HSC 2013 MX2 Marathon (archive) (1 Viewer)

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Yes the version in the form (a+ib) is correct, how did you get the De Moivere's theorem version from that though?

As far as how to get cos(pi/24) is concerned, you need to find how to get from cos(7pi/12) to cos(pi/24)

What I just put it in mod arg form :S but where did you get cos(7pi/12) from wtf

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

See this is why I never post itt because i can't even maths

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

MethewYan said:
lol what I actually equated the imaginary stuff to get sin 7pi/12 then used complementary (seems faster to me)

Ah yes my bad, this is what I did as well, because starting with cos(7pi/12) leads to sin(pi/12)

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Fawun said:
What I just put it in mod arg form :S but where did you get cos(7pi/12) from wtf

What did you put in mod-arg form?

How did you know the angle was -5pi/12? It is not an exact ratio like pi/3 or pi/6 that people remember

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
What did you put in mod-arg form?

How did you know the angle was -5pi/12? It is not an exact ratio like pi/3 or pi/6 that people remember

I put what the expansion was in mod-arg form because I saw that the second part had cos and angles in it so that's why I assumed that I was suppose to put it into mod-arg form. And lol there are calculators for a reason. arg = arctan(y/x) = arctan(1+root3/1-root3) = -5pi/12

RealiseNothing · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$a,b,c > 1$

$\frac{1}{a^2 -1} + \frac{1}{b^2 -1} + \frac{1}{c^2-1} = 1$

$\\ $Prove that$ \\ \\ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \leq 1$

My solution at the moment is way too long. I'm going to try make it shorter (I'm sure I can skip a lot of redundant steps) and post it.

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Fawun said:
I put what the expansion was in mod-arg form because I saw that the second part had cos and angles in it so that's why I assumed that I was suppose to put it into mod-arg form. And lol there are calculators for a reason. arg = arctan(y/x) = arctan(1+root3/1-root3) = -5pi/12

Therein lies the problem with relying on arctan, it forces the solution to be within the domain, $-\frac{\pi}{2} < x < \frac{\pi}{2}$ (this will become clearer when you learn inverse functions properly).

This is what is supposed to happen:

$\\ (1+i\sqrt{3})(1+i) = 2cis \left(\frac{\pi}{3} \right) \times \sqrt{2} cis \left(\frac{\pi}{4} \right) = 2\sqrt{2} cis \left(\frac{7\pi}{12} \right) \\ \\ \therefore \ \cos \frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}} \\ \\ \sin \frac{7\pi}{12}= \frac{1+\sqrt{3}}{2\sqrt{2}}$

And from there try to arrive at cos(pi/24)

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Therein lies the problem with relying on arctan, it forces the solution to be within the domain, $-\frac{\pi}{2} < x < \frac{\pi}{2}$ (this will become clearer when you learn inverse functions properly).

This is what is supposed to happen:

$\\ (1+i\sqrt{3})(1+i) = 2cis \left(\frac{\pi}{3} \right) \times \sqrt{2} cis \left(\frac{\pi}{4} \right) = 2\sqrt{2} cis \left(\frac{7\pi}{12} \right) \\ \\ \therefore \ \cos \frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}} \\ \\ \sin \frac{7\pi}{12}= \frac{1+\sqrt{3}}{2\sqrt{2}}$

And from there try to arrive at cos(pi/24)

lol wtf I was waaay off with my answer :L

umm i have no clue how you can arrive from cos (7pi/12) to cos(pi/24) but all i can see right now is to rationalise the denominator lol

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

this is like impossibru sigh

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Make sure to say so if this one is too hard/open

$\\ $Using complex numbers find the sum$ \\ \\ \cos(\theta) + \cos(\theta + \alpha) + \cos(\theta + 2\alpha) + \dots + \cos(\theta + (n-1)\alpha)$

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Fawun said:
lol wtf I was waaay off with my answer :L

umm i have no clue how you can arrive from cos (7pi/12) to cos(pi/24) but all i can see right now is to rationalise the denominator lol

First step

$\sin \frac{7\pi}{12} = \sin \left(\frac{\pi}{2} + \frac{\pi}{12} \right) = \cos \left(-\frac{\pi}{12} \right) = \cos \frac{\pi}{12}$

Next step is to get from cos(pi/12) to cos(pi/24)

Sy123 · Dec 12, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
First step

$\sin \frac{7\pi}{12} = \sin \left(\frac{\pi}{2} + \frac{\pi}{12} \right) = \cos \left(-\frac{\pi}{12} \right) = \cos \frac{\pi}{12}$

Next step is to get from cos(pi/12) to cos(pi/24)

$$Recall$ \ \ \cos(2\theta) = 2\cos^2 \theta - 1 \\ \\ \Rightarrow \ \cos \frac{\pi}{12} = 2\cos^2\frac{\pi}{24} - 1 \\ \\ \Rightarrow \ \cos \frac{\pi}{24} = \frac{1}{2} \sqrt{1+\frac{1+\sqrt{3}}{2\sqrt{2}}}$

Which after a bit of squaring and manipulation, should yield what the question wants to be shown.

-------------------

$i) Show that$ \ (1+ki)(k+i) \ $for real$ \ k \ $is purely imaginary$

$$ii) Hence describe the sketch of the function$ \ \ f(x) = \tan^{-1} (x) + \tan^{-1} \left(\frac{1}{x} \right)$

Carrotsticks · Dec 12, 2013

Re: HSC 2014 4U Marathon

Interesting approach to that identity!

Sy123 · Dec 12, 2013

Re: HSC 2014 4U Marathon

Carrotsticks said:
Interesting approach to that identity!

A similar method can be used with (1+Ai)(1+Bi) to generate the more general version;

$\tan^{-1} A + \tan^{-1}B = \tan^{-1} \frac{A+B}{1-AB}$

anomalousdecay · Dec 12, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
A similar method can be used with (1+Ai)(1+Bi) to generate the more general version;

$\tan^{-1} A + \tan^{-1}B = \tan^{-1} \frac{A+B}{1-AB}$

Yeah didn't the 4U q16 of 2012 ask to prove that?

RealiseNothing · Dec 12, 2013

Re: HSC 2014 4U Marathon

anomalousdecay said:
Yeah didn't the 4U q16 of 2012 ask to prove that?

Not using Sy's method though.

RealiseNothing · Dec 12, 2013

Re: HSC 2014 4U Marathon

Prove whether or not there are any integer pairs (a,b) such that both of the following are perfect squares:

$a^2+b^2$

$a^2+2b^2$

rumbleroar · Dec 12, 2013

ImageUploadedByTapatalk1386837444.575239.jpg

This was from our term 1 complex numbers assessment

Much tear
It got ripped off Sydney Grammar though so some of you may have seen it before
No one got iii

Have fun

)))

Sent from my iPhone using Tapatalk

RealiseNothing · Dec 12, 2013

Re: HSC 2014 4U Marathon

rumbleroar said:
View attachment 29383

This was from our term 1 complex numbers assessment

Much tear
It got ripped off Sydney Grammar though so some of you may have seen it before
No one got iii

Have fun )))

R is the centroid of triangle OPQ. By symmetry, triangle PQS is a reflection of triangle OPQ. Consider R' is the centroid of triangle PQS. A property of all triangles is the centroid divides the median (i.e. OM, QN, MS, etc) into the ratio 2:1. Thus OR:RM is 2:1 and similarly SR':MR' is 2:1.

Thus since OM=MS, then it follows that SR:OR is 2:1, hence $z=\frac{1}{3}(u+v)$

RealiseNothing · Dec 12, 2013

Re: HSC 2014 4U Marathon

rumbleroar said:
View attachment 29383

This was from our term 1 complex numbers assessment

Much tear
It got ripped off Sydney Grammar though so some of you may have seen it before
No one got iii

Have fun )))

I'm surprised VC didn't get it.

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