HSC 2015 Maths Marathon (archive) (2 Viewers)

Flop21 · Aug 26, 2015

Re: HSC 2015 2U Marathon

This is probably really simple, but I'm not sure how to do it.

Find all the values of 'theta' where 0* < theta < 360*, that satisfy the equation

cos(theta) - (2/5) = 0

***

Obviously you put the -2/5 on the other side, but then how do you find theta?

keepLooking · Aug 26, 2015

Re: HSC 2015 2U Marathon

$\\ \cos^{-1}{(\frac{2}{5})}=66.42. \\ \cos{\theta}$ is positive in the 1st and 4th quadrants. $\\ \therefore \theta =66.42, (360-66.42)$

Someone teach me how to properly use LaTeX pls.

[thanks rand_althor and Drsoccerball]

Drsoccerball · Aug 26, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
This is probably really simple, but I'm not sure how to do it.

Find all the values of 'theta' where 0* < theta < 360*, that satisfy the equation

cos(theta) - (2/5) = 0

***

Obviously you put the -2/5 on the other side, but then how do you find theta?

cos inverse both sides and one angle is the answer you see in your calculator and the other value is 360 degrees minus the first answer since cos is positive in quadrant 1 and 4

Drsoccerball · Aug 26, 2015

Re: HSC 2015 2U Marathon

keepLooking said:
$\\cos^{-1}(\frac{2}{5})=66.42\\\\ Since\, cos(\theta)\,is\,positive,\\ \textit{cos}\, is\, positive\,in\,the\,1^{st}\,and\,4^{th}\\ Therefore,\\\\ \theta \Rightarrow 66.42,\,360 - 66.42\\$

Someone teach me how to properly use LaTeX pls.

Omg wth is up with the code you used ahahaha

Drsoccerball · Aug 26, 2015

Re: HSC 2015 2U Marathon

$cos^{-1}(\frac{2}{5})=...$

$Hi my names ....$

$\therefore \rightarrow$

Click reply with quote and look at my code

keepLooking · Aug 26, 2015

Re: HSC 2015 2U Marathon

i just don't know what's going on . _ .

$\bold{holy shit}$
$OKAY, I HAVE BEEN A FUCKING RETARD$

Drsoccerball · Aug 26, 2015

Re: HSC 2015 2U Marathon

keepLooking said:
i just don't know what's going on . _ .

$\bold{holy shit}$
$OKAY, I HAVE BEEN A FUCKING RETARD$

Now try making mathematical expressions such as

$\frac{2}{3}, \binom{2}{3} \sum_{k=0}^n{n} \int{lnx}dx$

keepLooking · Aug 26, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Now try making mathematical expressions such as

$\frac{2}{3}, \binom{2}{3} \sum_{k=0}^n{n} \int{lnx}dx$

$$\int{uck}me$

rand_althor · Aug 26, 2015

Re: HSC 2015 2U Marathon

keepLooking said:
Someone teach me how to properly use LaTeX pls.

Here is a more cleaner version of what you wrote:

\cos^{-1}{(\frac{2}{5})}=66.42. \, \cos{\theta}$ is positive in the 1st and 4th quadrants. $ \therefore \theta =66.42, (360-66.42)

Use dollar signs ($) to enclose text which doesn't require mathematical characters. i.e. " is positive in the 1st and 4th quadrants". "\," can be used to create a blank space. Also have a read of this: http://community.boredofstudies.org/12/mathematics/234262/short-guide-latex.html. This website is a good online editor: https://www.codecogs.com/latex/eqneditor.php

keepLooking · Aug 26, 2015

Re: HSC 2015 2U Marathon

Thank you so much!

I have edited my post and will read the guide before posting with LaTeX again ._.

Flop21 · Aug 26, 2015

Re: HSC 2015 2U Marathon

keepLooking said:
$\\ \cos^{-1}{(\frac{2}{5})}=66.42. \\ \cos{\theta}$ is positive in the 1st and 4th quadrants. $\\ \therefore \theta =66.42, (360-66.42)$

Someone teach me how to properly use LaTeX pls.

[thanks rand_althor and Drsoccerball]

OhhH. I was doing that but wasn't getting that answer. Maybe I had it on radian mode or something.

Drsoccerball · Aug 26, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
OhhH. I was doing that but wasn't getting that answer. Maybe I had it on radian mode or something.

That feeling...

leehuan · Aug 27, 2015

Re: HSC 2015 2U Marathon

Some students at my school bring two calculators into the exam room. One for degrees, one for radians. Though I never do it, it seems to work on occasion.

Drsoccerball said:
Now try making mathematical expressions such as

$\frac{2}{3}, \binom{2}{3} \sum_{k=0}^n{n} \int{lnx}dx$

Oh, ok. Randomness time.

$f\left( x \right) =\prod _{ c=1 }^{ x }{ \int _{ c }^{ \infty }{ \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) { \left( 900-x \right) }^{ n-k }\left( { 4 }^{ k } \right) } } \log _{ 10 }{ \left( \frac { 4x+10 }{ 900{ x }^{ 2 }-1 } \right) } dn }$

leehuan · Aug 27, 2015

Re: HSC 2015 2U Marathon

Sy123 said:
$\\ $Consider the function$ \ f(x) = \tan x - x, \ 0 \leq x < \frac{\pi}{2} \\ \\ $i) Show that$ \ f(x) \ $has exactly one stationary point, and that it is a minimum$ \\ \\ $ii) Hence explain why$ \ f(x) \geq 0 \\ \\ $iii) Hence show that$ \ \frac{1}{\cos x} > \frac{x}{\sin x} \ $for$ \ 0 < x < \frac{\pi}{2}$

$\\ $iv) Explain with the use of a diagram or otherwise, that$ \ x > \sin x \ $for$ \ x > 0$

$\\ $v) Hence finally show that$ \ 1 < \frac{x}{\sin x} < \frac{1}{\cos x} \ $for$ \ 0 < x < \frac{\pi}{2}$

$\\ $i)$ f\left( x \right) =\tan { x } -x\quad \left[ \left 0,\frac { \pi }{ 2 } \right) \right \\ f^{ \prime }\left( x \right) =\sec ^{ 2 }{ x } -1 \\ $Let $f^{ \prime }\left( x \right) = 0$ for the stationary points. $\\ \sec ^{ 2 }{ x } -1=0\\ \sec { x } =\pm 1\\ x=0 \\ f\left( 0 \right) =\tan { 0 } -0\\ \therefore $There is a stationary point at $(0,0)$

$\\ f^{ \prime }\left( 0.001 \right) =\sec ^{ 2 }{ \left( -0.001 \right) } -1\approx 1.00\times { 10 }^{ -6 }>0 \\ $Hence, the stationary point is a local minimum. \\ (Note that it is actually a horizontal point of inflexion, but the domain $x<0$ is not included in the question.)$$

$\\ $ii) The curve is monotonic increasing for the domain. Hence, the global minimum of $f\left( x \right) $ is 0 and thus $f\left( x \right) \ge 0$$

$\\ $iii) $\tan { x } -x > 0\\ \tan { x } > x\\ \frac { \sin { x } }{ \cos { x } } > x\\ \frac { 1 }{ \cos { x } } > \frac { x }{ \sin { x } } \\ $Note that we excluded $x=0$ from the original domain. $$

$\\ $iv) We infer this result directly from this graph which I simulated because I can't be bothered taking photos. $$

$\\ $v) $\frac { x }{ \sin { x } } <\frac { 1 }{ \cos { x } } $ from iii)$ \\ x>\sin { x } $ from iv)$ \\ \frac { x }{ \sin { x } } >1\\ 1<\frac { x }{ \sin { x } } \\ \therefore 1<\frac { x }{ \sin { x } } <\frac { 1 }{ \cos { x } }$

Flop21 · Aug 28, 2015

Re: HSC 2015 2U Marathon

So length of AO I calculated to be (square root) 2.

Now how do I find that angle? I've seen the answers but still don't understand why and how they are doing what they're doing. Could someone try and explain it like I'm stupid. Thanks.

Drsoccerball · Aug 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
So length of AO I calculated to be (square root) 2.

Now how do I find that angle? I've seen the answers but still don't understand why and how they are doing what they're doing. Could someone try and explain it like I'm stupid. Thanks.

Tantheta= gradient where theta is angle between line and origin

InteGrand · Aug 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
So length of AO I calculated to be (square root) 2.

Now how do I find that angle? I've seen the answers but still don't understand why and how they are doing what they're doing. Could someone try and explain it like I'm stupid. Thanks.

$From part (ii), we know the gradient of $OA$ is $-1$. Therefore, the angle that the line $OA$ makes with the positive $x$-axis, $\theta$, satisfies $\tan \theta=-1\Rightarrow \theta = 135^\circ$. This angle is precisely $\angle AOC$. Therefore, $\angle AOC=135^\circ$.$

Flop21 · Aug 28, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
$From part (ii), we know the gradient of $OA$ is $-1$. Therefore, the angle that the line $OA$ makes with the positive $x$-axis, $\theta$, satisfies $\tan \theta=-1\Rightarrow \theta = 135^\circ$. This angle is precisely $\angle AOC$. Therefore, $\angle AOC=135^\circ$.$

Why does the gradient satisfy tan?? Sorry is this a property we should just know, or am I missing something?

Sorry I know I probably sound silly, but I've either missed this or forgotten it. Thanks!

InteGrand · Aug 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
Why does the gradient satisfy tan?? Sorry is this a property we should just know, or am I missing something?

Sorry I know I probably sound silly, but I've either missed this or forgotten it. Thanks!

$Yes, you are expected to know that if a line has slope $m \in \mathbb{R}$ and makes angle $\theta$ with the positive $x$-axis, then we have $m= \tan \theta$. In our example, the slope was $m =-1$.$

davidgoes4wce · Aug 29, 2015

Re: HSC 2015 2U Marathon

There has been a slight error in the solutions of Mathematics in Fitzpatrick, Aus

Megan still earns the most after 4 years.

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