Lesson 14
Transforming Trigonometric Functions
14.1: Translated Parabolas (5 minutes)
Warmup
The goal of this warmup is to recall how a quadratic equation changes when its graph is translated horizontally. This will prepare students for examining this same translation when it is applied to trigonometric functions in this lesson.
Student Facing
Match each equation with its graph. Be prepared to explain your reasoning.
 \(y = x^2\)
 \(y = (x1)^2\)
 \(y = (x+3)^2\)
Student Response
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Activity Synthesis
Invite students to share how they made their matches. Make sure students recall:
 The three graphs have the same shape.
 They are horizontal translations of one another.
 Adding a positive number to the input translates the graph to the left and adding a negative number to the input translates the graph to the right. Display the three graphs together for all to see to highlight this.
14.2: Windmills Everywhere (15 minutes)
Activity
In the previous lesson, students learned the definition of amplitude and midline by examining sine graphs produced to model the vertical position of a point on a rotating windmill. This activity keeps the windmill context and asks students to interpret both cosine and sine equations. They will need to focus on three different aspects of the equations:
 The amplitude of the sine graph (length of the windmill blade)
 The midline of the sine graph (height of the center of the windmill blades)
 The direction the windmill spins (clockwise or counterclockwise).
Launch
Arrange students in groups of 2. Display an image of the blades of a windmill for all to see, such as this one from a previous lesson:
Tell students that these blades extend 4 meters from the center and that the center of the windmill is 20 meters off the ground. The windmill rotates counterclockwise.
Ask students to write an equation to represent the height in meters of the point \(W\), relative to the ground, as a function of the angle \(\theta\) of rotation (\(h = 4\sin(\theta) + 20\)). Select 2–3 students to share how they wrote their equation. Tell them that in this activity they are going to work the opposite direction. That is, they are going to start with an equation giving the vertical position of a point \(W\) at the tip of a windmill blade and use it to describe the windmill.
Design Principle(s): Support sensemaking; Cultivate conversation
Supports accessibility for: Language; Memory
Student Facing
Here are three equations for three different windmills. Each equation describes the height \(h\), in feet above the ground, of a point at the tip of a blade of a different windmill. The point is at the far right when the angle \(\theta\) takes the value 0. Describe each windmill and how it is spinning.
 \(h = 2.5 \sin(\theta) + 10\)
 \(h = \frac{4}{5} \sin(\theta) + 3\)
 \(h = \text1.5 \sin(\theta) + 5\)
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
If students struggle with the negative amplitude in the final question, encourage them to substitute in a few angle values \(\theta\) and plot them.
Activity Synthesis
Display the 3 equations for all to see. Begin the discussion by asking, “What is the midline and amplitude of each function and what do these values mean in this context?” (The midline is the height of windmill blades’ center and the amplitude is the length of the blades.) Record student responses next to the equations.
Ask students how they interpreted the sign in front of the sine on the last equation (it means that the windmill is spinning in the clockwise direction as the angle \(\theta\) increases). To help students visualize the impact of the negative sign in front of the expression, note that while \(\sin(\theta)\) increases from 0 to 1 as \(\theta\) increases from 0 to \(\frac{\pi}{2}\), \(\text\sin(\theta)\) decreases from 0 to 1 as \(\theta\) increases from 0 to \(\frac{\pi}{2}\). In terms of the windmill, this would mean that instead of going toward the highest spot on its circle of rotation, the point \(W\) begins by going to the lowest spot on its circle of rotation. This means that the windmill is rotating in a clockwise direction instead of a counterclockwise direction.
14.3: Spinning Fan (15 minutes)
Activity
In this activity students investigate how adding a constant to the variable inside a trigonometric function influences the graph. They have done this work in an earlier unit, for polynomials and other nonperiodic functions, and the underlying structure is the same for periodic functions. The particular focus of this activity is interpreting a horizontal translation for a periodic function representing circular motion.
There are several strategies students may use for graphing the functions. Monitor for students who
 evaluate the functions using the classroom unit circle display or technology and plot individual values
 identify expressions like \(\cos\left(\frac{\pi}{6} \right)\) and \(\sin\left(\frac{\pi}{6}\right)\) for the vertical intercepts of the graphs
 connect the different starting position to a change of input for the sine and cosine functions
Identifying specific values is helpful for plotting points and producing the graphs. The expressions using cosine and sine help to identify that the graphs are horizontal translations and to write expressions representing these functions. Highlight the importance of both strategies during the wholeclass discussion, if they come up.
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Supports accessibility for: Language; Organization
Student Facing
A fan has radius 1 foot. A point \(P\) starts in the position shown in the picture. The center of the fan is at \((0,0)\) and the point \(P\) is at the \(\frac{\pi}{6}\) position on the circle. The fan turns in a counterclockwise direction.

Sketch a graph of the horizontal position \(h\), in feet, of \(P\) as a function of the angle of rotation \(\theta\) of the fan from its starting position.
 How does this graph compare to the graph of \(h = \cos(\theta)\)?

Sketch a graph of the vertical position \(v\), in feet, of \(P\) as a function of the angle of rotation \(\theta\) of the fan.
 How does this graph compare to the graph of \(v = \sin(\theta)\)?
Student Response
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Student Facing
Are you ready for more?
I attach a laser beam to the fan at the same point \(P\) at position \(\frac{\pi}{6}\) so that the laser points away from the center of the fan. At the starting position the laser beam hits the right wall near the ceiling and does so until position \(\frac{\pi}{4}\). It then moves along the ceiling followed by moving down the left wall of the room, then onto the floor, and finally back up the right wall. We move from the floor to the right wall at the position \(\frac{7\pi}{4}\).
 At what angles of rotation between \(0\) and \(4\pi\) from the fan's starting position will the beam hit the right wall?
 Measure the position of the beam on the right wall by marking evenly spaced points from 10 at the floor to 10 at the ceiling so that we hit the 0 mark when the beam is at position 0 on the circle. Write a function which gives the position \(y\) on the wall at angle of rotation \(\theta\) where \(\theta\) is in the domain you found in the first part.
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
For students who struggle to create a plot, or to identify the relationship between the graphs and the equations, consider asking them to complete a table with columns for \(\theta\), \(h\), and \(v\), where $\theta is from 0 to \(\frac{4\pi}{3}\) in increments of \(\frac{\pi}{3}\).
Activity Synthesis
Begin the discussion by inviting students to share strategies they used for graphing the functions. Some strategies may include:
 Create a table of values and plot them.
 Use technology to calculate values and then plot them.
 Use transformations on a function to determine the new function and graph it.
Ask students how the new graphs are similar to and different from the graphs of \(h = \cos(\theta)\) and \(v = \sin(\theta)\). Students should understand that while they have the same shape and also the same period (\(2\pi\)) and amplitude (1), they are different because they don’t “start” at the same place. The \(v\) graph starts at \(\sin\left(\frac{\pi}{6}\right)\) rather than at \(\sin(0)\) and the \(h\) graph starts at \(\cos\left(\frac{\pi}{6}\right)\) rather than at \(\cos(0)\). Connect this back to the graphs of \(y = x^2\) and \(y = (x+3)^2\) from the warm up: both cases represent a horizontal translation to the left (by \(\frac{\pi}{6}\) for the trigonometric functions and by 3 for the parabola).
Display the graphs of \(v=\sin(\theta)\) and \(v=\sin\left(\theta+\frac{\pi}{6}\right)\) (the equation for the vertical position of the point \(P\) in this activity) for all to see on the same axes to help visualize the horizontal translation.
If equations for the transformations are not brought by students, invite them to explain how changing the starting position of the blade influences the graph. They should identify that the graph for this problem is a horizontal translation of the graph of \(v = \sin(\theta)\) and that the translation is to the left by \(\frac{\pi}{6}\). Tell students that an equation defining the vertical position of the point \(P\) is \(v=\sin\left(\theta+\frac{\pi}{6}\right)\) and that the \(\frac{\pi}{6}\) comes from the point \(P\) starting at the \(\frac{\pi}{6}\) position.
Time permitting, ask students what the graphs and equations for the vertical position would look like if the point \(P\) started at \((0,1)\) (They would be translated to the left by \(\frac{\pi}{2}\) so the equation would be \(v=\sin\left(\theta+\frac{\pi}{2}\right)\).) What if \(P\) started at \((0,\text1)\)? (They would be translated to the left by \(\frac{3\pi}{2}\) or to the right by \(\frac{\pi}{2}\) so the equation would be \(v=\sin\left(\theta+\frac{3\pi}{2}\right)\) or \(v=\sin\left(\theta\frac{\pi}{2}\right)\), respectively.)
Lesson Synthesis
Lesson Synthesis
The purpose of this discussion is for students to consider a periodic function where 3 types of transformations are in effect at once, building on the work from the activities and looking ahead to the next lesson. Display for all to see the graph of the function \(y = 2\sin\left(\theta\frac{\pi}{4}\right) 1\):
Here are some questions for discussion:
 “How does the 2 from the equation show up on the graph?” (2 is the amplitude, the distance from the midline to the maximum value or to the minimum value.)
 “How does the 1 from the equation show up on the graph?” (1 is the vertical translation or midline, halfway between the maximum and minimum values.)
 “How does the \(\text\frac{\pi}{4}\) from the equation show up on the graph?” (It translates the graph to the right by \(\frac{\pi}{4}\).)
Lastly, display for all to see the graph of \(y = 2\sin(\theta)1\) on the same axes as \(y = 2\sin\left(\theta\frac{\pi}{4}\right) 1\) to highlight visually the translation to the right:
If time permits, ask students what nonzero value they could replace \(\frac{\pi}{4}\) by in order for its graph to look the same as the graph of \(y = 2\sin(\theta)1\). (Any integer multiple of \(2\pi\).)
14.4: Cooldown  Translating and Stretching (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
The graphs of cosine and sine functions can be translated vertically or horizontally and the size or height of their graphs can also be modified similar to how we transformed other types of functions in an earlier unit. Let’s look at the graphs of \(y = \sin(\theta)\) and \(y = 2\sin(\theta) + 3\).
The coefficient 2 stretches the graph vertically, doubling the amplitude of the sine graph. This means that the distance from the midline to the maximum or minimum value is now 2 instead of 1. Adding 3 to the equation translates the midline up by 3 units.
What if we want to translate the graph of \(y = \sin(\theta)\) to the left? We can do this by adding an angle to the input \(\theta\). Let’s look at the graph of \(y = \sin\left(\theta + \frac{\pi}{2}\right)\). The graph of this function looks just like the graph of \(y = \sin(\theta)\) translated to the left by \(\frac{\pi}{2}\). (It also looks a lot like \(y=\cos(\theta)\)!)