HSC 2018 MX2 Marathon

**jathu123** · 12 Feb 2018, 8:22 AM

Originally Posted by mrbunton

glad someone noticed my question
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)

**altSwift** · 12 Feb 2018, 4:27 PM

Originally Posted by mrbunton

glad someone noticed my question
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

**mrbunton** · 12 Feb 2018, 5:02 PM

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

**altSwift** · 12 Feb 2018, 8:56 PM

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:$

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

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

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

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

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

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

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

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

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

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

**mrbunton** · 12 Feb 2018, 9:13 PM

yup

**fan96** · 12 Feb 2018, 9:17 PM

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

**altSwift** · 12 Feb 2018, 10:35 PM

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})} $ ???$$

**InteGrand** · 12 Feb 2018, 11:32 PM

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!

**altSwift** · 13 Feb 2018, 5:21 PM

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)$

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

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

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

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

$I am never writing latex again$

**InteGrand** · 13 Feb 2018, 5:37 PM

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 $n$.$

**jathu123** · 13 Feb 2018, 5:44 PM

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

**altSwift** · 13 Feb 2018, 5:48 PM

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?

**Paradoxica** · 13 Feb 2018, 6:49 PM

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}} }$

**altSwift** · 13 Feb 2018, 7:36 PM

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:

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

**InteGrand** · 13 Feb 2018, 8:04 PM

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).

**Paradoxica** · 13 Feb 2018, 8:05 PM

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.

**altSwift** · 13 Feb 2018, 11:57 PM

Originally Posted by Paradoxica

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

**InteGrand** · 14 Feb 2018, 6:18 AM

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.)$

**altSwift** · 14 Feb 2018, 8:42 AM

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

**altSwift** · 14 Feb 2018, 8:47 AM

$New Question!$

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

**fan96** · 14 Feb 2018, 11:15 AM

Is the answer $m=4$ ?

(I intend to post a writeup later)

**jathu123** · 14 Feb 2018, 12:44 PM

Originally Posted by fan96

Is the answer $m=4$ ?

(I intend to post a writeup later)

yes it's 4

**fan96** · 14 Feb 2018, 3:22 PM

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

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

**altSwift** · 14 Feb 2018, 4:17 PM

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)$$

**mrbunton** · 16 Feb 2018, 8:57 PM

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