HSC 2012 MX2 Marathon (archive) (2 Viewers)

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
sin^2theta+cos^2theta=1 and sub in the values then solve? You in fact do get an exact answer.

Yep

Carrotsticks said:
A platonic one... until tan(x) comes along.

What? lol

Fizzy_Cyst · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

tan only comes if sin is on top

if cos is on top then we got a cot to lay tan in next time

I just realised how lame this attempt at making a lame attempt is.. Sorry.

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

It becomes cos(-1/4)+[root(15)/4]i and [2/3]pi-[root3/2]i

Can you confirm?

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
It becomes cos(-1/4)+[root(15)/4]i and [2/3]pi-root3/2

I don't think that's right. You should have:
$\cos\theta = -\frac{1}{4} \\\\ sin^2\theta = 1 - \frac{1}{16} \\\\ \sin\theta = \pm\frac{\sqrt{15}}{4}$

$$And with $ \cos\theta = \frac{1}{2} $ we get $ \sin\theta = \pm\frac{\sqrt{3}}{2}$

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

I meant those solutions are z=... --> I read it again and turns out I didn't work it very well.

I got those same values for sintheta though.

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
I meant those solutions are z=... --> I read it again and turns out I didn't work it very well.

I got those same values for sintheta though.

Lol yeah what you said didn't make much sense

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

But anyway, the solutions are: z=cos(-1/4)+[root(15)/4]i and z=[2/3]pi-[root3/2]i

Is that complete so I can get ready to sleep? haha

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
But anyway, the solutions are: z=cos(-1/4)+[root(15)/4]i and z=[2/3]pi-[root3/2]i

Is that complete so I can get ready to sleep? haha

It is very close lol, but not quite there. The solutions are
$z = \frac{-1}{2} \pm \frac{\sqrt{3}}{2}i, z = \frac{-1}{4} \pm \frac{\sqrt{15}}{4}i$

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

Oh, right, cistheta --> costheta=-1/4 FARK that's the mistake.

I get that bit now. The plus/minus is there because roots/solutions come in conjugate pairs?

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
Oh, right, cistheta --> costheta=-1/4 FARK that's the mistake.

I get that bit now. The plus/minus is there because roots/solutions come in conjugate pairs?

Yep, real coefficients so the roots come in conjugate pairs. And also, when you do the s^2 + c^2 = 1, remember when you take the sqrt its plus/minus.

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
And also, when you do the s^2 + c^2 = 1, remember when you take the sqrt its plus/minus.

Yeah, I totally forgot about that. Anyway, thanks for all the advice and help you've given me tonight.

Highly appreciated - good night!

IamBread · Feb 11, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
Yeah, I totally forgot about that. Anyway, thanks for all the advice and help you've given me tonight.

Highly appreciated - good night!

Haha your welcome! Goodnight.

study-freak · Feb 11, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
$\\ \begin{align*} z^{5}-1 &= \left [z-cis (0) \right ] \left [z-cis \left (\frac{\pi}{5} \right ) \right ] \left [z-cis \left (\frac{2\pi}{5} \right ) \right ] \left [z-cis \left (-\frac{\pi}{5} \right ) \right ] \left [z-cis \left ( -\frac{2\pi}{5} \right ) \right ] \\\\ &= \left [z-\cos (0) - i \sin (0) \right ] \left [z- \cos \left (\frac{\pi}{5} \right ) - i \sin \left (\frac{\pi}{5} \right ) \right ] \left [z- \cos \left (\frac{2\pi}{5} \right ) - i \sin \left (\frac{2\pi}{5} \right ) \right ] \left [z- \cos \left (-\frac{\pi}{5} \right ) - i \sin \left (-\frac{\pi}{5} \right ) \right ] \left [z- \cos \left ( -\frac{2\pi}{5} \right ) - i \sin \left ( -\frac{2\pi}{5} \right ) \right ] \\\\ \end{align*}$

$\\ $Using the identity $ \cos (- \theta ) = \cos (\theta ) $ and $ \sin(- \theta ) = - \sin(\theta): \\\\ \left [z - 1 \right ] \left [z- \cos \left (\frac{\pi}{5} \right ) - i \sin \left (\frac{\pi}{5} \right ) \right ] \left [z- \cos \left (\frac{2\pi}{5} \right ) - i \sin \left (\frac{2\pi}{5} \right ) \right ] \left [z- \cos \left (\frac{\pi}{5} \right ) + i \sin \left (\frac{\pi}{5} \right ) \right ] \left [z- \cos \left ( \frac{2\pi}{5} \right ) + i \sin \left ( \frac{2\pi}{5} \right ) \right ] \\\\$

$\\ $Re-arrange:$ \\\\ \left [z - 1 \right ] \left [z- \cos \left (\frac{\pi}{5} \right ) - i \sin \left (\frac{\pi}{5} \right ) \right ] \left [z- \cos \left (\frac{\pi}{5} \right ) + i \sin \left (\frac{\pi}{5} \right ) \right ] \left [z- \cos \left (\frac{2\pi}{5} \right ) - i \sin \left (\frac{2\pi}{5} \right ) \right ] \left [z- \cos \left ( \frac{2\pi}{5} \right ) + i \sin \left ( \frac{2\pi}{5} \right ) \right ] \\\\ $Recognising the situation $ (a-b)(a+b) = a^{2} - b^{2} : \\\\ \left [z - 1 \right ] \left [\left (z- \cos \left (\frac{\pi}{5} \right ) \right )^{2} + \sin ^{2} \left (\frac{\pi}{5} \right ) \right ] \left [\left (z- \cos \left (\frac{2\pi}{5} \right ) \right )^{2} + \sin ^{2} \left (\frac{2\pi}{5} \right ) \right ] \\\\$

$\\ $As you can see, the $ i $ has disappeared, meaning that we no longer have complex factors. We now have real quadratic and linear factors, but we must simplify them: $ \\\\ \left [z - 1 \right ] \left [z^{2}- 2 \cos \left (\frac{\pi}{5} \right ) + \cos^{2} \left (\frac{\pi}{5} \right )+ \sin ^{2} \left (\frac{\pi}{5} \right ) \right ] \left [z^{2}- 2 \cos \left (\frac{2 \pi}{5} \right ) + \cos^{2} \left (\frac{2 \pi}{5} \right )+ \sin ^{2} \left (\frac{2 \pi}{5} \right ) \right ] \\\\ = \left [z - 1 \right ] \left [z^{2}- 2 \cos \left (\frac{\pi}{5} \right ) + 1 \right ] \left [z^{2}- 2 \cos \left (\frac{2 \pi}{5} \right ) + 1 \right ]$

You forgot to put z in the quadratic factors after 2cos(pi/5) and 2cos(2pi/5).
Just to let you know, it's easier if you:
$\\$let $a=cis\left(\frac{\pi}{5}\right)$, $b=cis\left(\frac{2\pi}{5}\right)$, then$\\(z-1)(z-a)(z-\overline{a})(z-b)(z-\overline{b}) \\=(z-1)(z^2-(a+\overline{a})z+a\overline{a})(z^2-(b+\overline{b})z+b\overline{b}) \\=(z-1)(z^2-2Re(a)z+\left | a \right |^2)(z^2-2Re(b)z+\left | b \right |^2) \\=(z-1)\left [ z^2-2cos\left(\frac{\pi}{5}\right)z+1 \right ]\left [ z^2-2cos\left(\frac{2\pi}{5}\right)z+1 \right ]$

IamBread · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

$\alpha $ and $ \beta $ are the roots of $ z^2 - 2z +2 = 0. \\\\ $If $ \cot\theta = x + 1, $ prove that$ \\\\ \frac{(x+\alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin n\theta}{(\sin\theta)^n}$

SpiralFlex · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

Lol Patel.

SpiralFlex · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

Let the roots be $\alpha, \beta$

$\cot \theta = x+1$

$x=\cot \theta - 1$

We need to prove,

$\frac{(\cot \theta - 1+\alpha)^n - (\cot \theta - 1+\beta)^}{\alpha - \beta}=\frac{\sin (n \theta)}{\sin^n \theta}$

By solving the equation,

$\alpha = 1+i, \beta =1-i$

$\therefore LHS=\frac{(\cot \theta - 1+1+i)^n - (\cot \theta - 1+1-i)^n}{\alpha - \beta}$

$=\frac{(\cot \theta +i)^n - (\cot \theta -i)^n}{2i}$

$=\frac{cis (n \theta)-cis(-n\theta)}{2i\sin (n\theta)}$

$=\frac{\sin (n \theta)}{\sin^n \theta}$

Drongoski · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$\alpha $ and $ \beta $ are the roots of $ z^2 - 2z +2 = 0. \\\\ $If $ \cot\theta = x + 1, $ prove that$ \\\\ \frac{(x+\alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin n\theta}{(\sin\theta)^n}$

$\therefore z^2-2z+1 = -1 \implies (z-1)^2 = -1 \implies z = 1 \pm i $ $ \implies \alpha = 1+i $ & $ \beta = i-i$

$\therefore x+\alpha = cot \theta -1 + 1+ i = cot \theta +i = \frac {cos \theta}{sin \theta} + i = \frac {cos \theta + isin \theta}{sin \theta} = \frac {cis \theta} {sin \theta}$

Similarly $x+\beta = \frac {cis(- \theta)}{sin \theta}$

$\therefore \frac {(x+\alpha)^n - (x+\beta)^n}{\alpha-\beta} = \frac {\frac {(cis \theta)^n}{sin^n \theta} - \frac {(cis(- \theta))^n}{(sin \theta)^n}}{2i} = \frac {cis (n\theta) - cis (-n\theta)}{2i(sin \theta)^n}$

$= \frac {cos(n\theta)+isin(n\theta) - cos(-n\theta)-isin(-n\theta)}{2i(sin \theta)^n}$

$= \frac{cos(n\theta)+isin(n\theta) - cos(n\theta)+isin(n\theta)}{ditto} = \frac {2isin(n\theta)}{2i(sin \theta)^n} = RHS$

IamBread · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

$$The polynomial $ 2x^3 - x^2 + 5 = 0 $ has roots $ \alpha, \beta, \gamma. \\\\ $Find the polynomial with integer coefficients whose roots are $ \alpha^3, \beta^3, \gamma^3$

nightweaver066 · Feb 12, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$The polynomial $ 2x^3 - x^2 + 5 = 0 $ has roots $ \alpha, \beta, \gamma. \\\\ $Find the polynomial with integer coefficients whose roots are $ \alpha^3, \beta^3, \gamma^3$

$2x^3 - x^2 + 5 = 0$

$$Related roots: y = x^3$

$x = ^3\sqrt{y}$

$\therefore 2y - y^{\frac{2}{3}} + 5 = 0$

$(2y + 5)^3 = y^2$

$8y^3 + 60y^2 + 150y + 125 = y^2$

$8y^3 + 59y^2 + 150y + 125 = 0$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

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