# Thread: HSC 2018 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)

2. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

ill write it up in latex later this afternoon

3. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)
Originally Posted by altSwift
I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

ill write it up in latex later this afternoon
soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
i apologise for any inconvenience this may have caused

4. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2-2z+2=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n
Originally Posted by mrbunton
soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
i apologise for any inconvenience this may have caused
$Roots (\alpha and \beta): 1 \pm i$
$Subbing in roots and y + 1 = \cot(x) into the LHS:$

$LHS = \frac{(cotx + i)^n - (cotx - i)^n}{2i}$

$= \frac{(\frac{cosx}{sinx} + i)^n - (\frac{cosx}{sinx} - i)^n}{2i}$

$= \frac{\frac{1}{sin^nx}(cosx + isinx)^n - \frac{1}{sin^nx}(cosx - isinx)^n}{2i}$

$= \frac{(cosx + isinx)^n - (\frac{(cosx - isinx)(cosx + isinx)}{cosx + sinx})^n}{2isin^nx}$

$= \frac{(cosx + isinx)^n - \frac{1}{(cosx + isinx)^n}}{2isin^nx}$

$= \frac{(cosx + isinx)^n - (cosx + isinx)^{-n}}{2isin^nx}$

$= \frac{cos{(nx)} + isin{(nx)} - (cos{(-nx)} + isin{(-nx)})}{2isin^nx}$

$= \frac{cos{(nx)} + isin{(nx)} - (cos{(nx)} - isin{(nx)})}{2isin^nx}$

$= \frac{2isin{(nx)}}{2isin^nx}$

$= \frac{sin{(nx)}}{sin^nx} = RHS$

yup

6. ## Re: HSC 2018 MX2 Marathon

Oh whoops... I read that as $\sin(x^n)$.

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
A classic one

$Simplify \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$
Before I write up the latex, is the answer

$\frac{sin(\frac{nx}{2})(sin(x\frac{n+1}{2}))}{sin( \frac{x}{2})} ???$

8. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
Before I write up the latex, is the answer

$\frac{sin(\frac{nx}{2})(sin(x\frac{n+1}{2}))}{sin( \frac{x}{2})} ???$
Correct!

9. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
A classic one

$Simplify \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$
Originally Posted by InteGrand
Correct!
$Let z = cosx + isinx$

$sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}(z + z^2 + z^3 + ... + z^n)$

$GP with a = z and r = z$

$\frac{z(1-z^n)}{1-z}$

$= \frac{(cosx + isinx)(1-(cosx + isinx)^n)}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- cosnx - isinnx))}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- (1 - 2sin^2\frac{nx}{2}) - 2isin\frac{nx}{2}cos\frac{nx}{2}))}{1- (1 - 2sin^2\frac{x}{2}) - 2isin\frac{x}{2})cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin^2\frac{nx}{2} - 2isin\frac{nx}{2}cos\frac{nx}{2})}{2sin^2\frac{x}{ 2} - 2isin\frac{x}{2}cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})}{(2sin\frac{x}{2})(sin\frac{x}{2 } - icos\frac{x}{2})}$

$= \frac{(cosx + isinx)(sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{(sin\frac{x}{2})(sin\frac{x}{2} - icos\frac{x}{2})(sin\frac{x}{2} + icos\frac{x}{2})}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{nx}{2}cos\frac{x}{2} + sin\frac{nx}{2}sin\frac{x}{2} + i(sin\frac{nx}{2}cos\frac{x}{2} - cos\frac{nx}{2}sin\frac{x}{2}))}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$Now sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}\left(\frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}\right)$

$= \frac{(sin\frac{nx}{2})(cosx)(sin\frac{x(n-1)}{2}) + (sin\frac{nx}{2})(sinx)(cos\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})\left((sin\frac{x(n-1)}{2})(cosx) + (cos\frac{x(n-1)}{2})(sinx)\right)}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{2x + xn - x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{x(n + 1)}{2})}{sin\frac{x}{2}}$

$I am never writing latex again$

10. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$Let z = cosx + isinx$

$sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}(z + z^2 + z^3 + ... + z^n)$

$GP with a = z and r = z$

$\frac{z(1-z^n)}{1-z}$

$= \frac{(cosx + isinx)(1-(cosx + isinx)^n)}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- cosnx - isinnx))}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- (1 - 2sin^2\frac{nx}{2}) - 2isin\frac{nx}{2}cos\frac{nx}{2}))}{1- (1 - 2sin^2\frac{x}{2}) - 2isin\frac{x}{2})cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin^2\frac{nx}{2} - 2isin\frac{nx}{2}cos\frac{nx}{2})}{2sin^2\frac{x}{ 2} - 2isin\frac{x}{2}cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})}{(2sin\frac{x}{2})(sin\frac{x}{2 } - icos\frac{x}{2})}$

$= \frac{(cosx + isinx)(sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{(sin\frac{x}{2})(sin\frac{x}{2} - icos\frac{x}{2})(sin\frac{x}{2} + icos\frac{x}{2})}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{nx}{2}cos\frac{x}{2} + sin\frac{nx}{2}sin\frac{x}{2} + i(sin\frac{nx}{2}cos\frac{x}{2} - cos\frac{nx}{2}sin\frac{x}{2}))}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$Now sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}\left(\frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}\right)$

$= \frac{(sin\frac{nx}{2})(cosx)(sin\frac{x(n-1)}{2}) + (sin\frac{nx}{2})(sinx)(cos\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})\left((sin\frac{x(n-1)}{2})(cosx) + (cos\frac{x(n-1)}{2})(sinx)\right)}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{2x + xn - x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{x(n + 1)}{2})}{sin\frac{x}{2}}$

$I am never writing latex again$
$\noindent Well done! Incidentially, another method is to consider$

$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from k=1 to n.$

11. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent Well done! Incidentially, another method is to consider$

$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from k=1 to n.$
Question 7 Of Ekman's Compilation
for the working out of integrands clever solution if anyones interested

12. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent Well done! Incidentially, another method is to consider$

$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from k=1 to n.$
Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?

13. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$I am never writing latex again$
Alternatively, you can use the fact that the Imaginary part of 1 is 0 to do this:

$\noindent S \equiv \Im{\left( \sum_{k=0}^n z^k \right)} \equiv \Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{2i\sin{\frac{(n+1)x}{2}}}{2i\sin{\frac{x}{2} }} \right )} \\\\ \equiv \Im{ \left( \left( \cos{\frac{nx}{2}}+i\sin{\frac{nx}{2}} \right ) \frac{\sin{\frac{(n+1)x}{2}}}{\sin{\frac{x}{2}}} \right )} \equiv \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

14. ## Re: HSC 2018 MX2 Marathon

I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0

$\Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$

Also, did you mean:

$\Im{\left( \frac{z^{n+1}-z}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$

15. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?
One thing (probably not the only thing) you can try is searching up past years' HSC 4U marathons here on BOS (including the Advanced Level ones, though these are typically harder than HSC style Q's or would be broken into multiple steps if asked in the HSC exam).

16. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
Also, did you mean:

$\Im{\left( \frac{z^{n+1}-z}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$
That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.

17. ## Re: HSC 2018 MX2 Marathon

That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.
Sorry if I'm missing something obvious, but just to clarify

$\noindent S = \Im{\left( \sum_{k=0}^n z^k \right)} = \Im(z^0 + z^1 + z^2 + ... + z^{n+1}) = \Im(1 + z^1 + z^2 + ... + z^{n+1})$

$= \Im(\frac{(1)(z^{n+1} - 1)}{z - 1}) = ... = \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha

18. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
Sorry if I'm missing something obvious, but just to clarify

$\noindent S = \Im{\left( \sum_{k=0}^n z^k \right)} = \Im(z^0 + z^1 + z^2 + ... + z^{n+1}) = \Im(1 + z^1 + z^2 + ... + z^{n+1})$

$= \Im(\frac{(1)(z^{n+1} - 1)}{z - 1}) = ... = \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha
$\noindent Remember, adding 1 (or any real number) to something won't change its imaginary part! (You can probably prove this fact easily if you're unsure of it.) So to compute$

$\mathrm{Im}\left(z+z^{2} +\cdots + z^{n}\right),$

$\noindent you can just compute$

$\mathrm{Im}\left(\color{red}{1}\color{black}+z+z ^{2} + \cdots + z^{n}\right).$

$\noindent This is what Paradoxica did. (The advantage of doing this is it makes the resulting GP formula a bit simpler compared to if we leave out the \color{red}{1}\color{black}, so it's easier to compute the imaginary part of.)$

19. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent Remember, adding 1 (or any real number) to something won't change its imaginary part! (You can probably prove this fact easily if you're unsure of it.) So to compute$

$\mathrm{Im}\left(z+z^{2} +\cdots + z^{n}\right),$

$\noindent you can just compute$

$\mathrm{Im}\left(\color{red}{1}\color{black}+z+z ^{2} + \cdots + z^{n}\right).$

$\noindent This is what Paradoxica did. (The advantage of doing this is it makes the resulting GP formula a bit simpler compared to if we leave out the \color{red}{1}\color{black}, so it's easier to compute the imaginary part of.)$
Yeah that makes more sense now, its pretty creative. Thank you

20. ## Re: HSC 2018 MX2 Marathon

$New Question!$

$Find the smallest \textbf{positive} value of m if (1 + i)^m = (1 - i)^m$

21. ## Re: HSC 2018 MX2 Marathon

Is the answer $m=4$?

(I intend to post a writeup later)

22. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
Is the answer $m=4$?

(I intend to post a writeup later)
yes it's 4

23. ## Re: HSC 2018 MX2 Marathon

$(1+i)^m = (1-i)^m$

Converting to polar form,

$m\sqrt{2}\left(\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4} \right)=m\sqrt{2}\left(\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4} \right)$

Since $m \neq 0$,

$\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4}=\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4}$

The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,

$\sin \frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

Since the sine function is odd, $\sin x = - \sin -x$ for all $x$.

$-\sin -\frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

And clearly this is only true when $\sin \frac{m\pi}{4} = 0$.

So therefore, $\frac{m\pi}{4} = \pi \implies m = 4$

24. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
$(1+i)^m = (1-i)^m$

Converting to polar form,

$m\sqrt{2}\left(\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4} \right)=m\sqrt{2}\left(\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4} \right)$

Since $m \neq 0$,

$\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4}=\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4}$

The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,

$\sin \frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

Since the sine function is odd, $\sin x = - \sin -x$ for all $x$.

$-\sin -\frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

And clearly this is only true when $\sin \frac{m\pi}{4} = 0$.

So therefore, $\frac{m\pi}{4} = \pi \implies m = 4$
Nice!

$Alternatively (1+i)^m = (1-i)^m$

$\noindent \frac{(1+i)^m}{(1-i)^m} = 1$

$\frac{m\sqrt{2}(cis \frac{m\pi}{4})}{m\sqrt{2}(cis \frac{-m\pi}{4})} = 1$

$cis \frac{m\pi}{2} = cis0$

$\frac{m\pi}{2} = 0 + 2k\pi where k = 0, -1, 1, -2, 2...$

$m = 4k \therefore smallest positive value when k = 1 (ie m=4)$

25. ## Re: HSC 2018 MX2 Marathon

2u integration+circle geo
i made this question myself in paint so dont mind the bad production quality.
edit: look for the question which has the circle with one edge on -5 instead of -3. ignore the question where the circle touches -3.
integral composite graph.png

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