HSC 2012 MX2 Marathon (archive) (3 Viewers)

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

I got costheta=-1/4 and costheta=-1/2 [then solve normally]

Is this correct?

SpiralFlex · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Hint: Factorise/divide z^2 from both sides.

Spoiler alert. Spoiler alert.

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
I got costheta=-1/4 and costheta=-1/2 [then solve normally]

Is this correct?

Sounds good. Just need need to find the value of $\sin\theta$ now.

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

By that, do you mean sub in the value of theta as sintheta to make it cistheta?

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
By that, do you mean sub in the value of theta as sintheta to make it cistheta?

You could do it that way but with $\cos\theta = \frac{1}{4}$ , you aren't going to get a very exact answer. Whats the relationship between $\cos\theta $ and $ \sin\theta$ ?

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

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

So yeah, I understand it now

Carrotsticks · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
You could do it that way but with $\cos\theta = \frac{1}{4}$ , you aren't going to get a very exact answer. Whats the relationship between $\cos\theta $ and $ \sin\theta$ ?

A platonic one... until tan(x) comes along.

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$\\(z-1)(z-a)(z-\overline{a})(z-b)(z-\overline{b}) \\=(z-1)(z^2-(a+\overline{a})z+a\overline{a})(z^2-(b+\overline{b})z+b\overline{b}) \\=(z-1)(z^2-2Re(a)z+\left | a \right |^2)(z^2-2Re(b)z+\left | b \right |^2) \\=(z-1)\left [ z^2-2cos\left(\frac{\pi}{5}\right)z+1 \right ]\left [ z^2-2cos\left(\frac{2\pi}{5}\right)z+1 \right ]$

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