Lesson 15

Here are pictures of sound waves for two different musical notes:

What do you notice? What do you wonder?

Match each equation with its graph. More than 1 equation can match the same graph.

Equations:

1. \(y = \text-\cos(\theta)\)
2. \(y = 2\sin(\theta)-3\)
3. \(y = \cos\left(\theta + \frac{\pi}{2}\right)\)
4. \(y = 3\sin(\theta) - 2\)
5. \(y = \sin(\theta-\frac\pi 2)\)
6. \(y = \sin(\theta+\pi)\)

 

1. Complete the table of values for the expression \(\sin(2\theta)\)

   \(\theta\)	0	\(\frac{\pi}{12}\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{2}\)	\(\frac{3\pi}{4}\)	\(\pi\)	\(\frac{5\pi}{4}\)	\(\frac{3\pi}{2}\)	\(\frac{7\pi}{4}\)	\(2\pi\)
   \(\sin(2\theta)\)

2. Plot the values and sketch a graph of the equation \(y = \sin(2\theta)\). How does the graph of \(y = \sin(2\theta)\) compare to the graph of \(y = \sin(\theta)\)?

   

   

3. Predict what the graph of \(y = \cos(4\theta)\) will look like and make a sketch. Explain your reasoning.

   

   

We can find the amplitude and midline of a trigonometric function using the graph or from an equation. For example, let’s look at the function given by the equation \(y = 3\cos\left(\theta+\frac{\pi}{4}\right) + 2\). We can see that the midline of this function is 2 because of the vertical translation up by 2. This means the horizontal line \(y = 2\) goes through the middle of the graph. The amplitude of the function is 3. This means the maximum value it takes is 5, 3 more than the midline value, and the minimum value it takes is -1, 3 less than the midline value. The horizontal translation is \(\frac{\pi}{4}\) to the left, so instead of having, for example, a minimum at \(\pi\), the minimum is at \(\frac{3\pi}{4}\). Here is what the graph looks like:

Another type of transformation is one that affects the period and that is when a horizontal scale factor is used. For example, let's look at the equation \(y = \cos(2\theta)\) where the variable \(\theta\) is multiplied by a number. Here, 2 is the scale factor affecting \(\theta\). When \(\theta = 0\), we have \(2\theta = 0\) so the graph of this cosine equation starts at \((0,1)\), just like the graph of \(y = \cos(\theta)\). When \(x = \pi\), we have \(2\theta = 2\pi\) so the graph of \(y = \cos(2\theta)\) goes through two full periods in the same horizontal span it takes \(y = \cos(\theta)\) to complete one full period, as shown in their graphs.