proofs (maths)

1. Yes. Why wouldn't it be ok? If you were really concerned you could just write "read backwards" on your paper

2. I highly doubt there's a defined amount. They're easy to figure out though, just practice a lot of manipulating trig formulae.

Well its quite obvious there is an infinite amount of trig identities. Take cos(a+b) for example if you allow a nd b to be any interger, of which there is an infinite number, then you have an infinite number, Or sin(a+b) sp there is another infinity, of course you still only have infinity(Geore Cantor called this aleph null) Of course if a and b are reals then you have aleph one. And so on. But there are other methods of proving them then sheer manipulation. e.g maclaruin and de moivre.

PurpleFistMixer

Well I think (s)he meant general identities, as in... not using any numbers. The method for proving cos(a-b) is gonna be the same as cos(c-d)...

md99

The trig proofs are mostly fairly simple, only 4 with a bit of a kick (ie requires a little learning) are proof of the Sine Rule, Cosine Rule, Sin (A-B)/A+B and Cos (A-B)/A+B, the rest can be done with the log tables really, as long as you know the first step, ie

Prove: Cos 3A = ....
Step 1: Cos 3A = Cos (2A+A), etc (work from the tables)

sd123 wrote:

2 things bout trig q's part (b) proofs. (not learn off proofs)

1. is it ok to work from the rhs to get your answer equal to the lhs and vice versa

Although I may seem pedantic, just a point I'd like to stress: this only works if all of your steps are reversible. For example:
4=3
4x0=3x0
0=0
Which is clearly true.

Just because the last line is true, and because only elementary operations have been used (multiplication), doesn't mean that the first line is. That is, if we try to work backwards, how do we go from 4x0=3x0 to 4=3? Divide by 0. But of course this isn't valid.

This is a pretty basic example that is easy to see as incorrect, but the same problem can sneak up in more subtle ways! LC examiners will know whether the steps are reversible or not, so if you're steps can be worked backwards, you'll get full marks. But it can be easy to make mistakes that way, and if you have time it can be worthwhile to go back through the steps yourself, or even write them out backwards (and as most LC proof problems are only a few steps, this would only take a minute). Not for the examiners sake, but to make sure your steps are okay.

Just something to keep in mind.

PurpleFistMixer wrote:

Well I think (s)he meant general identities, as in... not using any numbers. The method for proving cos(a-b) is gonna be the same as cos(c-d)...

I disagree because the proofs for identities such as cos(3a) are examined quite regularly also.(note these are also general-in such that the angle A is not defined) And I was actually more making the point to JC 2k3 that there do exist an infinite number of different trig identities. e.g

Sin(3a)+sin(a) -sin(2a)
Divide by
Cos(3a)+Cos(a) - Cos(2a)
is equal to
Tan(2a)

Which can be further generelised again.

PurpleFistMixer

dan719 wrote:

I disagree because the proofs for identities such as cos(3a) are examined quite regularly also.(note these are also general-in such that the angle A is not defined) And I was actually more making the point to JC 2k3 that there do exist an infinite number of different trig identities. e.g

Aye but when you get into it, cos(4a) is just cos(2b) where b = 2a, so you're gonna start getting repetition and that kind of stuff. I don't think it's right to say that just because there are an infinite number of numbers, that there's an infinite number of trig identities. There may well be, I don't know, but I wouldn't base a claim on such logic as that.

rjt wrote:

Although I may seem pedantic, just a point I'd like to stress: this only works if all of your steps are reversible. For example:
4=3
4x0=3x0
0=0
Which is clearly true.

Just because the last line is true, and because only elementary operations have been used (multiplication), doesn't mean that the first line is. That is, if we try to work backwards, how do we go from 4x0=3x0 to 4=3? Divide by 0. But of course this isn't valid.

This is a pretty basic example that is easy to see as incorrect, but the same problem can sneak up in more subtle ways! LC examiners will know whether the steps are reversible or not, so if you're steps can be worked backwards, you'll get full marks. But it can be easy to make mistakes that way, and if you have time it can be worthwhile to go back through the steps yourself, or even write them out backwards (and as most LC proof problems are only a few steps, this would only take a minute). Not for the examiners sake, but to make sure your steps are okay.

Just something to keep in mind.

That's retarded. You started off with 4=3, which isn't true. EVERYTHING is reversible in maths unless it involves dividing by 0.

dan719 wrote:

And I was actually more making the point to JC 2k3 that there do exist an infinite number of different trig identities

Where the hell did I say there wasn't????

PurpleFistMixer wrote:

Aye but when you get into it, cos(4a) is just cos(2b) where b = 2a, so you're gonna start getting repetition and that kind of stuff. I don't think it's right to say that just because there are an infinite number of numbers, that there's an infinite number of trig identities. There may well be, I don't know, but I wouldn't base a claim on such logic as that.

Define "trig identity".

There are 3 basic trig ratios: sine, cosine and tangent. If you say well what about (sinA)^2 or sin2, well then you can say (sinA)^n or sin(nA) with lim n->infinity, and therefore there are an infinite amount of trig identities.

JC 2K3 are you agreeing with me? Of course we need to define an idendity. I think this is suitable.

An identity is defined as any expression which can be logically proven from another area of mathematics(e.g cosine rule or demoivre)

Therefore using above logic(infinite numbers.. ) there exist an infinite number of identities.

dan719 wrote:

JC 2K3 are you agreeing with me?

Yes. When you said you were trying to prove a point to me all I had said in the thread was I didn't think that SEC had a definate defined amount(ie. a list) of trig equalities they could ask students to prove in the exam.....

JC 2K3 wrote:

Yes. When you said you were trying to prove a point to me

You mistake my intention. I was trying to give your intuition a rigourous logical foundation. Not 'prove a point' in the weak, petty sort of way!:D

Fair enough.

PurpleFistMixer

JC 2K3 wrote:

Define "trig identity".

I wouldn't, but I'd understand it to be some kind of relationship between the ratios, you know, like cos^2 + sin^2 = 1. Though in that respect I guess it'll all boil down to the essential three ratios because all the trig identities (I've seen so far, anyway) can be derived from each other.

ZorbaTehZ

SD123 wrote:

1. is it ok to work from the rhs to get your answer equal to the lhs and vice versa?

JC 2K3 wrote:

1. Yes. Why wouldn't it be ok? If you were really concerned you could just write "read backwards" on your paper

Wrong.
You should take it as a rule of thumb that you are to manipulate the LHS so that you come out with the RHS.
I thought that too, that it doesn't matter whether you manipulate the LHS/RHS but it seems it does - I got docked marks in my mocks for doing it and when I challenged my teacher on it she showed me the marking scheme and it stated clearly that I was to have used the LHS. Saw at least two other marking schemes from past papers that said that too in the notes.

mmk, if I do it in the Lc i'll write it out backwards quickly, problem solved.

sd123

one other thing:
in the proof of de moivre's theorem how does it go from CosKxCosx - sinKxsinx + i(sinKxCosx + CosKxSinx) to Cos(k+1)X +isin (k+1)x

md99

sd123 wrote:

one other thing:
in the proof of de moivre's theorem how does it go from CosKxCosx - sinKxsinx + i(sinKxCosx + CosKxSinx) to Cos(k+1)X +isin (k+1)x

always wondered that actually, but i'm happy skipping it if it's allowed?

sd123 wrote:

one other thing:
in the proof of de moivre's theorem how does it go from CosKxCosx - sinKxsinx + i(sinKxCosx + CosKxSinx) to Cos(k+1)X +isin (k+1)x

Basic formula from the maths tables.....

Cos(A+B) = cosAcosB - sinAsinB
Sin(A+B) = sinAcosB + sinBcosA

md99 wrote:

always wondered that actually, but i'm happy skipping it if it's allowed?

lol, are you saying you simply learn each line in a proof off by heart??

sd123

but it is ok to skip that and give the answer tho?

As long as it's clear what you're doing I think it's fine..

You'd probably be fine with:
(CosKxCosx - sinKxsinx) + i(sinKxCosx + CosKxSinx)
Cos(kx+x) + iSin(kx+x)
Cos(x(k+1)) + iSin(x(k+1))

But why not just do:
CosKxCosx - sinKxsinx + i(sinKxCosx + CosKxSinx)

(CosKxCosx - sinKxsinx) = Cos(kx+x)
(sinKxCosx + CosKxSinx) = Sin(kx+x)

therefore:
CosKxCosx - sinKxsinx + i(sinKxCosx + CosKxSinx) =
Cos(kx+x) + iSin(kx+x)
Cos(x(k+1)) + iSin(x(k+1))

JC 2K3 wrote:

That's retarded. You started off with 4=3, which isn't true. EVERYTHING is reversible in maths unless it involves dividing by 0.

Firstly, there are plenty of irreversible operations. Take a look at this (and there are more too). Secondly, I know 4!=3, but I was making a point. When working backwards, you can 'prove' anything=anything. So although you start with the trig identity/inequality/whatever you have in the exam, you can take steps to reach a true statement, regardless of whether or not the first statement is correct or not. So if your steps aren't reversible, you've proved nothing. In the example, the mistake was obvious. But as I said, there are more subtle examples. It comes up a lot of different areas in the LC, not just problems involving working backwards (for example, inequalities&squaring, taking a periodic function of two sides, eg. sin, and adding extra roots, as the link talks about, etc.).

Anyway, all I said was to check your work. No harm in stressing this, especially when a lot of people don't get it, eh?

rjt wrote:

Firstly, there are plenty of irreversible operations. Take a look at this (and there are more too). Secondly, I know 4!=3, but I was making a point. When working backwards, you can 'prove' anything=anything. So although you start with the trig identity/inequality/whatever you have in the exam, you can take steps to reach a true statement, regardless of whether or not the first statement is correct or not. So if your steps aren't reversible, you've proved nothing. In the example, the mistake was obvious. But as I said, there are more subtle examples. It comes up a lot of different areas in the LC, not just problems involving working backwards (for example, inequalities&squaring, taking a periodic function of two sides, eg. sin, and adding extra roots, as the link talks about, etc.).

Anyway, all I said was to check your work. No harm in stressing this, especially when a lot of people don't get it, eh?

Your example was completely unrelated to your point and unrelated to LC trig proofs where there's never any squaring involved and it's always stated that no trig function which is a divisor is equal to zero.

JC 2K3 wrote:

Your example was completely unrelated to your point and unrelated to LC trig proofs where there's never any squaring involved and it's always stated that no trig function which is a divisor is equal to zero.

My original point was about working backwards in general, not specific to trig proofs, and certainly not specific to LC questions. No harm in being rigorous, you know?

MathsManiac

JC 2K3 wrote:

Your example was completely unrelated to your point and unrelated to LC trig proofs where there's never any squaring involved and it's always stated that no trig function which is a divisor is equal to zero.

Entertaining as it is to watch JC 2K3 attempting to extricate him/her-self from the error that rjt's flawless logic has identified, I would like to point out that this is not the question originally asked.

rjt correctly asserts that if you want to prove an identity by reducing it through a series of steps to a true statement, you must ensure that all steps are reversible. The original question did not describe such a process however. It instead asked whether you could start with the rhs and show that it's equal to something else, which is equal to something else, etc. ... equal to the lhs. This is of course perfectly acceptable, since at each step you are not "doing something to both sides of an equation", but rather transforming an expression in some way. (Obviously you need to be sure you're transforming the expression correctly at each step.)

i.e. You're being asked to prove that A=B, and you do it by showing that:
B=C
=D
=E
=F
...
=A.

No problem.

MathsManiac wrote:

Entertaining as it is to watch JC 2K3 attempting to extricate him/her-self from the error that rjt's flawless logic has identified, I would like to point out that this is not the question originally asked.

rjt correctly asserts that if you want to prove an identity by reducing it through a series of steps to a true statement, you must ensure that all steps are reversible. The original question did not describe such a process however. It instead asked whether you could start with the rhs and show that it's equal to something else, which is equal to something else, etc. ... equal to the lhs. This is of course perfectly acceptable, since at each step you are not "doing something to both sides of an equation", but rather transforming an expression in some way. (Obviously you need to be sure you're transforming the expression correctly at each step.)

i.e. You're being asked to prove that A=B, and you do it by showing that:
B=C
=D
=E
=F
...
=A.

No problem.

Ah, indeed. I misunderstood the initial question.

MathsManiac

rjt wrote:

Ah, indeed. I misunderstood the initial question.

But if you hadn't started this interesting conversation, I wouldn't have had the opportunity to try to think up a trig inequality that you can prove by squaring both sides of something, like this one:

cos^4 (A) + sin^4 (A) = 1 - 1/2 sin^2 (2A)

MathsManiac wrote:

cos^4 (A) + sin^4 (A) = 1 - 1/2 sin^2 (2A)

Except that the easiest way to prove that is to use de moivre!:D

MathsManiac

dan719 wrote:

Except that the easiest way to prove that is to use de moivre!:D

You reckon that's easier than squaring cos^2 (x) + sin^2 (x) = 1 ???

Well, whatever floats your boat.

I also thought of supporting rjt's reversibility point with a trig inequality that MIGHT land one in trouble if trying to work backwards, like this, maybe:

cos^4 (x) > sin(2x)cos(x)-1