MX2 Marathon (1 Viewer)

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$From a ruse paper:$

$If $z = \cos\theta + i\sin\theta$, prove that $\frac{2}{1 + z} = 1 - i\tan\left(\frac{\theta}{2}\right)$$

An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$\text{Let}\,t=\tan\left(\frac{\theta}{2}\right)$

$=1 - i\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{1+t^2}{1+t^2}}$

$=1 - i\frac{2t}{2}$

$=1 - i\tan\left(\frac{\theta}{2}\right),\,\theta \neq \pm\pi$

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

For anyone interested, the Patel question I was talking about is:

$\text{Prove that}\,\, (1-\cos\theta + 2i\sin\theta)^{-1} = \frac{1-2i\cot\left(\frac{\theta}{2}\right)}{5+3\cos\theta}$

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$\text{Let}\,t=\tan\left(\frac{\theta}{2}\right)$

$=1 - i\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{1+t^2}{1+t^2}}$

$=1 - i\frac{2t}{2}$

$=1 - i\tan\left(\frac{\theta}{2}\right),\,\theta \neq \pm\pi$

Nice! Here is my solution:

$\frac{2}{1+z}$

$= \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{(\cos\theta+1)^2+\sin^2\theta}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{\cos^2\theta+2\cos\theta+1+(1-\cos^2\theta)}$

$=\frac{2\cos\theta + 2-2i\sin\theta}{2\cos\theta+2}$

$=1 - i\frac{\sin\theta}{\cos\theta+1}$

$Using the double angle formulae...$

$=1 -i\frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{2\cos^2(\frac{\theta}{2})-1+1}$

$=1 - i\frac{\sin(\frac{\theta}{2})}{\cos(\frac{\theta}{2})}$

$= 1 -i\tan\left(\frac{\theta}{2}\right)$

$Where \theta \neq \pm\pi$ $ (not gunna lie I completely forgot about this haha)$$

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
For anyone interested, the Patel question I was talking about is:

$\text{Prove that}\,\, (1-\cos\theta + 2i\sin\theta)^{-1} = \frac{1-2i\cot\left(\frac{\theta}{2}\right)}{5+3\cos\theta}$

$This is probably more convoluted than using the t-method$

$$LHS $ = \frac{1}{1-\cos\theta + 2i\sin\theta} \times \frac{1-cos\theta -2i\sin\theta}{1-cos\theta -2i\sin\theta}$

$= \frac{1-\cos\theta -2i\sin\theta}{1-2\cos\theta + cos^2\theta + 4\sin^2\theta}$

$Using \sin^2\theta = 1 -\cos^2\theta$

$=\frac{1-\cos\theta -2i\sin\theta}{5 - 3cos^2\theta -2\cos\theta}$

$=\frac{1-\cos\theta -2i\sin\theta}{(3\cos\theta + 5)(1-\cos\theta)}$

$Splitting fractions and applying double angle formulae...$

$= \frac{1}{3\cos\theta + 5} - \frac{4i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{(3\cos\theta + 5)(2-2\cos^2(\frac{\theta}{2}))}$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{(3\cos\theta + 5)(1-\cos^2(\frac{\theta}{2}))}$

$Again using \sin^2\theta = 1 -\cos^2\theta$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{(3\cos\theta + 5)(\sin^2(\frac{\theta}{2}))}$

$= \frac{1}{3\cos\theta + 5} - \frac{2i\cot(\frac{\theta}{2}))}{3\cos\theta + 5}$

$= \frac{1-2i\cot(\frac{\theta}{2})}{5+3\cos\theta}$

$= RHS$

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)

Ooh these hints

jathu123 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
This is an interesting question I found in the SGS notes.

If $z$ is a complex number, find the locus of $z$ if

$\mathrm{Re}\left(z-\frac{1}{z}\right) = 0$

let z= rcosθ+isinθ
z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ
Re(z-1/z) = (r-1/r)cosθ

If LHS = 0, then (r-1/r) = 0 or cosθ = 0
hence, r=1 or θ= ±(2n+1)π/2

Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))

altSwift · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

jathu123 said:
let z= rcosθ+isinθ
z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ
Re(z-1/z) = (r-1/r)cosθ

If LHS = 0, then (r-1/r) = 0 or cosθ = 0
hence, r=1 or θ= ±(2n+1)π/2

Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))

ahahaha I forgot to consider x = 0 jesus

jathu123 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

A classic one

$$Simplify $ \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$

fan96 · Feb 10, 2018

Re: HSC 2018 MX2 Marathon

jathu123 said:
let z= rcosθ+isinθ
z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ
Re(z-1/z) = (r-1/r)cosθ

If LHS = 0, then (r-1/r) = 0 or cosθ = 0
hence, r=1 or θ= ±(2n+1)π/2

Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))

Yep.

If you used $z = x + iy$ , you ended up with the locus

$x^3 + xy^2 = x$

Which is the correct answer.

It's tempting to divide both sides by $x$ , but if you performed the division you were implicitly assuming that $x \neq 0$ , which lost solutions.

Clearly if $x=0$ , any value of $y$ is a solution, except for zero since $z \neq 0$ .

That can also be shown by considering
$\text{Re}\left(iy + \frac{1}{iy}\right)= \text{Re}\left(iy - i\frac{1}{y}\right)$
which is zero for all $y \neq 0$ .

To graph $x^3 + xy^2 = x$ , you could consider two separate cases: $x=0$ , and $x\neq 0$ .

This is the solution from the SGS notes:

altSwift · Feb 11, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2+2z-1=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n

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?

jathu123 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
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 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
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 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

jathu123 said:
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 said:
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 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
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

mrbunton said:
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:$

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

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

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

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

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

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

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

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

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

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

fan96 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

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

altSwift · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

jathu123 said:
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 · Feb 12, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
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 · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

jathu123 said:
A classic one

$$Simplify $ \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$

InteGrand said:
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 · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$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 $n$.$

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