Algebra for Equivalence
Date: 11/12/2011 at 17:48:42 From: Steven Subject: Why do I lose solutions when solbing trig equations? My teacher explained in class that I would lose possible solutions to trigonometric equations if I divided by a function or took the square root of a function. I want to avoid any kind of loss. What are all the operations that do this in trigonometry? And could you please explain to me how they destroy solutions?
Date: 11/12/2011 at 23:39:55 From: Doctor Peterson Subject: Re: Why do I lose solutions when solbing trig equations? Hi, Steven. In general (and this has nothing to do with trig, really), you should avoid doing anything that you don't KNOW will produce an equivalent equation. Don't just do things that feel reasonable; make a list of what has been proved to be valid in this context. When in doubt, don't; or at least, go check it out and make sure you can do it. The examples you give (which may be the only ones at your level) are the opposite of the actions that produce extraneous solutions -- but these ideas are closely related. The actions you're asking about accidentally reduce the solution set of a problem so that you miss solutions; the opposite actions accidentally expand the solution set so that you find solutions that are not solutions of the original. So searching for information on extraneous solutions may give you some useful insights here. The following pages, in particular, may be of interest: Extraneous Roots http://mathforum.org/library/drmath/view/52659.html Equality Properties and What They Really Mean http://mathforum.org/library/drmath/view/72802.html Fractions and Trig Functions http://mathforum.org/library/drmath/view/54188.html First, if you divide both sides of an equation by an expression containing a variable, you are making an assumption: that what you are dividing by is not zero. If, for some value of the variable, the divisor is zero, then since division by zero is not defined, that value may not be a solution of the new equation, even though it was a solution of the original. It's as if you were searching for a criminal hiding in your neighborhood, and you don't bother to look in your own house because that possibility doesn't even occur to you. If he's there, it's your own fault that you didn't find him. For example, take the simple equation x^2 = x If you see that you can divide both sides by x, and do so, you get x = 1 But in dividing by x, you were assuming that x was not zero. If x were zero, then when you had ... x^2 x --- = --- x x ... you would really have 0/0 on both sides; the left side is not x, and the right side is not 1. This case is just quietly swept under the rug when you simplify. So you really have to make a separate check for that case. Here, it turns out that x = 0 is in fact a solution. You missed it because you didn't consider the possibility. The better way to do this, which doesn't require a separate check, is to factor rather than divide: x^2 - x = 0 x(x - 1) = 0 x = 0 or x - 1 = 0 x = 0 or x = 1 This will always work if you could have divided; it takes that x you divided by in the other method, and keeps it out in the open, so you won't miss it. As for taking a square root, I'd like to see an example where you've done that, but the case I have in mind is like this: x^2 = 1 sqrt(x^2) = sqrt(1) x = 1 Here, the issue is a little different: nothing is undefined; you just forgot that any positive number has TWO square roots, not just one. The radical symbol represents only the principal root (the non-negative one), so you have to explicitly put a plus-or-minus symbol before it to tell you to use both signs: x^2 = 1 sqrt(x^2) = +/-sqrt(1) x = +/-1 x = 1 or x = -1 (By the way, technically we should put the +/- on the left side as well, but that wouldn't gain us anything, since changing the sign on one side has the same effect as changing the sign on the other.) Does that help? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 11/13/2011 at 02:09:24 From: Steven Subject: Thank you (Why do I lose solutions when solbing trig equations?) Thank you for explaining all of this to me! It really helps. :D.
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