Your mission, should you choose to accept it, as Agent Trigonometry is to graph the function \begin{align*}y=2 \cos x\end{align*}. What are the minimum and maximum of your graph?

### 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 \begin{align*}(\cos \theta, \sin \theta)\end{align*} where \begin{align*}\theta\end{align*} is the central angle. To graph, \begin{align*}y=\sin x\end{align*} rewrite the coordinates as \begin{align*}(x, \sin x)\end{align*} where \begin{align*}x\end{align*} is the central angle, in radians. Below we expanded the sine coordinates for \begin{align*}\frac{3\pi}{4}\end{align*}.

Notice that the curve ranges from 1 to -1. The maximum value is 1, which is at \begin{align*}x=\frac{\pi}{2}\end{align*}. The minimum value is -1 at \begin{align*}x=\frac{3 \pi}{2}\end{align*}. This “height” of the sine function is called the **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 \begin{align*}2 \pi\end{align*}. After \begin{align*}2 \pi\end{align*}, the sine values repeat. Therefore, the curve above will repeat every \begin{align*}2 \pi\end{align*} units, making the **period** \begin{align*}2 \pi\end{align*}. The domain is all real numbers.

Similarly, when we expand the cosine curve, \begin{align*}y=\cos x\end{align*}, from the unit circle, we have:

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 \begin{align*}2 \pi\end{align*}. If we draw the graph past \begin{align*}2\pi\end{align*}, it would look like:

Comparing \begin{align*}{\color{red}y=\sin x}\end{align*} and \begin{align*}{\color{blue}y=\cos x}\end{align*} (below), we see that the curves are almost identical, except that the sine curve starts at \begin{align*}y = 0\end{align*} and the cosine curve starts at \begin{align*}y = 1\end{align*}.

If we shift either curve \begin{align*}\frac{\pi}{2}\end{align*} units to the left or right, they will overlap. Any horizontal shift of a trigonometric function is called a **phase shift**. We will discuss phase shifts more in the upcoming concepts.

#### Solve the following problem

Identify the highlighted points on \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*} below.

For each point, think about what the sine or cosine value is at those values. For point \begin{align*}A\end{align*}, \begin{align*}\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}\end{align*}, therefore the point is \begin{align*}\left(\frac{\pi}{4},\frac{\sqrt{2}}{2}\right)\end{align*}. For point \begin{align*}B\end{align*}, we have to work backwards because it is not exactly on a vertical line, but it is on a horizontal one. When is \begin{align*}\sin x=-\frac{1}{2}\end{align*}? When \begin{align*}x=\frac{7 \pi}{6}\end{align*} or \begin{align*}\frac{11 \pi}{6}\end{align*}. By looking at point \begin{align*}B\end{align*}’s location, we know it is the second option. Therefore, the point is \begin{align*}\left(\frac{11 \pi}{6},\frac{1}{2}\right)\end{align*}.

For the cosine curve, point \begin{align*}C\end{align*} is the same as point \begin{align*}A\end{align*} because the sine and cosine for \begin{align*}\frac{\pi}{4}\end{align*} is the same. As for point \begin{align*}D\end{align*}, we use the same logic as we did for point \begin{align*}B\end{align*}. When does \begin{align*}\cos x = -\frac{1}{2}\end{align*}? When \begin{align*}x=\frac{2 \pi}{3}\end{align*} or \begin{align*}\frac{4\pi}{3}\end{align*}. Again, looking at the location of point \begin{align*}D\end{align*}, we know it is the second option. The point is \begin{align*}\left(\frac{4\pi}{3},\frac{1}{2}\right)\end{align*}.

### Amplitude

In addition to graphing \begin{align*}y=\sin x\end{align*} and \begin{align*}y=\cos x\end{align*}, we can stretch the graphs by placing a number in front of the sine or cosine, such as \begin{align*}y=a\sin x\end{align*} or \begin{align*}y=a\cos x\end{align*}. \begin{align*}|a|\end{align*} is the amplitude of the curve. In the next concept, we will shift the curves up, down, to the left and right.

#### Graph the functions

Graph \begin{align*}y=3 \sin x\end{align*} over two periods.

Start with the basic sine curve. Recall that one period of the parent graph, \begin{align*}y=\sin x\end{align*}, is \begin{align*}2 \pi\end{align*}. Therefore, two periods will be \begin{align*}4 \pi \end{align*}. The 3 indicates that the range will now be from 3 to -3 and the curve will be stretched so that the maximum is 3 and the minimum is -3. The red curve is \begin{align*}y=3 \sin x\end{align*}.

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.