• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2013 MX2 Marathon (archive) (2 Viewers)

Status
Not open for further replies.

Fawun

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Re: HSC 2014 4U Marathon

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

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Re: HSC 2014 4U Marathon

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

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Re: HSC 2014 4U Marathon

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
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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, (this will become clearer when you learn inverse functions properly).

This is what is supposed to happen:



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

Fawun

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Re: HSC 2014 4U Marathon

Therein lies the problem with relying on arctan, it forces the solution to be within the domain, (this will become clearer when you learn inverse functions properly).

This is what is supposed to happen:



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

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Re: HSC 2014 4U Marathon

this is like impossibru sigh
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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

 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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



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

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

First step



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


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

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



 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
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:



 

rumbleroar

Survivor of the HSC
Joined
Nov 30, 2011
Messages
2,271
Gender
Female
HSC
2014
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

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon

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
 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top