trig question (2 Viewers)

[natman87]

prove:

(1+c-s) / (1+c+s) = c/(1+s)

where
s = sinA
c = cosA

i can't do it..........

 

[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.

Note: Theres probably a more formal way to do it, i just felt like an algebra bash

 

[100percent]

(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.

Note: Theres probably a more formal way to do it, i just felt like an algebra bash

i think the examiners would prefer it if you just played with one side instead of both LHS & RHS

LHS
1+c-s
--------
1+c+s

1+c-s 1-c-s
------- x -------
1+c+s 1-c-s

1-2s-c²+s²
--------------
1-c²-2cs-s²

1-2s-(1-s²) +s²
-------------------
1 - (c²+s²)-2cs

2s² -2s
---------
-2cs

-2s(1-s)
----------
-2cs

1-s 1+s
---- x -----
c 1+s

1-s²
-------
c(1+s)

c²
-------
c(1+s)

c
-----
1+s

RHS

 

[acmilan]

Why? It doesnt say anything like that, and you often get questions where you have to play with both sides to get to some sort of equality. They cant drop marks for using a correct method unless they specify which method to use.

 

[100percent]

Why? It doesnt say anything like that, and you often get questions where you have to play with both sides to get to some sort of equality. They cant drop marks for using a correct method unless they specify which method to use.

ok ok, i said 'think', wasn't implying you were wrong

 

[Riviet]

acmilan, there is an acceptable way to prove using both LHS and RHS. For example, you can start with:
LHS= this
=that
=something that you can't simplify
Then RHS= this
=that
=the last line that you couldn't simplify on the LHS
However, in my opinion it doesn't work very well especially if one side is really simple like x+y or sinx.

 

[香港!]

Prove: (1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)

let t=tan(A\2)
(1+cosA-sinA) / (1+cosA+sinA)
=(1+(1-t²)\(1+t²)-2t\(1+t²))\(1+(1-t²)\(1+t²)+2t\(1+t²))
=(1+t²+(1-t²)-2t)\(1+t²) \ (1+t²+(1-t²)+2t)\(1+t²)
=(2-2t)\(2+2t)
=(1-t)\(1+t)

cosA/(1+sinA)=(1-t²)\(1+t²) \ (1+t²+2t)\(1+t²)
=(1-t²)\(t+1)²
=(1-t)(1+t)\(1+t)²
=(1-t)\(1+t)
.: (1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)

 

[acmilan]

^Mr hong kong just showed you how using both LHS and RHS is a good method. Sometimes you come across questions where it is almost impossible, if not very hard, to prove both sides are equal without manipulating both.

 

[Riviet]

Ah yes... you can use t as a tool for playing around with both sides. It provides an alternative approach to these trig proofs n_n

 

[ngai]

(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.

Note: Theres probably a more formal way to do it, i just felt like an algebra bash

that will probably get u 0 in an exam
i know what u meant to say...or at least what u shouldve said
but never say "if the question is true, then ..., which is true, so the question was true"

anyway, a straight forward method is using LHS - RHS and make common denominator and etc
nothing too tricky, and no need for t

 

[Slidey]

Why would it get 0, ngai? He did not use the question to 'prove' the question, after all.

In fact what he has done is essentially equivalent to proving LHS-RHS=0 (the manipulations and concept are the same), which is definitely an acceptable method of proof.

Let me format his proof a little differently:

To prove: (1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)

LHS-RHS = (1+cosA-sinA) / (1+cosA+sinA) - cosA/(1+sinA) [now find common denominator]
= {(1+cosA-sinA)(1+sinA)-cosA(1+cosA+sinA)}/{(1+sinA)(1+cosA+sinA)}
= (1+sinA+cosA+cosAsinA-sinA-sin^2A-cosA-cos^2A-cosAsinA)/(1+sinAcosA+sin^2A)
= (1-sin^2A-cos^2A)/(1+sinAcosA+sin^2A)
= 0 #

I know people get pretty annoyed when you start manipulating LHS and RHS all over the place (and that markers might be inclined to give you zero), but that doesn't mean your manipulations are incorrect - it means that those markers are incorrect.

 

[ngai]

Why would it get 0, ngai? He did not use the question to 'prove' the question, after all.

because he said "If that is true, then:"

 

[acmilan]

Well you could always replace it, assume it isnt true and prove by contradiction.

 

[KFunk]

because he said "If that is true, then:"

I've never fully understood the issues with the direction of a proof. It seems to be an issue of semantics 'cause both are logically saying the same thing (aren't they?):

Case 1: Assume p. p implies q. q is true, ∴ p is true. (as acmillan did)

Case 2: q is true. q implies p. ∴ p is true. (conventional method)

Is there an inconsistency between the two cases?

 

[who_loves_maths]

^ some proofs in maths are "iff" truths, and some are not necessarily. your case 1 KFunk differentiates between these two types of mathematical proofs, hence is not logically universal.

however, luckily, in the case of this problem, the identity (1+c-s) / (1+c+s) = c/(1+s) is an "iff" truth. as Ngai said, acmilan's "proof" would score him zero in a formal test, but if he were to simply add a second side to the proof by reversing the steps of his method to start from the last line and derive the identity, then the proof would be complete and irrefutable.


alternatively, usually in mathematics in order to overcome the logical slip in the statement of your case 1, mathematicians tends to write:

Case 1: Assume the opposite of p {i.e. assume p is not true}. the opposite of p implies q. q is false. ∴ initial assumption is wrong. p must be true.

but of course since the 30s, thanks to Godel, we know that that argument is no longer valid in its entirety - it is "incomplete".
the logic only works on statement 'p' IFF the condition that "p is only either true or false" is satisfied. {which it is for this trig. problem.}

 

[KFunk]

I think I've missed what the logical slip is with my case 1. For the sake of argument let q be something like sin²θ+cos²θ = 1 which is, at least in the MX1 forum, an irrefutable truth.

 

[KFunk]

i think I might get it. Is it the difference between say:

'a' represents sin²θ+cos²θ = 1

'b' represents statement x² + y² = 1

where if a is true then not(a) is untrue but if b is true then that does not logically imply that not(b) is untrue?

 

[acmilan]

I was wrong in that proof because I just assumed that the steps can be reversed without directly doing it. For this question, basically, you cannot prove something is true by supposing it is true and showing this assumption implies something which is obviously true. You have to check that each step could be reversed. This is what is called an 'all' statement, and a proof has to come from a contradiction or if you can get a double implication, which is implied in the answer.

 

[who_loves_maths]

Originally Posted by KFunk
i think I might get it. Is it the difference between say:

'a' represents sin2θ+cos2θ = 1

'b' represents statement x2 + y2 = 1

where if a is true then not(a) is untrue but if b is true then that does not logically imply that not(b) is untrue?

^ yes the idea is similar, but not the same... your examples play on the fact that the parameters of the equations are not set, so it's not really what i was talking about before - in fact, your statements themselves are not even complete. you have not defined what the symbols "theta", "x", or "y" mean.
however, i understand what you are saying.

put it this way; in your initial case 1

Case 1: Assume p. p implies q. q is true, ∴ p is true. (as acmillan did)

the logical slip is between "p implies q" and "q is true, ∴ p is true."; you needed to have also stated that "q conversely implies p". but because you didn't, the case is not complete.
eg. if there is a statement p such that it implied a statement q, but q cannot be conversely used to imply p, then your argument would fail.
{and don't say there's no such thing, because there actually is that's why there are 'iff' proofs in maths, i just can't think of any example off the top of my head right now sorry.}

so if we apply this to acmilan's method, we see that in trying to prove that LHS = RHS, he 1) assumed its truth, and then 2) found an implication of it, and 3) proved the implication (i.e. NOT the initial statement!), and then 4) loosely made an intuitive jump to say that "because the implication is true, then the initial assumption must also be true".

but notice that the logical direction of the argument "because the implication is true, then the initial assumption must also be true" goes from implication -> assumption.
SO, unless he can prove that the implication can indeed go backwards and be used to arrive at the initial statement of RHS = LHS again, then the direction of argument is based on an airy assumption that the implication can go backward - i.e. acmilan's proof holds this implicit assumption which he has not proven to be true!

i know that for this trivial trig. problem, it's obvious to see that sin^2 + cos^2 = 1 can be made to form RHS = LHS again by reversing acmilan's steps, but the point is that he didn't actually do it - so his "proof" was incomplete.

remember, the questions asks to prove the statement RHS = LHS, not to merely prove the implications of that statement. logically, it's not enough to simply do that.

 

[who_loves_maths]

getting back to the initial discussions of methods of solving this problem, here's one:

- it's not as elegant as other methods potentially are, but i did it to illustrate my point with acmilan's method:
at the end of using the quadratic formula to solve it, you get two potential answers for 'x'. now, you can very well say that both answers are derived implications of the initial statement LHS = x.
so since they are derived from the initial statement, then are they both correct? no, because the initial statement only had one equality, not two. hence, to check which one is right, we need to refer the implications backward to the initial statement.


p.s. also, the method of elimination of one of the 'x' values is done through contradiction.