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:
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
[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
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.
Original URL: http://www.purplemath.com/modules/proving3.htm Copyright 2009 Elizabeth Stapel; All Rights Reserved.
Copyright 2009 Elizabeth Stapel; All Rights Reserved.