cos (\theta )} = \frac {h}{1} \]

\[ \tan (\theta ) = h\]

$\tan (\theta )$ can be represented as the length of the tangent line segment, from the tangent point to the $x$-axis.

Is this important? Not really. But, it is nice to collect all of our Trigonometric functions together on the unit circle. It explains their names.

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2025-08-08 17:39:34