Step II 2000 Q4...

prove 3arctan(1/4) + arctan(1/20) + arctan(1/1985) = pi/4

in the solutions it tells you to consider [(4+i)^3 . (20+i)]/(1+i)

the numerator i can 'see' from the first two terms in the original expression. but how are you meant to come up with there being a (1+i) in the denominator? why not in the numerator like in the previous part?

Reply 1

Speleo

God knows what that complex number stuff is about. All the question needs is 5 minutes of trigonometry.

prove arctan(1/20) + arctan(1/1985) = pi/4 - 3arctan(1/4)
prove tan(arctan(1/20) + arctan(1/1985)) = tan(pi/4 - 3arctan(1/4))

LHS -> [(1/20) + (1/1985)]/[1 - 1/(20*1985)]
= [(397 + 4)/(7940)]/[39,699/39,700]
= 401*5/39,699
= 2005/39,699
= 5/99

RHS -> [1 - tan(3arctan(1/4))]/[1 + tan(3arctan(1/4)]
tan3x identity
tan3x = [tan2x + tanx]/[1 - tanxtan2x]
= [2tanx/(1-tan^2x) + tanx]/[1 - 2tan^2x/1-tan^2x]
= [2tanx + tanx - tan^3x]/[1 - tan^2x - 2tan^2x]
tan3x = [3tanx - tan^3x]/[1 - 3tan^2x]
tan(3arctan(1/4)) = [3/4 - 1/64]/[1 - 3/16]
= (47/64)/(52/64)
= 47/52

RHS-> [1 - 47/52]/[1 + 47/52]
= (5/52)/(99/52)
= 5/99

EDIT: I've loaded up the paper and will see what it's talking about.
EDIT2: I'd do the other part by the same method aswell.

Reply 2

Dirac Delta Function

chewwy

so strange, speleos method is bleeding obvious..

anyway, you need to subtract pi/4, ie multiply by (1-i). dividing by 1+i has the same effect: 1/(1+i) * (1-i)/(1-i)

Reply 3

Dirac Delta Function

by the way, where on earth are you getting solutions from??

Reply 4

Dez

The AEA paper usually lists about four different methods for each question, giving equal marks for each of them. If STEP mark schemes were publicly available, I'm sure they'd follow a similar method.

In short, if you're answer's right, it doesn't matter if you used the textbook method or not.

Reply 5

nota bene

Dirac Delta Function

by the way, where on earth are you getting solutions from??

I have seen worked solutions to all 1999-2004 on the net I think...but not any official markscemes.

Reply 6

Dez

nota bene

but not any official markscemes.

They don't exist.

Reply 7

DFranklin

Dez

In short, if you're answer's right, it doesn't matter if you used the textbook method or not.

From the examiners' reports I've seen for STEP, if the question says "Hence" (as distinct from "hence or otherwise") and your proof is "otherwise", you will lose marks. I've not seen it, but I've heard mention of a STEP question that has you prove an identity and then "hence deduce 7 + 4 = 11"! (might not be the exact numbers, but you get the idea...)

Reply 8

Rabite

I remember doing this one.

The first part shows that arg(wz) = arg w + arg z, which is what we need.

3arctan(1/4) + arctan(1/20) + arctan(1/1985) = pi/4
--> arctan(1/1985) = pi/4 - 3arctan(1/4) - arctan(1/20)
=pi/4 + 3arctan(-1/4)+arctan(-1/20)
By looking at what happened to the stuff in the given example (that you worked out, but it's practically given)...
The power goes infront of an arctan, and the pi/4 comes from arg(1+i).

So we consider the thing that has the argument of that sum:
(1+i)(4-i)³(20-i)=(1+i)(52-47i)(20-i)
=(1+i)(993-992i)=1985+i.
So arg(1985+i) [which happens to equal arctan(1/1985)] is equal to the sum of the arguments of each bit in (1+i)(4-i)³(20-i).

Seems I'm a bit late, but whatever. xD

Reply 9

Speleo

Good thinking Rabite :smile:

The paper uses equivalent wording to 'hence or otherwise', so the only problem I see with my working is that tan(x) = tan(y) does not necessarily imply that x = y, so you might have to fiddle about showing that both sides are less than pi/2 (and greater than -pi/2).