Proving Trigonometric Identities (page 3 of 3)
When you were back in algebra, you rationalized complex and radical denominators by multiplying by the conjugate; that is, by the same values, but with the opposite sign in the middle. If the denominator was a complex value, like 3 + 4i, you would rationalize by multiplying, top and bottom, by 3 – 4i. In this way, you'd create a difference of squares, and the "i" terms would drop out, leaving you with the rational denominator 9 – 12i + 12i – 16i2 = 9 – 16(–1) = 9 + 16 = 25.
Every once in a very great while, you'll need to do something similar in other contexts, such as the following:
- Prove the identity:
This is just a mess! The only stuff I have with 1's in them are the Pythagorean identities, and they have squared stuff in them. So they won't work here. But what will happen if I multiply the LHS, top and bottom, by the "conjugate" of the denominator?
The denominator can be stated as [sin(θ) + cos(θ)] – 1; then the conjugate would be
[sin(θ) + cos(θ)] + 1. I'll multiply the bottom by this; since this creates a difference of squares, the result is:
[sin(θ) + cos(θ)]2 – 1 = sin2(θ) + 2sin(θ)cos(θ) + cos2(θ) – 1
The two squared terms simplify to just 1, so I get:
sin2(θ) + cos2(θ) + 2sin(θ)cos(θ) – 1
1 + 2sin(θ)cos(θ) – 1
2sin(θ)cos(θ) Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
Now for the numerator. Just as when I was working with complexes and radicals back in algebra, the multiplication across the top is going to get pretty nasty!
Well, while the denominator sure simplified, I've still got some work to do with the numerator. I'll move the sine out in front of the squared terms, and then restate the 1 using the Pythagorean identity:
2sin(θ) + sin2(θ) – cos2(θ) + 1
2sin(θ) + sin2(θ) – cos2(θ) + sin2(θ) + cos2(θ)
2sin(θ) + 2sin2(θ)
...because the squared cosine terms cancelled out. So this is my fraction for the LHS:
I can factor and then cancel:
Don't expect always, or even usually, to be able to "see" the solution when you start. Be willing to try different things. If one attempt isn't working, try a different approach. Identities usually work out, if you give yourself enough time.
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Cite this article as:
Stapel, Elizabeth. "Proving Trigonometric Identities." Purplemath. Available from
In today's lesson it is important to take some time at the beginning of class to set the stage for students. I will talk about where we have been, where we are now, and where we are going in Trigonometry. I want students to believe that they now have a lot of knowledge from our prior units (Trigonometry Basics and Trigonometric Functions). In this unit we will pull everything together. Our first goal is to learn about Trig Identities. Later, we will apply these tools to solve equations.
Next, I will remind students what an identity is and how they are used in mathematics. I plan to have students add the definition of an identity to their notes for reference. Then, in teams, students will work on the three Warm Up problems. These problems are on Pages 3-5 of the Discovering Trig Identities Flipchart.
- Problem 1: Students are asked to text in one thing they already know about reciprocal trig functions.
- Problem 2: Students are asked to text in one thing they do not understand about reciprocal identities.
- Problem 3: Students are asked to text in one question they have about Trig Identities.
Depending on how quickly the lesson is moving, I may address some of these questions now or save them for down time later in the unit.
Teacher's Note: While students are responding to the warm-up prompts, I will call students up group-by-group to transfer a Ti-Navigator calculator file to their calculators.