HSC 2016 Maths Marathon (archive) (1 Viewer)

Speed6 · Oct 30, 2015

Hi there 2016'ers, including myself!

Welcome to the 2016 HSC 2U Marathon,

Some simple housekeeping rules:

1. Only post questions within the level and scope of 2U HSC Mathematics

2. Provide neat working out when possible

3. Don't flood the thread with multiple unanswered questions

4. Since HSC 2U Mathematics can be tested up to 30% of the maths you did in Preliminary, feel free to post the often odd and challenging questions

5. And remember to have fun and enjoy the beauty of mathematics (I don't really know if this is a rule lol)

Q1/

$$Find the equation of the tangent to the curve y=sin3x at the point where $x=\frac{\pi}{4}$.$

davidgoes4wce · Oct 31, 2015

Re: HSC 2016 2U Marathon

Speed6 said:
Hi there 2016'ers, including myself!

Welcome to the 2016 HSC 2U Marathon,

Some simple housekeeping rules:

$$Find the equation of the tangent to the curve y=sin3x at the point where $x=\frac{\pi}{4}$.$

leehuan · Oct 31, 2015

Re: HSC 2016 2U Marathon

$$Show that $\frac { \left( \sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } \right) \left( 1-\sin { x } \cos { x } \right) }{ \cos { x } \left( \sec { x } -\csc { x } \right) \left( \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } \right) } =\sin { x }$

calamebe · Oct 31, 2015

Re: HSC 2016 2U Marathon

leehuan said:
$$Show that $\frac { \left( \sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } \right) \left( 1-\sin { x } \cos { x } \right) }{ \cos { x } \left( \sec { x } -\csc { x } \right) \left( \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } \right) } =\sin { x }$

Sorry for bad quality, I should learn LateX, anyway: http://m.imgur.com/j827gtK

leehuan · Oct 31, 2015

Re: HSC 2016 2U Marathon

calamebe said:
Sorry for bad quality, I should learn LateX, anyway: http://m.imgur.com/j827gtK

I never learnt LaTeX. I just used Daum Equation editor

______________________________
$$By choosing an appropriate curve, use integration calculus to find the formula of a sphere of radius $r$

leehuan · Oct 31, 2015

Re: HSC 2016 2U Marathon

Alternate question because if I think about it now it's a bit too early for integration.

$\\ $By considering right-angled trigonometry, prove the cosine rule.$$

Hint: Right-angled trigonometry implies you will need to construct at least one altitude on any arbitrary triangle.

EDIT: Extra - Prove the sine rule whilst you're here. (EASIER)

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

You know someone's notes are not good if they make them for maths... Also you can't trade/sell independent/CSSA/NEAP it's illegal.

kawaiipotato · Oct 31, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
You know someone's notes are not good if they make them for maths... Also you can't trade/sell independent/CSSA/NEAP it's illegal.

Why does having notes for maths mean it's not good?

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

leehuan said:
I never learnt LaTeX. I just used Daum Equation editor
______________________________
$$By choosing an appropriate curve, use integration calculus to find the formula of a sphere of radius $r$

$Rotating a semicircle around a line will give you a sphere$

$\therefore V=\pi \int_{-a}^a{a^2-x^2}dx$

$V= 2 \pi [a^2x -\frac{x^3}{3}]_0^a$

$V= \frac{4 \pi a^3}{3} $ Where $ a $ is the radius of the sphere.$

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

kawaiipotato said:
Why does having notes for maths mean it's not good?

Notes for maths means that you wear raincoats in showers.

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

leehuan said:
Alternate question because if I think about it now it's a bit too early for integration.

$\\ $By considering right-angled trigonometry, prove the cosine rule.$$

Hint: Right-angled trigonometry implies you will need to construct at least one altitude on any arbitrary triangle.

EDIT: Extra - Prove the sine rule whilst you're here. (EASIER)

$$ Construct an arbitrary triangle letting $ cos \theta =x$

$$ This implies (Via soh cah toa) $ sin\theta = \sqrt{1-x^2}$

$c^2=a^2+b^2 -2ab cos C$

$1-x^2= 1+x^2 -2x^2$

$LHS=RHS$

[]

$\frac{sin\theta}{\sqrt{1-x^2}} = \frac{1}{1} = \frac{cos\theta}{x}$

[]

kawaiipotato · Oct 31, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
Notes for maths means that you wear raincoats in showers.

I think notes would still be useful for revision. Obviously 100% reliance on them wouldn't be efficient but it's not counter-productive (if that's what you're suggesting).

calamebe · Oct 31, 2015

Re: HSC 2016 2U Marathon

leehuan said:
Alternate question because if I think about it now it's a bit too early for integration.

$\\ $By considering right-angled trigonometry, prove the cosine rule.$$

Hint: Right-angled trigonometry implies you will need to construct at least one altitude on any arbitrary triangle.

EDIT: Extra - Prove the sine rule whilst you're here. (EASIER)

http://m.imgur.com/vAg4Rff for cos rule

calamebe · Oct 31, 2015

Re: HSC 2016 2U Marathon

leehuan said:
Alternate question because if I think about it now it's a bit too early for integration.

$\\ $By considering right-angled trigonometry, prove the cosine rule.$$

Hint: Right-angled trigonometry implies you will need to construct at least one altitude on any arbitrary triangle.

EDIT: Extra - Prove the sine rule whilst you're here. (EASIER)

Sine rule: http://m.imgur.com/ffJ2aal

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

calamebe said:
http://m.imgur.com/vAg4Rff for cos rule

I think my ways significantly easier

Drsoccerball said:
$$ Construct an arbitrary triangle letting $ cos \theta =x$

$$ This implies (Via soh cah toa) $ sin\theta = \sqrt{1-x^2}$

$c^2=a^2+b^2 -2ab cos C$

$1-x^2= 1+x^2 -2x^2$

$LHS=RHS$

[]

$\frac{sin\theta}{\sqrt{1-x^2}} = \frac{1}{1} = \frac{cos\theta}{x}$

[]

calamebe · Oct 31, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
I think my ways significantly easier

me too, I just wanted to do it a different way lol

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

calamebe said:
me too, I just wanted to do it a different way lol

ahah my bad... Why are you on the 2U marathon :L ? You should spend heeps of time on extension 2

Sien · Oct 31, 2015

Re: HSC 2016 2U Marathon

Drsoccerball said:
ahah my bad... Why are you on the 2U marathon :L ? You should spend heeps of time on extension 2

Cuz mx2 is too ez pz for him

Sent from my D6503 using Tapatalk

Speed6 · Oct 31, 2015

Re: HSC 2016 2U Marathon

His over here because MX2 rated G and 2U is R 18+

Drsoccerball · Oct 31, 2015

Re: HSC 2016 2U Marathon

Sien said:
Cuz mx2 is too ez pz for him

Sent from my D6503 using Tapatalk

Then why would he be doing 2U now lol...

HSC 2016 Maths Marathon (archive) (1 Viewer)

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