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Re: HSC 2018 MX2 Marathon I tried taking out a factor of \frac{1}{sin^n(x)} from the top two brackets but I still cant get to the LHS... any hints?

Re: HSC 2018 MX2 Marathon ahahaha I forgot to consider x = 0 jesus

Re: HSC 2018 MX2 Marathon Ooh these hints

Re: HSC 2018 MX2 Marathon $This is probably more convoluted than using the t-method$ $LHS $ = \frac{1}{1-\cos\theta + 2i\sin\theta} \times \frac{1-cos\theta -2i\sin\theta}{1-cos\theta -2i\sin\theta} = \frac{1-\cos\theta -2i\sin\theta}{1-2\cos\theta + cos^2\theta + 4\sin^2\theta}...

Re: HSC 2018 MX2 Marathon Nice! Here is my solution: \frac{2}{1+z} = \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta} =\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}...

Re: HSC 2018 MX2 Marathon rip, I'll have another go tonight

Re: HSC 2018 MX2 Marathon $From a ruse paper:$ $If $z = \cos\theta + i\sin\theta$, prove that $\frac{2}{1 + z} = 1 - i\tan\left(\frac{\theta}{2}\right)$ $

Re: HSC 2018 MX2 Marathon $Noting that z \neq 0 $ z - \frac{1}{z} = z - \frac{\bar z}{z\bar z} = x + iy - \frac{x - iy}{x^2 + y^2} $$ = \frac{x(x^2 + y^2 )+ iy(x^2 + y^2) -x + iy}{x^2 + y^2}$$ = \frac{x(x^2 + y^2 -1) + iy(x^2 + y^2 + 1)}{x^2 + y^2} $If...

Alright cool, thanks again!

$\noindent Ah I see now, starting from $(x-1)^2 \geq 0$ $(x > 0)$ makes more sense because I always thought messing with inequalities instead of equating LHS and RHS wasn't allowed or isn't a real proof. Thank you.$ $\noindent I'll also look into the AM-GM inequality, although terry lee...

$ I mangaed to get this after lots of tries $ $$\left|x + \frac{1}{x}\right| = \left|x + \frac{1}{x} -2 + 2\right| $$ $$= \left|\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} + 2\right|$$ $ since $\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2}$ will always be $\geq 0$, $$\left|x +...

yeah sorry I meant confused with iii), will fix now and have another try

Confused with iii). Any help would be appreciated!

I was thinking this... I wonder why they didn't include it in the answers, thanks! All good, camel just explained it

100 as in raw mark? Apparently Anthony Henderson from USYD got 100% in 93'

rip, are you from ruse as well?

ahh I see it now haha, yeah that confused me for a while. so would they give a specific condition for z =/ a for some locus question like w = \frac{z - ib}{z - a}

Did they make a mistake on line two, with x(x-a)?? [When I multiplied out, i got x(x+a), ending up with the locus as a circle with centre (-a/2, b/2) instead of (a/2, b/2)] also, is there supposed to be a restriction on the locus? like excluding (a,0) (because z =/ a on the denominator) Thanks!

that makes much mores sense now, I had to go back over the complex conjugate properties to wrap my head around it. Thanks!!

In the answer to the question above on line 5 they seem to substitute an abs term inside of an abs term itself. A similar thing occurs on line 2, ie given that |x| = 1, |x - 2| can be written as |1 - 2|. However I thought that x = 1, -1, and so it can be written as either that or |-1-2|. Am I...