HSC 2012 MX2 Marathon (archive) (2 Viewers)

kingkong123 · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

$\textup{Prove that the equation of the chord joining }P(acos\theta, bsin\theta) \textup{ and } Q(acos\alpha, bsin\alpha)\textup{ of the ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \textup{ is } bcos\frac{\theta + \alpha}{2}x + asin(\frac{\theta + \alpha}{2})y = abcos(\frac{\theta - \alpha}{2}).$

i did it but i want to know if there's a neater/easier solution.
go wild

deswa1 · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
$\textup{Prove that the equation of the chord joining }P(acos\theta, bsin\theta) \textup{ and } Q(acos\alpha, bsin\alpha)\textup{ of the ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \textup{ is } bcos\frac{\theta + \alpha}{2}x + asin(\frac{\theta + \alpha}{2})y = abcos(\frac{\theta - \alpha}{2}).$

i did it but i want to know if there's a neater/easier solution.
go wild

From memory, the only way to do this involves a lot of trig manipulation so I don't think there's a faster way. I'm happy to be corrected though

someth1ng · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:
factorise Z⁵-1=0, Into real linear and quadratic roots.

Since it's z^5, there are 5 roots.
z^5-1=0
.`. z^5=1
Hence, z0=1+0i

This can be shown as cis0
In complex roots of unity, the other 5 roots will be shown equal angles around the unit circle.

Since 360/5=72
.`. z0=cis0
z1=cis72
z2=cis144
z3=cis216
z4=cis288

You can factorise this as (z-z0)(z-z1)(z-z2)(z-z3)(z-z4)=0

Now, this equals to (z-cis0)(z-cis144)...(z-cis288)=0

Is this right?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

someth1ing, try to avoid using degrees. In Extension 2, we use radians now.

Trebla · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
$\textup{Prove that the equation of the chord joining }P(acos\theta, bsin\theta) \textup{ and } Q(acos\alpha, bsin\alpha)\textup{ of the ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \textup{ is } bcos\frac{\theta + \alpha}{2}x + asin(\frac{\theta + \alpha}{2})y = abcos(\frac{\theta - \alpha}{2}).$

i did it but i want to know if there's a neater/easier solution.
go wild

Just sub it in to verify. For point P

$LHS$\\\\=bx\cos\left(\dfrac{\theta+\alpha}{2}\right)+ay\sin\left(\dfrac{\theta+\alpha}{2}\right)\\\\=ab\cos\theta\cos\left(\dfrac{\theta+\alpha}{2}\right)+ab\sin\theta\sin\left(\dfrac{\theta+\alpha}{2}\right)\\\\=ab\cos\left(\theta-\dfrac{\theta+\alpha}{2}\right)\\\\=ab\cos\left(\dfrac{\theta-\alpha}{2}\right)\\\\=$ RHS$

For point Q, one can argue that similarly (with theta and alpha interchanged) the LHS expression reduces to

$ab\cos\left(\dfrac{\alpha-\theta}{2}\right)$

but using the property that the cosine is an even function i.e.

$\cos x = \cos(-x)$

we can see that it becomes the RHS as well

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

Perhaps I'm just very picky, but I've never really been a fan of 'proving' expressions using the LHS = XXXX = RHS method.

I prefer derivations. Maybe it's just me.

Trebla · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

The actual derivation of the equation would be more appropriate if you were asked to find the equation rather than prove it.

someth1ng · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
someth1ing, try to avoid using degrees. In Extension 2, we use radians now.

Yeah, you're right - I shouldn't be using degrees but it was abit of an inconvenience for me to use radiance since I would have to type "pi" or something and it would get abit messy. It's still technically right, right?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
The actual derivation of the equation would be more appropriate if you were asked to find the equation rather than prove it.

Oh of course, substituting in and proving LHS = XXXX = RHS is perfectly valid and takes up very little time.

What I was trying to convey is that I feel uncomfortable accepting a formula, then using it without understanding where it came from (unless deriving it requires higher maths of course).

someth1ng said:
Yeah, you're right - I shouldn't be using degrees but it was abit of an inconvenience for me to use radiance since I would have to type "pi" or something and it would get abit messy. It's still technically right, right?

Yes it is still 'technically' right.

However, your answer is incorrect because the question asked for real linear and quadratic factors. You had linear complex factors. You need to pair up the solutions with the appropriate conjugate in order to acquire this.

IamBread · Feb 9, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Perhaps I'm just very picky, but I've never really been a fan of 'proving' expressions using the LHS = XXXX = RHS method.

I prefer derivations. Maybe it's just me.

Yeah you're picky

. I like that method, it is generally a lot easier ha

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
However, your answer is incorrect because the question asked for real linear and quadratic factors. You had linear complex factors. You need to pair up the solutions with the appropriate conjugate in order to acquire this.

What do you mean?

Carrotsticks · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

$z^{5}-1 = (z-cis0)(z-cis \frac{\pi}{5})(z-cis \frac{2\pi}{5})(z-cis \frac{-\pi}{5})(z-cis \frac{-2\pi}{5})$

This was your solution. However, they are linear complex factors, since if I expand out the cis, I will get an i, which makes the factor complex.

Expand the cis (leave in exact form) and simplify the expressions. If you do this correctly, you will notice a 'difference of two squares' situation.

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Expand the cis (leave in exact form) and simplify the expressions. If you do this correctly, you will notice a 'difference of two squares' situation.

Errr...I don't know what you mean.

D94 · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
Errr...I don't know what you mean.

Group the conjugate pairs and then you'll notice that (e.g. z1 and z2 are conjugates) (z-z1)(z-z2) will equal z^2 -(z1 + z2)z + z1.z2

So, adding conjugates, you'll end up with the 'cos' part of the expression, and multiplying conjugates, you'll get 1. Hence now all parts are linear expressions, and not complex.

Carrotsticks · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
Errr...I don't know what you mean.

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

$\\ $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 ]$

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

Oh, I'm assuming that I need to actually know how to do that?

Okay, I understand how that works now.

IamBread · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

$$If $ z = \cos\theta + i\sin\theta, $ prove that $ z^n + z^{-n} = 2cos n\theta $ and hence solve$ \\\\ 2z^4 + 3z^3 +5z^2 + 3z^2 + 2 = 0$

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$If $ z = \cos\theta + i\sin\theta, $ prove that $ z^n + z^{-n} = 2cos n\theta $ and hence solve$ \\\\ 2z^4 + 3z^3 +5z^2 + 3z^2 + 2 = 0$

I can do the first bit but not sure on the second bit.

Carrotsticks · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
I can do the first bit but not sure on the second bit.

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

someth1ng · Feb 10, 2012

Re: 2012 HSC MX2 Marathon

OH, THE SECOND QUESTION IS "SOLVE"...I THOUGHT IT WAS "PROVE" HAHAHAH

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