# 2.9: Sine and Cosecant Graphs

**At Grade**Created by: CK-12

**Practice**Sine and Cosecant Graphs

Imagine for a moment that you have a clock that has only one hand - that rotates counterclockwise! However, the hand is very slim all the way until the tip, where there is a ball on the end. In fact, the hand is so slim you won't notice it. You only notice the ball on the end of the rotating hand. This hand is rotating faster than normal. Here is a picture of the clock:

Consider what it would be like if you put a light next to the clock and let the shadow of the hands fall on the far wall.

What pattern would that shadow trace out? If you think about it, you might realize that the shadow would make an up and down motion, over and over as the hand of the clock rotated. Now imagine that instead of a wall, there was a large piece of paper for the shadow to fall on. And wherever the shadow fell, there would be a mark on the paper. Finally, imagine moving the paper as the clock rotates. Can you imagine sort of pattern this would trace out?

By the end of this Concept, you'll understand just how this relates to trigonometric functions in general, and the sine graph in particular.

### Watch This

James Sousa Animation: Graphing the Sine Function Using the Unit Circle

### Guidance

By now, you have become very familiar with the specific values of sine, cosine, and tangents for certain angles of rotation around the coordinate grid. In mathematics, we can often learn a lot by looking at how one quantity changes as we consistently vary another. We will be looking at the sine value **as a function** of the angle of rotation around the coordinate plane. We refer to any such function as a **circular function**, because they can be defined using the unit circle. Recall from earlier sections that the sine of an angle in standard position is the ratio of \begin{align*}\frac{y}{r}\end{align*}

Because the ratios are the same for a given angle, regardless of the length of the radius \begin{align*}r\end{align*}**unit circle** as a basis for all calculations.

The denominator is now 1, so we have the simpler expression, \begin{align*}\sin x=y\end{align*}

Through Quadrant I that height gets larger, starting at 0, increasing quickly at first, then slower until the angle reaches \begin{align*}90^\circ\end{align*}

As you rotate into the second quadrant, the height starts to decrease towards zero.

When you start to rotate into the third and fourth quadrants, the length of the segment increases, but this time in a negative direction, growing to -1 at \begin{align*}270^\circ\end{align*}

After one complete rotation, even though the angle continues to increase, the sine values will repeat themselves. The same would have been true if we chose to rotate clockwise to investigate negative angles, and this is why the sine function is a periodic function. The period is \begin{align*}2\pi\end{align*}

Let’s translate this circular motion into a graph of the sine value vs. the angle of rotation. The following sequence of pictures demonstrates the connection. These pictures plot \begin{align*}(\theta, \sin \theta)\end{align*}

After we rotate around the circle once, the values start repeating. Therefore, the sine curve, or “wave,” also continues to repeat. The easiest way to sketch a sine curve is to plot the points for the quadrant angles. The value of \begin{align*}\sin \theta\end{align*}

Filling in the gaps in between and allowing for multiple rotations as well as negative angles results in the graph of \begin{align*}y = \sin x\end{align*}

As we have already mentioned, \begin{align*}\sin x\end{align*}**range** of a sine curve is \begin{align*}\left \{-1 \le y \le 1 \right \}\end{align*}**domain** of \begin{align*}\sin x\end{align*}

Cosecant is the reciprocal of sine, or \begin{align*}\frac{1}{y}\end{align*}

The period of the function is \begin{align*}2\pi\end{align*}*except* where sine is defined (other than the points at the top and bottom of the sine curve).

Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs at 1 and -1. Many students are reminded of parabolas when they look at the half-period of the cosecant graph. While they are similar in that they each have a local minimum or maximum and they have the same beginning and ending behavior, the comparisons end there. Parabolas are not restricted by asymptotes, whereas the cosecant curve is.

#### Example A

Graph the following function:

\begin{align*}g(x)=\frac{1}{2} \sin \left( 3x \right)\end{align*}

**Solution:**

As you can see from the graph, the \begin{align*}\frac{1}{2}\end{align*}

#### Example B

Graph the following function:

\begin{align*}f(x)=\frac{1}{3} \csc \left( \frac{1}{2}x \right)\end{align*}.

#### Example C

Graph \begin{align*}f(x)=5 \sin \left( 2(x+\frac{\pi}{3}) \right)\end{align*}.

### Vocabulary

**Circular Function:** A ** circular function** is a function that is measured by examining the angle of rotation around the coordinate plane.

### Guided Practice

1. Graph \begin{align*}g(x)=5 \csc \left( \frac{1}{4}(x+\pi \right))\end{align*}.

2. Determine the function creating this graph:

3. Graph \begin{align*}h(x)=3 \sin \left( \frac{1}{2}(x+\frac{\pi}{2} \right))\end{align*}.

**Solutions:**

1.

2.

This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical shift is -1. The period is the difference between the two given \begin{align*}x-\end{align*}values, \begin{align*}\frac{7\pi}{4}-\frac{3\pi}{4}=\pi\end{align*}, so the frequency is \begin{align*}\frac{2\pi}{\pi}=2\end{align*}. The horizontal shift incorporates the frequency, so in \begin{align*}y=\csc x\end{align*} the corresponding \begin{align*}x-\end{align*}value to \begin{align*}\left(\frac{3\pi}{4}, 0\right)\end{align*} is \begin{align*}\left(\frac{\pi}{2}, 1\right)\end{align*}. The difference between the \begin{align*}x-\end{align*}values is \begin{align*}\frac{3\pi}{4}-\frac{\pi}{2}=\frac{3\pi}{4}-\frac{2\pi}{4}=\frac{\pi}{4}\end{align*} and then multiply it by the frequency, \begin{align*}2 \cdot \frac{\pi}{4}=\frac{\pi}{2}\end{align*}. The equation is \begin{align*}y=-1+ \csc \left(2(x-\frac{\pi}{2}\right))\end{align*}.

3.

### Concept Problem Solution

As you have seen in this Concept, the shadow of a light applied vertically to a rotating clock hand would trace out a sine graph. The graph would begin at zero when the hand is lying flat along the positive "x" axis. It would then increase until the hand was vertical. It would then decrease until the rotating hand was pointing straight down. Finally, the graph would increase again to zero when the hand is returning to the positive "x" axis.

### Practice

Graph each of the following functions.

- \begin{align*}f(x)=\sin(x)\end{align*}.
- \begin{align*}h(x)=\sin(2x)\end{align*}.
- \begin{align*}k(x)=\sin(2x+\pi)\end{align*}.
- \begin{align*}m(x)=2\sin(2x+\pi)\end{align*}.
- \begin{align*}g(x)=2\sin(2x+\pi)+2\end{align*}.
- \begin{align*}f(x)=\csc(x)\end{align*}.
- \begin{align*}h(x)=\csc(2x)\end{align*}.
- \begin{align*}k(x)=\csc(2x+\pi)\end{align*}.
- \begin{align*}m(x)=2\csc(2x+\pi)\end{align*}.
- \begin{align*}g(x)=2\csc(2x+\pi)+2\end{align*}.
- \begin{align*}h(x)=\sin(3x)\end{align*}.
- \begin{align*}k(x)=\sin(3x+\frac{\pi}{2})\end{align*}.
- \begin{align*}m(x)=3\sin(3x+\frac{\pi}{2})\end{align*}.
- \begin{align*}g(x)=3\sin(3x+\frac{\pi}{2})+3\end{align*}.
- \begin{align*}h(x)=\csc(3x)\end{align*}.
- \begin{align*}k(x)=\csc(3x+\frac{3\pi}{2})\end{align*}.
- \begin{align*}m(x)=4\csc(3x+\frac{3\pi}{2})\end{align*}.
- \begin{align*}g(x)=4\csc(3x+\frac{3\pi}{2})-3\end{align*}.

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### Image Attributions

Here you'll learn how to draw the graphs of the sine and cosecant functions.