Your mission, should you choose to accept it, as Agent Trigonometry is to graph the function

### Graphing Sine and Cosine

In this concept, we will take the unit circle, introduced in the previous chapter, and graph it on the Cartesian plane.

To do this, we are going to “unravel” the unit circle. Recall that for the unit circle the coordinates are

Notice that the curve ranges from 1 to -1. The maximum value is 1, which is at **amplitude**. The amplitude is the absolute value of average between the highest and lowest points on the curve.

Now, look at the domain. It seems that, if we had continued the curve, it would repeat. This means that the sine curve is **periodic**. Look back at the unit circle, the sine value changes until it reaches **period**

Similarly, when we expand the cosine curve,

Notice that the range is also between 1 and -1 and the domain will be all real numbers. The cosine curve is also periodic, with a period of

Comparing

If we shift either curve **phase shift**. We will discuss phase shifts more in the upcoming concepts.

#### Solve the following problem

Identify the highlighted points on

For each point, think about what the sine or cosine value is at those values. For point

For the cosine curve, point

### Amplitude

In addition to graphing

#### Graph the functions

Graph

Start with the basic sine curve. Recall that one period of the parent graph,

Notice that the \begin{align*}x\end{align*}-intercepts are the same as the parent graph. Typically, when we graph a trigonometric function, we always show two full periods of the function to indicate that it does repeat.

Graph \begin{align*}y=\frac{1}{2}\cos x\end{align*} over two periods.

Now, the amplitude will be \begin{align*}\frac{1}{2}\end{align*} and the function will be “smooshed” rather than stretched.

Graph \begin{align*}y=-\sin x\end{align*} over two periods.

The last two examples dealt with changing \begin{align*}a\end{align*} and \begin{align*}a\end{align*} was positive. Now, \begin{align*}a\end{align*} is negative. Just like with other functions, when the leading coefficient is negative, the function is reflected over the \begin{align*}x\end{align*}-axis. \begin{align*}y=-\sin x\end{align*} is in red.

### Examples

#### Example 1

Earlier, you were asked what are the minimum and maximum of your graph.

The 2 in front of the cosine function indicates that the range will now be from 2 to –2 and the curve will be stretched so that the maximum is 2 and the minimum is –2.

#### Example 2

Is the point \begin{align*}\left(\frac{5 \pi}{6},\frac{1}{2}\right)\end{align*} on \begin{align*}y=\sin x\end{align*}? How do you know?

Substitute in the point for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and see if the equation holds true.

\begin{align*}\frac{1}{2}=\sin\left(\frac{5 \pi}{6}\right)\end{align*}

This is true, so \begin{align*}\left(\frac{5 \pi}{6},\frac{1}{2}\right)\end{align*} is on the graph.

Graph the following functions for two full periods.

#### Example 3

\begin{align*}y=6 \cos x\end{align*}

Stretch the cosine curve so that the maximum is 6 and the minimum is -6.

#### Example 4

\begin{align*}y=-3 \cos x\end{align*}

The graph is reflected over the \begin{align*}x\end{align*}-axis and stretched so that the amplitude is 3.

#### Example 5

\begin{align*}y=\frac{3}{2} \sin x\end{align*}

The fraction is equivalent to 1.5, making 1.5 the amplitude.

### Review

- Determine the exact value of each point on \begin{align*}y=\sin x\end{align*} or \begin{align*}y=\cos x\end{align*}.
- List all the points in the interval \begin{align*}[0,4 \pi]\end{align*} where \begin{align*}\sin x=\cos x\end{align*}. Use the graph from #1 to help you.
- Draw from \begin{align*}y=\sin x\end{align*} from \begin{align*}[0, 2 \pi]\end{align*}. Find \begin{align*}f \left(\frac{\pi}{3}\right)\end{align*} and \begin{align*}f \left(\frac{5 \pi}{3}\right)\end{align*}. Plot these values on the curve.

For questions 4-12, graph the sine or cosine curve over two periods.

- \begin{align*}y=2 \sin x\end{align*}
- \begin{align*}y=-5 \cos x\end{align*}
- \begin{align*}y=\frac{1}{4} \cos x\end{align*}
- \begin{align*}y=- \frac{2}{3} \sin x\end{align*}
- \begin{align*}y=4 \sin x\end{align*}
- \begin{align*}y=-1.5 \cos x\end{align*}
- \begin{align*}y=\frac{5}{3} \cos x\end{align*}
- \begin{align*}y=10 \sin x\end{align*}
- \begin{align*}y=-7.2 \sin x\end{align*}
- Graph \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*} on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps the cosine curve?
- Graph \begin{align*}y=\sin x\end{align*} and \begin{align*}y=-\cos x\end{align*} on the same set of axes. How many units would you have to shift the sine curve (to the left or right) so that it perfectly overlaps \begin{align*}y=- \cos x\end{align*}?

Write the equation for each sine or cosine curve below. \begin{align*}a>0\end{align*} for both questions.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.1.