HSC 2016 MX1 Marathon (archive) (1 Viewer)

leehuan · Feb 18, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Need to memorise that, that's one of the formulas not listed on the 2016 Formulae Sheet.

She's doing a bridging course. Not her HSC. I think that was a bit disadvantaging.

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

Just realised it's on the standard integrals sheet which is good.

wu345 · Feb 20, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$Evaluate $ \dfrac{\displaystyle \sum_{n=1}^{44}\cos n^{\circ}}{\displaystyle \sum_{n=1}^{44}\sin n^{\circ} }$

I still have no idea how to do this without sums and products - could someone please explain?

InteGrand · Feb 20, 2016

Re: HSC 2016 3U Marathon

wu345 said:
I still have no idea how to do this without sums and products - could someone please explain?

Blast1 posted a method using complex number vectors in the 4U marathon: http://community.boredofstudies.org...4755/hsc-2016-4u-marathon-11.html#post7100224

(Check his quote of a Paradoxica post there too, as it contains the question with a hint of considering complex number vectors. Blast1 answered two questions in that post by the way, the relevant one is his answer to Question 1 there.)

wu345 · Feb 20, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Blast1 posted a method using complex number vectors in the 4U marathon: http://community.boredofstudies.org...4755/hsc-2016-4u-marathon-11.html#post7100224

(Check his quote of a Paradoxica post there too, as it contains the question with a hint of considering complex number vectors. Blast1 answered two questions in that post by the way, the relevant one is his answer to Question 1 there.)

Paradoxica said there was a way to do it with complimentary angles though, how would you do that?

Paradoxica · Feb 20, 2016

Re: HSC 2016 3U Marathon

wu345 said:
Paradoxica said there was a way to do it with complimentary angles though, how would you do that?

$$\noindent $\sin{A} \equiv \cos{(90^{\circ}-A)}\\ \sqrt{2}\sin{(A+45^\circ)} \equiv \sin{A}+\cos{A}$

leehuan · Feb 20, 2016

Re: HSC 2016 3U Marathon

I would so not have considered auxiliary angle method for this. Obviously complementary angles was required.

Paradoxica · Feb 20, 2016

Re: HSC 2016 3U Marathon

leehuan said:
I would so not have considered auxiliary angle method for this. Obviously complementary angles was required.

When I first saw the problem, it was the first idea that popped in my head.

Also, you do not go from right to left, you go from left to right. This is not a compactification of trig, this is an expansion of trig. Try it and see for yourself.

davidgoes4wce · Feb 21, 2016

Re: HSC 2016 3U Marathon

$Was going through the Year 12 textbook on trigonometry, has there ever been a question in HSC in relation to small-angle theory?$

$sin 61\degree = sin (60 \degree + 1 \degree)=sin 60 \degree cos 1 \degree + cos 60 \degree sin 1\degree$

$cos \theta \approx =1 , sin \theta = \frac{\pi}{180}$

Drsoccerball · Feb 21, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$Was going through the Year 12 textbook on trigonometry, has there ever been a question in HSC in relation to small-angle theory?$

$sin 61\degree = sin (60 \degree + 1 \degree)=sin 60 \degree cos 1 \degree + cos 60 \degree sin 1\degree$

$cos \theta \approx =1 , sin \theta = \frac{\pi}{180}$

It's in astrophysics $d= \frac{1}{p} $ Since $ \tan{x} \rightarrow x $ as $ x \rightarrow 0$

KingOfActing · Feb 21, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$Was going through the Year 12 textbook on trigonometry, has there ever been a question in HSC in relation to small-angle theory?$

$sin 61\degree = sin (60 \degree + 1 \degree)=sin 60 \degree cos 1 \degree + cos 60 \degree sin 1\degree$

$cos \theta \approx =1 , sin \theta = \frac{\pi}{180}$

I don't know if there's been one before, but I think they would say something like "Using the approximations sin(x) ~= x, cos(x) ~= 1" etc.

Or at least something like "Considering lim x->0 sin(x)/x = 1" or something like that

leehuan · Feb 21, 2016

Re: HSC 2016 3U Marathon

lim x->0 sin(x) = x
and
lim x->0 cos(x) = 1
are introduced in MIF funnily enough

Paradoxica · Feb 21, 2016

Re: HSC 2016 3U Marathon

leehuan said:
lim x->0 sin(x) = x
and
lim x->0 cos(x) = 1
are introduced in MIF funnily enough

Why would the second one need a limit? There are no issues with it.

wu345 · Feb 25, 2016

Re: HSC 2016 3U Marathon

$8(i)\quad $Simplify$\quad sin(A+B)-sin(A-B)\\ (ii)\quad $Hence\quad prove\quad by\quad mathematical\quad induction\quad for\quad all\quad positive\quad integers\quad n\quad that$\\ \sin { \theta } +\sin { 2\theta } +\sin { 3\theta } +...+\sin { n\theta } \frac { \sin { \frac { (n+1)\theta }{ 2 } } \sin { \frac { n\theta }{ 2 } } }{ \sin { \frac { \theta }{ 2 } } }$

Part ii please

Drsoccerball · Feb 25, 2016

Re: HSC 2016 3U Marathon

wu345 said:
$8(i)\quad $Simplify$\quad sin(A+B)-sin(A-B)\\ (ii)\quad $Hence\quad prove\quad by\quad mathematical\quad induction\quad for\quad all\quad positive\quad integers\quad n\quad that$\\ \sin { \theta } +\sin { 2\theta } +\sin { 3\theta } +...+\sin { n\theta } \frac { \sin { \frac { (n+1)\theta }{ 2 } } \sin { \frac { n\theta }{ 2 } } }{ \sin { \frac { \theta }{ 2 } } }$

Part ii please

$$ (iii) Prove that $ \sin{\frac{\pi}{2015}} +\sin{\frac{2\pi}{2015}}+...+\sin{\frac{2014\pi}{2015}} = \cot{\frac{\pi}{4030}}$

wu345 · Feb 26, 2016

Re: HSC 2016 3U Marathon

dw got it now

leehuan · Feb 28, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Why would the second one need a limit? There are no issues with it.

It doesn't technically. Probably used as an aid, however, for the derivative of sin(x) by first principles without S2P or P2S (I forgot which).

lim h->0 (cosh-1)/h

KingOfActing · Feb 28, 2016

Re: HSC 2016 3U Marathon

leehuan said:
It doesn't technically. Probably used as an aid, however, for the derivative of sin(x) by first principles without S2P or P2S (I forgot which).

lim h->0 (cosh-1)/h

You only need h->0 sin(h)/h if you do it this way:

$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}\\(\sin{x})' = \lim_{h\to 0}\frac{\sin{(x+h)}-\sin{(x-h)}}{2h} = \lim_{h\to 0}\frac{\sin{h}\cos{x}}{h} = \cos{x}\\$The first definition of the derivative may be justified by the substitutions $ x \to x - \frac12 h$, $ h \to 2h$

InteGrand · Feb 28, 2016

Re: HSC 2016 3U Marathon

leehuan said:
It doesn't technically. Probably used as an aid, however, for the derivative of sin(x) by first principles without S2P or P2S (I forgot which).

lim h->0 (cosh-1)/h

What is S2P and P2S?

KingOfActing · Feb 28, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
What is S2P and P2S?

Sum to Product and Product to Sum

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