trig question (2 Viewers)

[KFunk]

Hmm, thanks for the explanation. I still think one could argue, however, that there's nothing wrong with using such logic when the issues you describe are 'trivial' (as you put it). If all the operations are perfectly reversible the p implies q also necesitates q implies p , does it not (correct me on any slip ups)?

The main thing which bugs me is it seems that this kind of proof isn't accepted due to a hang up from more complex mathematics where such logical slip ups are an issue. I understand if mathematicians want to teach good practise for later on but otherwise, in these trivial cases, it would seem that the logic works.

--> insert your proof of my incorrectness here

 

[who_loves_maths]

^ lol, yes i think i already agreed in my other post that "If all the operations are perfectly reversible the p implies q also necesitates q implies p."
you are correct in saying this, but the point of my other post was to show that this step in logic is necessary. it does not matter if the problem is "trivial" - since triviality is a subjective notion.

like i had said, i know you can immediately see that the steps are reversible in this case, but the point is to put it on paper.

 

[KFunk]

There should be some statement you can make to justify a backwards proof. It feels like this is a grammatical (not a mathematical) issue. It's like the way good is favoured over bad, first before last and over before under ---> likewise forwards is favoured over backwards. In conclusion: I hold a certain bitterness over things like this, I have been subjected to too much post modernism and I think acmillan's original solution is perfectly valid. [/rant]

 

[who_loves_maths]

^ well i disagree with you on this. acmilan's solution is, by his own admission, not valid. and this is certainly not a grammatical error - it is a logical one. and that makes it mathematical.

anyways, onto something else. here's another form (1+cosA-sinA) / (1+cosA+sinA) can be written as in half angles:

- this is like a half-angle formula 'supplement' to 香港!'s other post in which he cleverly solved the problem using 't' variables.

 

[KFunk]

Never mind, you've explained it enough as is. I think it's an agree to disagree situation .

 

[who_loves_maths]

^ like i said, the flaw is that he didn't connect the implication back to the initial statement after arriving at sin^2 + cos^2 = 1. this is the logical flaw - which at the time, he did not realise.
and even if he did or does realise now, the fact remains that his proof is still incomplete in its current form:

Originally Posted by acmilan
(1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)

If that is true, then:

(1+cosA-sinA)(1+sinA) = cosA(1+cosA+sinA)
1+cosA-sinA + sinA+sinAcosA-sin2A = cosA+cos2+sinAcosA
1-sin2 = cos2
sin2+cos2 = 1

Which we know to be true, hence the original equation is true.

 

[KFunk]

I geuss the way I'm looking at it is that all of his operations are reversible, so a proof in one direction implies both.

 

[who_loves_maths]

^ i understand your reasoning. but in maths it's not enough to "see" or feel that something is intuitively correct. it's actually putting it down on paper as a proof. imagine if this was not a trivial case, then even if you can see a connection, it may still be hard to put in mathematical equations when you get to writing it all out again.

logic isn't something you can just decide to impose on one thing in maths and let another thing, trivial or not, slip by... it has to be applied consistently.

but anyways, yes i do 'agree' that it's time to leave it at "we disagree" now .