Maths problem

Kwekubo

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?

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.

tolosenc

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

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.

JC 2K3

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

Fremen

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

...and employing me

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

tolosenc

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

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.

tolosenc

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

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)

Fremen

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

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.

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.

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.

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.

JC 2K3

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

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.

tolosenc

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

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.

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).

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.

JC 2K3

I disagree.

Plenty of LC Proofs require assumptions.

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....

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.

cocoa

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

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

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

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!

cocoa

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

JC 2K3

^Yup.

Tomlowe

^i concur