A Trick for Proving Trig Identities
Purplemath
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.
For instance, 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 proof:
- Prove the following identity:
![[sin(theta) - cos(theta) + 1] / [sin(theta) + cos(theta) - 1] = [sin(theta) + 1] / cos(theta)](trig/prove17.gif){kind=small}
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 top and bottom by this expression. Since this creates a difference of squares in the denominator, 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)
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. So I'll do the multiplication vertically.
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)]](trig/prove19.gif){kind=small}
I can factor a 2sin(θ) from the two terms in the numerator, and then cancel:
![[2sin(theta) * (1 + sin(theta))] / [2sin(theta)cos(theta)] = [1 + sin(theta)] / cos(theta)](trig/prove20.gif)
Putting it all together, starting on the lhs, my proof is:
Don't expect always, or even usually, to be able to "see" the solution steps to a proof when you start working on the proof. Expect to be confused, at least a little, at least some of the time. 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.
You can do this.
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