5.4: Describing Relationships with the Tangent Function

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In this section, we will explore the graphs of the other four trigonometric functions. We’ll begin with the tangent function. Recall that in Chapter 5 we defined tangent as $y$/$x$ or sine/cosine, so you can think of the tangent as the slope of a line through the origin making the given angle with the positive $x$ axis.

At an angle of 0, the line would be horizontal with a slope of zero. As the angle increases towards $\pi $/2, the slo $A graph of tangent of theta. Dashed lines showing vertical asymptotes are drawn at theta equals negative pi over 2 and at pi over 2. Between these asymptotes, the graph increases, concave down to the left of the origin, and concave up to the right of the origin, approaching infinity as theta approches pi over 2 from the left, and approaching negative infinity as theta approachs negative pi over 2 from the right. This shape is repeated to the left and right, so another segment to the right of pi over 2 passes through pi comma 0.$ pe increases more and more. At an angle of $\pi $/2, the line would be vertical and the slope would be undefined. Immediately past $\pi $/2, the line would have a steep negative slope, giving a large negative tangent value. There is a break in the function at $\pi $/2, where the tangent value jumps from large positive to large negative.

We can use these ideas along with the definition of tangent to sketch a graph. Since tangent is defined as sine/cosine, we can determine that tangent will be zero when sine is zero: at -$\pi $, 0, $\pi $, and so on. Likewise, tangent will be undefined when cosine is zero: at -$\pi $/2, $\pi $/2, and so on.

The tangent is positive from 0 to $\pi $/2 and $\pi $ to 3$\pi $/2, corresponding to quadrants 1 and 3 of the unit circle.

Using technology, we can obtain a graph of tangent on a standard grid.

$A graph of tangent of theta on a standard axis$

Notice that the graph appears to repeat itself. For any angle on the circle, there is a second angle with the same slope and tangent value halfway around the circle, so the graph repeats itself with a period of $\pi $; we can see one continuous cycle from - $\pi $/2 to $\pi $/2, before it jumps and repeats itself.

The graph has vertical asymptotes and the tangent is undefined wherever a line at that angle would be vertical: at $\pi $/2, 3$\pi $/2, and so on. While the domain of the function is limited in this way, the range of the function is all real numbers.

FEATURES OF THE GRAPH OF TANGENT

The graph of the tangent function $m(\theta )=\tan (\theta )$

The period of the tangent function is $\pi $
The domain of the tangent function is $\theta \ne \dfrac{\pi }{2} +k\pi$, where $k$ is an integer
The range of the tangent function is all real numbers, $(-\infty ,\infty )$

With the tangent function, like the sine and cosine functions, horizontal stretches/compressions are distinct from vertical stretches/compressions. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph.

Example $\PageIndex{1}$

Find a formula for the function graphed here.

$A graph showing three increasing segments, each concave down at first, then switching to concave up. The first segment passes through negative 8 comma 0, the second through the origin, and the third through 8 comma 0. The middle segment approaches negative infinity as theta approaches negative 4 from the right, and approaches infinity as theta approaches 4 from the left.$

Solution

The graph has the shape of a tangent function, however the period appears to be 8. We can see one full continuous cycle from -4 to 4, suggesting a horizontal stretch. To stretch $\pi $ to 8, the input values would have to be multiplied by$\dfrac{8}{\pi }$. Since the constant $k$ in $f(\theta )=a\tan \left(k\theta \right)$is the reciprocal of the horizontal stretch $\dfrac{8}{\pi }$, the equation must have form

\[f(\theta )=a\tan \left(\dfrac{\pi }{8} \theta \right).\nonumber\]

We can also think of this the same way we did with sine and cosine. The period of the tangent function is $\pi$ but it has been transformed and now it is 8; remember the ratio of the “normal period” to the “new period” is $\dfrac{\pi }{8}$and so this becomes the value on the inside of the function that tells us how it was horizontally stretched.

To find the vertical stretch a, we can use a point on the graph. Using the point (2, 2)

\[2=a\tan \left(\dfrac{\pi }{8} \cdot 2\right)=a\tan \left(\dfrac{\pi }{4} \right)\nonumber\] Since \[\tan \left(\dfrac{\pi }{4} \right)=1,\quad a = 2\nonumber\]

This function would have a formula\[f(\theta )=2\tan \left(\dfrac{\pi }{8} \theta \right)\nonumber\]

Exercise $\PageIndex{1}$

Sketch a graph of $f(\theta )=3\tan \left(\dfrac{\pi }{6} \theta \right)$.

Answer: $A tangent graph, with vertical asysmptoes at negative 3 and 3. There are increasing segments between and on either side of the asymptotes. The middle one passes through the origin, the next passes through 6 comma 0.$

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FEATURES OF THE GRAPH OF TANGENT

Example \(\PageIndex{1}\)

Exercise \(\PageIndex{1}\)