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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:

    [sin(theta) - cos(theta) + 1] / [sin(theta) + cos(theta) - 1] = [sin(theta) + 1] / cos(theta)

  • 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!

      multiplication, showing result of sin^2(theta) - cos^2(theta) + 2sin(theta) + 1

    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:

      [2sin(theta) + 2sin^2(theta)] / [2sin(theta)cos(theta)]

    I can factor and then cancel:

      [2sin(theta) * (1 + sin(theta))] / [2sin(theta)cos(theta)] = [1 + sin(theta)] / cos(theta)

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
    http://www.purplemath.com/modules/proving3.htm. Accessed
 

 



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