Lesson 11 of 15
Objective: SWBAT solve trigonometric equations by squaring both sides.
To go over yesterday’s homework, I have students present their work on the document camera for any problems that students could not complete. While looking at their work, it is nice to have another student explain what happens from one step to the next. Asking “what did this student do to go from step 3 to step 4?” usually starts a rich conversation of why they chose the path they did.
After this conversation, it is a good idea to recap the different strategies, techniques, and tools that we have used to solve these equations. Factoring, trigonometric formulas, taking square roots, and inverses are the big ideas that students will want to think of as they solve these equations.
There is nothing I love more in math class than a surprising result. Today we will encounter a big surprise – solving the equation algebraically with legitimate steps will produce a solution that does not work when plugged back in to the original equation. Once students discover this, you can feel the increased determination in the room – they know that their algebraic work was correct and they will want to know why their answer will not work.
I give students the task worksheet and have them work on it with their table group. After a few minutes, I discuss questions 1 and 2 so that students are focused on the need to square both sides so they can use a Pythagorean substitution. When giving them instruction for the rest of the worksheet, I play off checking their solutions on a graphing calculator as a way for them to prepare for their upcoming test. I tell students that they could use a graphing calculator on the exam and I wanted them to practice checking their solutions – they have no idea that I am really setting them up!
Share and Summarize
The groups of students will likely get 90°, 180°, and 270° as their solutions and will be convinced that these are correct since all of their work is correct. When they plug them back into their equation or solve by graphing, they will see that 270° is not a solution. I let them struggle with this for a little while; automatically intervening will crush their flow. Some students may graph the squared version of the equation and will get 270° as a solution. This will be a great conversation – show students the graphs of the original equation and the squared version. More on this in the video below.
If no one realizes that the squaring produced the extraneous solution, I will give them an analogous equation that does not involve trig functions. Ask a student what the solution to the equation x = 3 is and then square both sides and ask what the solutions to the new equation are. They will see that -3 is a solution to the squared version but not the original. I end the lesson by summing up our work – dividing an equation by a variable my cause you to lose solutions and multiplying by a variable may cause you to gain extraneous solutions.
Here is an assignment for students to work on what we learned today. Each of these problems could be solving by squaring both sides of the equation, but there may be other solution pathways as well. For #1, students may notice that (1 + cos x)/(sin x) is the reciprocal of tan(½x), so they may replace (1 + cos x)/(sin x) with cot(½x). If students compare their solutions they will see that if they use cot(½x) they will not have an extraneous solution since they did not square both sides.