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Maths problem

  • 31-05-2007 8:24pm
    #1
    Registered Users, Registered Users 2 Posts: 1,238 ✭✭✭


    My sister, who's doing higher level maths in fifth year, was given this question on a handout by her teacher before her classes ended this year. She can't do it, I can't do it and her grind can't do it! Can anyone out there work it out, or is the problem flawed?


Comments

  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    I think you just have to multiply across by the denominator, then use the formulas for SinACosB and CosASinB. Two of the terms will cancel and you'll be left with the numerator.

    Post back if you still have trouble, I'll try to clarify. Boards doesn't have an equation editor so it's a nightmare to type this stuff up.


  • Registered Users, Registered Users 2 Posts: 6,889 ✭✭✭tolosenc


    Log tables Page 9. I got it first! I'll type it up now.


  • Closed Accounts Posts: 4,147 ✭✭✭E92


    Its pretty much what Fremen said.

    Heres the answer.


    sin 5x + sin 3x
    = 2 sin 2x
    cos 3x + cos x

    from the log tables sin A + sin B = 2 sin (1 ∕2)(A+B) cos (1/2)(A−B)

    and cos A + cos B = 2 cos (1∕2)(A+B) cos (1∕2)(A−B)
    .˙. we can say that

    2 sin 4x cos x
    = 2 sin 2x
    2 cos 2x cos x


    sin 4x
    = 2 sin 2x
    cos 2x

    multiply across the denominator and you get
    sin 4x = 2 sin 2x cos 2x

    which proves it as from the log tables sin 2A = 2sinAcosA and replacing A with 2x and its right.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    I'd consider firing that grind teacher if they couldn't do that.....


  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    I'd consider firing that grind teacher if they couldn't do that.....

    ...and employing me :D

    It's funny, I had to remind myself that multiplication was commutative. Stupid college maths.


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  • Registered Users, Registered Users 2 Posts: 6,889 ✭✭✭tolosenc


    Eh, you can't use the RHS if you have to prove it. Sorta defeats the purpose.


  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Eh, you can't use the RHS if you have to prove it. Sorta defeats the purpose.

    Of course you can. LHS = RHS as shown. Multiply both sides by something and the equality holds. That's all that needs to be done.


  • Registered Users, Registered Users 2 Posts: 6,889 ✭✭✭tolosenc


    WHAT?! I'm gonna kill my physics teacher...


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    You can, sure with practically all inequality proofs you have to or else it's very difficult.

    In any case, when he's got it equal to sin(4x)/cos(2x) he just has to go:

    sin(4x)/cos(2x)
    = 2sin(2x)cos(2x)/cos(2x)
    = 2sin(2x)


  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Maybe your physics teacher was talking about a different kind of problem. Post it up and I'll take a look.


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  • Registered Users, Registered Users 2 Posts: 6,889 ✭✭✭tolosenc


    Ah, 'twas back in 5th year, I don't remember what it was. I finished the test like 10 mins before everyone else and he said it was because that's not how you prove stuff. You have to go through steps to make the LHS=the RHS. See my workings in the attached file above.


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    if you have to proove something, you are free to manipulate both sides of the equation.
    If you have to derive something, you are more limited and probably have to work only with the LHS.


  • Registered Users, Registered Users 2 Posts: 1,238 ✭✭✭Kwekubo


    Sorted. Thanks everybody for chipping in. My sister is slightly surprised at how willing people are to do maths online :rolleyes:
    cocoa wrote:
    if you have to proove something, you are free to manipulate both sides of the equation.
    If you have to derive something, you are more limited and probably have to work only with the LHS.
    That makes sense. I'm guessing that's what obl's physics teacher was referring to, since derivations are what feature on the physics paper.


  • Closed Accounts Posts: 4,147 ✭✭✭E92


    Derive is the same word as prove though! Like I remeber in 1997 you had to derive the formula for a² in terms of b, c and A. Then in 2001 they asked you to prove that a² = b² + c² - 2bc cos A. The same question, written a different way! Sometime the proofs in Physics are called derivations for instance. @obl Where does it say that you cant do what I did above to show that a proof is true?Maybe you're thinking of Inequialities, where you can't multiply both sides by say -2 as it changes the statement eg 3<5 , both sides by -2 -> -6<-10 which is a complete lie since -6 is actually greater than -10.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Well you can, you just have to reverse the sign if multiplying by a negative number....


  • Closed Accounts Posts: 4,147 ✭✭✭E92


    I know that what I was trying to say(though not very well;) ) is that you cant just simply multiply or divide for that matter an inequality and just leave it at that. I completely admit that I should have given a better example, like say what do you do if you have to multiply across by say x, when xER? Square it of course, but with an unknown how do you know if you need to change the signor not, which is the point I wanted to get across.


  • Registered Users, Registered Users 2 Posts: 6,889 ✭✭✭tolosenc


    Oh I dunno anymore
    *Maths World comes crashing down around me only a week before my Leaving Cert*


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    E92 wrote:
    Derive is the same word as prove though!
    no, it's not... Prove means show that it's true, so manipulating it until you have a statement you know is true is perfectly valid. Derive means show where it came from and doesn't necessarily involve showing that it is inherently true.


  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    cocoa wrote:
    no, it's not... Prove means show that it's true, so manipulating it until you have a statement you know is true is perfectly valid. Derive means show where it came from and doesn't necessarily involve showing that it is inherently true.

    I'd agree almost entirely! My point of disagreement is that I would consider that deriving the result does show that it's inherently true.

    I agree with your main point however: derive certainly implies that you should start from known stuff and progress to the result required; so, it would not be satisfactory to start with the result and just extablish that it's true.

    I think a good informal definition of "derive" might be: "Show how you could logically extablish this result if you didn't already know it".

    On the LC, it's often used in "Derive the MacLaurin series for..." which means that you're supposed to use the general definition of MacLaurin series to establish the result required, rather than just writing it down (or even writing it down and then proving it by other means).


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    no, derivation still doesn't necessarily show it is true... For example, derive from first principals. It is not necessary to show what first principals are or that they are true, you are allowed to work from assumptions. When proving something, you have to go the extra step.


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  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    I disagree.

    Plenty of LC Proofs require assumptions.


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    yes, and i think we'll all agree LC maths is plenty full of holes. In the real world, the distinction between proofs and derivations is as such...

    I'm not sure what the argument in LC maths for calling them proofs is, they might claim the assumptions are axiomatic, or just plain lie....


  • Closed Accounts Posts: 1,597 ✭✭✭dan719


    But in LC maths one has to start somewhere, by rights when proving inequalities about real numbers a and b, one should state the laws of asociativity and commuativity etc and begin from there, we can however assume them, due to the use of the definition real numbers.

    Also in the proof by induction of de moivre I think one should have to proof the sum of the two angles(when multiplying two complex numbers in cis form)(for completion) but examiners do not expect it.


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    i agree totally about the angles thing, it's not even difficult to prove...


  • Registered Users, Registered Users 2 Posts: 3,620 ✭✭✭Grudaire


    I physics you actually have to make the formula from scratch, If you prove it by making one side equal the other, then just write it all out backwards to get full marks eg to prove v=u+at

    F=(mv-mu)/t (Newtons law)

    ma=m((v-u)/t)

    a=(v-u)/t

    at=v-u

    v=u+at


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    F=(mv-mu)/t (Newtons law)

    ma=m((v-u)/t)
    Rather fallable logic in that step....

    I thought to "derive" v = u + at all you did was:

    a = (v - u)/t
    at = v - u
    v = u + at

    The only reason F = ma in the first place is because accelleration is defined as (v - u)/t


  • Registered Users, Registered Users 2 Posts: 3,620 ✭✭✭Grudaire


    But where did the logic that a=(v-u)/t come from?

    It's what I've been thought, and its not wrong so I'll not change!


  • Registered Users, Registered Users 2 Posts: 1,269 ✭✭✭cocoa


    it's the definition of acceleration... change in velocity over time.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    ^Yup.


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  • Closed Accounts Posts: 131 ✭✭Tomlowe


    ^i concur


  • Closed Accounts Posts: 4,147 ✭✭✭E92


    Why not do the 3 straight line montion formulas in Physics by integration? no learning, once you understand how to do integration.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    ^Because LC Physics doesn't teach integration.


  • Closed Accounts Posts: 4,147 ✭✭✭E92


    Just cause you're not thought it doesn't mean u can't use it! Thats the way we learned them. I know they dont show it on the marking scheme that way but just like L'Opital's rule for Trig Limits in Maths(which is also not on the Maths Syllabus and we're not supposed to be thought that either) you can use it, and rightly so, its still right(and a lot easier)


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    I dunno about easier, linear motion formulas are pretty damn easy to prove anyway.

    With so much else you have to assume in LC Physics, I think that for this subject I'll just use the rote learning method of proving things.


  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    cocoa wrote:
    no, derivation still doesn't necessarily show it is true... For example, derive from first principals. It is not necessary to show what first principals are or that they are true, you are allowed to work from assumptions. When proving something, you have to go the extra step.

    Not so. Every proof requires assumptions. Even in the most formal of maths logic contexts, (which we weren't talking about anyway,) proofs always involve either axioms or previously established theorems (or both).

    Indeed, a proof in such formal contexts is defined as a finite sequence of well-formed expressions, each of which is either an instance of an axiom or follows from earlier expressions in the sequence by a rule of inference of the system. Clearly, no proof can exist without axioms.

    Stepping back from this formal context, your example doesn't stand up either, (assuming you're talking about differentiating from first principles). "Derive from first principles" is an example of something that would be equivalent to "prove from first principles". Since "first principles" means that you are operating solely from the very definition of the derivative and not using any other more advanced results, it's as rock-solid a proof as you can get in calculus.


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  • Closed Accounts Posts: 1,597 ✭✭✭dan719


    Eh taught and not 'thought' yeah?

    Also newtons second law actually states that f is proportional to the rate of change of momentum, we make it equal by allowing k=1 because of the definition of the newton, so you will certainly lose marks. Go look at the marking scheme.


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    ^Well that's if you're asked to state Newton's second law.

    It's not necessary at all in the derviation of v = u + at.


  • Closed Accounts Posts: 1,597 ✭✭✭dan719


    I know that, the derivation of v=u+at is merely manipulation of the definition of acceleration. a=(v-u)all over t blah blah blah.


  • Registered Users, Registered Users 2 Posts: 786 ✭✭✭spudington16


    obl wrote:
    Eh, you can't use the RHS if you have to prove it. Sorta defeats the purpose.

    Indeedy... If it says "prove", I'd be inclined to tackle one side (here, the LHS) to arrive at the RHS. Can't recommend multiplying across by the denominator....


  • Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Indeedy... If it says "prove", I'd be inclined to tackle one side (here, the LHS) to arrive at the RHS. Can't recommend multiplying across by the denominator....

    If you multiply across by the denominator, and apply the sum/product formulas, what you get reduces to a true statement. You probably get something like sin5x + sin3x = sin5x + sin3x, though I haven't done it.

    Essentially, this gives the proof "in reverse". If you wanted to be really pedantic, you could reverse the order of the steps, starting with a true statement and working up to the equation given in the problem. That would be a correct proof.


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