Proving Trig Identities vs. Solving Trig EquationsDate: 10/08/2006 at 13:25:51 From: Angel Subject: Trigonometry: Prove that a proof = 0 I need to prove that: sinx/(1 - cosx) - (1 + cosx)/sinx = 0. I am told that I need to show that the left side of the proof is equal to zero. I thought that I had done this so now I am really confused. Here's what I did: First: Use the additive property to add (1 + cosx)/sinx to both sides of the equation to give sinx/(1 - cosx) = (1 + cosx)/sinx. Next: Multiply both sides of the equation by sinx and also by (1 - cosx) which will give sin^2x = (1 + cosx)(1 - cosx) or 1 - cos^2x so we have sin^2x = 1 - cos^2x. Then: Use the additive property and add cos^2x to both sides, giving sin^2x + cos^2x = 1 This is the Pythagorean identity. It holds for all real values of x. This proves that the original equation is also true for all values of x except for where either sinx = 0 or where 1 - cosx = 0. This means that the equation is true except for where x = n(pi), where n is any integer. Sin(x) = 0 when x = n(pi) Cos(x) = 1 when x = 2n(pi) Therefore x =/= n(pi) Date: 10/08/2006 at 22:55:02 From: Doctor Peterson Subject: Re: Trigonometry: Prove that a proof = 0 Hi, Angel. I think you are saying that you submitted this proof, and were told it was wrong, without explanation. Is that right? It is basically valid, but is not in the form that many teachers require, namely working only on one side of the equation at a time to transform one side into the other. You should have been told that, rather than only being told that it is wrong, so that you would know how to correct it. What you have done DOES consitute a proof, IF you state it very carefully. You have done what you would do to SOLVE an equation, which is a somewhat different process. What you need to do here is to explicitly state that you are transforming the equation into an EQUIVALENT equation--one that is true when the original equation is true, and vice versa. This is not strictly true, since multiplying an equation by an expression containing the variable does not necessarily result in an equivalent equation, since that expression might be zero. You haven't ignored this problem, as many students would do, but have pointed out that you would have multiplied by zero only if the variable is a multiple of pi. The one thing you didn't mention is that in those cases, the equation you are trying to prove can't be evaluated in the first place, since you would be dividing by zero; so you have shown that it is true over its entire domain. With that one little exception, I think you've got a fine proof. The trouble, as I said, is that many teachers don't accept this-- because, I think, they want to avoid the subtleties involved in this sort of thinking (essentially protecting their students from the errors that could creep in if they were not as careful as you) by only allowing them to take the "safe" path, which consists of a single chain of equal expressions. You can usually turn an "unsafe" proof such as yours into this kind of proof by sort of turning it on its side, doing to the left side only, the operations equivalent to what you did with the whole equation. Here's how: Your first step just moved one fraction to the other side; we have to keep the whole left side together, so this can just be ignored. The second step is to multiply by the LCD. To do the same thing without changing the value of the expression, you can instead rewrite the fractions with a common denominator (the LCD), and then add them. That is, rewrite them both with denominator (sinx)(1 - cosx). Your third step just moved part back to the other side; the equivalent is just to simplify the combined fraction you now have. Then you can use the Pythagorean identity to simplify it further, replacing sin^2x + cos^2x with 1, and you will be done. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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