HSC 2012 MX2 Marathon (archive) (1 Viewer)

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

HSC 2012 MX2 Marathon (archive) (1 Viewer)

Retired Nov '14

Well-Known Member

Member

Retired Nov '14

Member

Retired Nov '14

Retired

Member

Well-Known Member

Retired Nov '14

Member

Retired Nov '14

Member

Retired Nov '14

Member

Retired Nov '14

Member

Retired Nov '14

Member

Bored of

Users Who Are Viewing This Thread (Users: 0, Guests: 1)