The Student Room Group

STEP III 2007, Q1, nice approach for 1st bit

I think it would be pretty stunning to spot this approach under exam conditions, but this came to me while sitting in the barber's chair, of all things...

We are asked to find tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) in terms of ti=tan(θk)t_i = \tan(\theta_k). Use De Moivre:

cos(θ1+θ2+θ3+θ4)+isin(θ1+θ2+θ3+θ4)\cos(\theta_1+\theta_2+\theta_3+\theta_4)+i \sin(\theta_1+\theta_2+\theta_3+\theta_4)

=k(cosθk+isinθk)=(cosθk)((1+itk))\displaystyle =\prod_k (\cos \theta_k + i \sin \theta_k) =\left(\prod \cos \theta_k\right) \left(\prod (1+i t_k)\right)

Then tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) is just the ratio of the real and imaginary parts of (1+itk)\prod (1+i t_k); the most difficult part at this point is simply writing out the answer - excuse me if I don't bother LaTeXing it.

Comparing (1+itk)\prod (1+i t_k) with (xtk)\prod (x-t_k) also lets you more or less write down the relation between tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) and the coefficients of the quartic with roots t1,t2,t3,t4t_1, t_2, t_3, t_4 as well.

Of course, this approach generalises nicely to arbitrary numbers of θk\theta_k as well.
Reply 1
Awesome. However you are right...I don't think it's an easy spot at all, especially under exam conditions. If I remember correctly, I just applied the compound angle formula and it took at least a page to come to that point, so clearly your method is infinitely superior :biggrin: . I'm now starting to wonder how many marks they can possibly give to this part of the question...eek
DFranklin
I think it would be pretty stunning to spot this approach under exam conditions, but this came to me while sitting in the barber's chair, of all things...

We are asked to find tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) in terms of ti=tan(θk)t_i = \tan(\theta_k). Use De Moivre:

cos(θ1+θ2+θ3+θ4)+isin(θ1+θ2+θ3+θ4)\cos(\theta_1+\theta_2+\theta_3+\theta_4)+i \sin(\theta_1+\theta_2+\theta_3+\theta_4)

=k(cosθk+isinθk)=(cosθk)((1+itk))\displaystyle =\prod_k (\cos \theta_k + i \sin \theta_k) =\left(\prod \cos \theta_k\right) \left(\prod (1+i t_k)\right)

Then tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) is just the ratio of the real and imaginary parts of (1+itk)\prod (1+i t_k); the most difficult part at this point is simply writing out the answer - excuse me if I don't bother LaTeXing it.

Comparing (1+itk)\prod (1+i t_k) with (xtk)\prod (x-t_k) also lets you more or less write down the relation between tan(θ1+θ2+θ3+θ4)\tan(\theta_1+\theta_2+\theta_3+\theta_4) and the coefficients of the quartic with roots t1,t2,t3,t4t_1, t_2, t_3, t_4 as well.

Of course, this approach generalises nicely to arbitrary numbers of θk\theta_k as well.



very nice
Reply 3
coffeym
Awesome. However you are right...I don't think it's an easy spot at all, especially under exam conditions. If I remember correctly, I just applied the compound angle formula and it took at least a page to come to that point, so clearly your method is infinitely superior :biggrin: . I'm now starting to wonder how many marks they can possibly give to this part of the question...eek
There's no way you were "supposed" to spot this; in an exam I'd have done exactly what you did - you can tell that the compound angle approach will work (even if messy), so you go for the bird in the hand.

It's only really once you've done the question once that you think "OK, there's obviously some relationship between the expression for tan(...) and the roots of that polynomial. It all comes out far too neatly for it to be coincidence." And that was what I lead me to think about whether there was a way of seeing that relationship without grinding through all the algebra.

So I'm absolutely certain that the fact you can solve it in a few lines won't affect the number of marks it's worth.

Having already done the question is "cheating" in a sense. In a similar way, I think you can solve the 2nd part in pretty short order once you've already done it once. Because it turns out nearly all the terms end up being the "denominator" parts of the expression for tan(...), so they don't matter. Pretty impossible to know that in advance though.
DFranklin
There's no way you were "supposed" to spot this; in an exam I'd have done exactly what you did - you can tell that the compound angle approach will work (even if messy), so you go for the bird in the hand.

It's only really once you've done the question once that you think "OK, there's obviously some relationship between the expression for tan(...) and the roots of that polynomial. It all comes out far too neatly for it to be coincidence." And that was what I lead me to think about whether there was a way of seeing that relationship without grinding through all the algebra.

So I'm absolutely certain that the fact you can solve it in a few lines won't affect the number of marks it's worth.

Having already done the question is "cheating" in a sense. In a similar way, I think you can solve the 2nd part in pretty short order once you've already done it once. Because it turns out nearly all the terms end up being the "denominator" parts of the expression for tan(...), so they don't matter. Pretty impossible to know that in advance though.

Would you agree that doing a ridiculous number of difficult/"novel" questions in advance is the surest way of scoring good results? Because in my so-many years of doing maths as a subject, I realise that there are really few times that "creative genius" would struck you in the middle of the exam hall, and having previously seen some pattern would be the best way. It is like an average person who have seen the treasure map would be better than a top-detective without it.
Reply 5
spencer11111
Would you agree that doing a ridiculous number of difficult/"novel" questions in advance is the surest way of scoring good results? Because in my so-many years of doing maths as a subject, I realise that there are really few times that "creative genius" would struck you in the middle of the exam hall, and having previously seen some pattern would be the best way. It is like an average person who have seen the treasure map would be better than a top-detective without it.
Obviously it helps. The more questions you've done, the more likely it is that the questions on the actual exam. But realistically, you're very unlikely to get the exact same question that you've seen before; what you hope is that you've seen something similar and so you can adapt some techniques you already know might be fruitful. Where "creative genius" comes in is in how good you are at the adaptation.

Taking the map analogy; imagine the map has various landmarks you can use to fix your position. If the only way you can know you're at the chest is by digging, you're going to need to find a landmark right on top of the chest (which is unlikely). If you have a cheap metal detector, you can get away with a landmark that's a few feet away. With a really good metal detector, you might not need any landmarks at all.

But certainly in my experience, 99% of the time people have thought I've come up with something brilliant, it's largely been because I've been able to relate it to a previous problem that I know how to do.
spencer11111
Would you agree that doing a ridiculous number of difficult/"novel" questions in advance is the surest way of scoring good results? Because in my so-many years of doing maths as a subject, I realise that there are really few times that "creative genius" would struck you in the middle of the exam hall, and having previously seen some pattern would be the best way. It is like an average person who have seen the treasure map would be better than a top-detective without it.

To put things in perspective, this kind of "creative genius" would never hit anyone in an exam. This is the kind of creativity that almost exclusively strikes after having done a question and seen your way into it. I ended up expanding it using the double angle formula twice: tan((1+2)+(3+4)), and then tan(1+2) and tan(3+4) in turn. (Excuse blatant breach of notation, I'm tired. :p:) You need no creative thought whatsoever to do that and it's simple enough to do, and probably takes twice as long as DFranklin's method... but it works, and if you're careful it won't take long.
Reply 7
generalebriety
To put things in perspective, this kind of "creative genius" would never hit anyone in an exam. This is the kind of creativity that almost exclusively strikes after having done a question and seen your way into it.You're absolutely right for this particular case (and I should have made that clearer).

One important thing to realise with the approach I took is that it could easily not have worked. You could try something like this and end up with a mess of sines and cosines that won't simplify. It's only once you know the answer that you can have any real confidence that it's going to simplify in a nice way.

Whereas:
I ended up expanding it using the double angle formula twice: tan((1+2)+(3+4)), and then tan(1+2) and tan(3+4) in turn. (Excuse blatant breach of notation, I'm tired. :p:) You need no creative thought whatsoever to do that and it's simple enough to do, and probably takes twice as long as DFranklin's method... but it works,

That's exactly what I did in the immediate post-exam analysis. The point is, not only does it work, but it's obvious that it's going to work. This is always good in an actual exam.

Semi-spoiler for last part:

Spoiler

DFranklin

Spoiler


I actually wrote the examiner a kind of "I don't know how to do it but I know what to do" note here... :p:
DFranklin

But certainly in my experience, 99% of the time people have thought I've come up with something brilliant, it's largely been because I've been able to relate it to a previous problem that I know how to do.

Yup I guess this is my experience too, but not 99%. May be 80%.

Sometimes it seems like solving extremely difficult maths problem is a bit like trial and error. You don't know how a certain way is doing to arrive at the answer until you have done it. Of course some genius would say you can do some insightful observations, see the powers, coefficients and so on. But I guess experience helps A LOT.
spencer11111
Yup I guess this is my experience too, but not 99%. May be 80%.

Sometimes it seems like solving extremely difficult maths problem is a bit like trial and error. You don't know how a certain way is doing to arrive at the answer until you have done it. Of course some genius would say you can do some insightful observations, see the powers, coefficients and so on. But I guess experience helps A LOT.

Course it does, yeah. No Cambridge graduate will tell you they found STEP easy to start off with. I doubt any one of them would say they could've passed it without any practice of anything harder than the normal further maths A-level.
This is a really nice approach!

Spoiler

Reply 12
Brilliant.
guess you derive this result from the inspiration of 1988.A.5. haha. Two months ago I struggled to explain the wield phenomenon between the tannθ and the binomial expansion.(just forget to use complex numbers----though it's just the knowledge of FP2...
(edited 5 years ago)