HSC 2013 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Mar 29, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work guys,

$Simplify$

$\sin \theta \sin(2\theta) + \sin (2\theta)\sin(3\theta) + \dots + \sin(n\theta) \sin((n+1)\theta)$

First method I used was:

$cos(A+B)=cosAcosB-sinAsinB$

$cos(A-B)=cosAcosB+sinAsinB$

Subtracting the two gives:

$2sinAsinB=cos(A-B)+cos(A+B)$

$sinAsinB=\frac{1}{2}[cos(A-B)+cos(A+B)]$

Using this on each term in the series gives:

$\frac{ncos\theta}{2}-\frac{1}{2}[cos3\theta +cos5\theta + ... + cos(2n+1)\theta]$

Now consider:

$cis^3\theta + cis^5\theta + ... + cis^{2n+1}\theta = \frac{cis^3\theta(cis^{2n}\theta-1)}{cis^2\theta-1}$

Taking the real part of this gives the sum of the cosines we need, so since I cbf evaluating the real part of this, your series is equal to:

$\frac{ncos\theta}{2}-\frac{1}{2}Re(\frac{cis^{2n+3}\theta-cis^3\theta}{cis^2\theta-1})$

Sy123 · Mar 29, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
First method I used was:

$cos(A+B)=cosAcosB-sinAsinB$

$cos(A-B)=cosAcosB+sinAsinB$

Subtracting the two gives:

$2sinAsinB=cos(A-B)+cos(A+B)$

$sinAsinB=\frac{1}{2}[cos(A-B)+cos(A+B)]$

Using this on each term in the series gives:

$\frac{ncos\theta}{2}-\frac{1}{2}[cos3\theta +cos5\theta + ... + cos(2n+1)\theta]$

Now consider:

$cis^3\theta + cis^5\theta + ... + cis^{2n+1}\theta = \frac{cis^3\theta(cis^{2n}\theta-1)}{cis^2\theta-1}$

Taking the real part of this gives the sum of the cosines we need, so since I cbf evaluating the real part of this, your series is equal to:

$\frac{ncos\theta}{2}-\frac{1}{2}Re(\frac{cis^{2n+3}\theta-cis^3\theta}{cis^2\theta-1})$

That is the method I had in mind, nice work.

This is the last series question I will post in a while, I want to ask more different questions. It is one that I'm trying to solve now but I haven't cracked it yet.

$Prove that$

$\tan \theta + \tan \left (\theta + \frac{\pi}{n} \right ) + \tan \left (\theta + \frac{2\pi}{n} \right ) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right ) = -n \cot \left (\frac{n\pi}{2} + n\theta \right )$

EDIT: I found a solution for it (not mine), using De Moivre to get a polynomial for tan(nx) (when making LHS zero) then using sum of roots.

Secant · Mar 30, 2013

Re: HSC 2013 4U Marathon

Totally an MX2 question:

Suppose p is an odd prime. Prove that:

$\sum_{j=0}^{p}\binom{p}{j}\binom{p+j}{j}\equiv 2^{p}+1$

$(mod$ $p^{2})$

~peace out homie~

twinklegal19 · Mar 30, 2013

Re: HSC 2013 4U Marathon

Evaluate

$\int_{0}^{1}\frac{x^2+3x+2}{x^2-x+1}dx$

Secant · Mar 30, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
Evaluate

$\int_{0}^{1}\frac{x^2+3x+2}{x^2-x+1}dx$

$\int_{0}^{1}\frac{x^2+3x+2}{x^2-x+1}dx$

$=\int \frac{x^2-x+1}{x^2-x+1}+2\int \frac{2x-1}{x^2-x+1}+\int \frac{3}{(x-\frac{1}{2})^{2}+\frac{3}{4}}$

$= x+2ln(x^2-x+1)+2\sqrt{3}tan^{-1}\left ( \frac{(2 x-1)}{\sqrt{3}} \right )$

The rest is easy

~peace out homie~

Secant · Mar 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That is the method I had in mind, nice work.

This is the last series question I will post in a while, I want to ask more different questions. It is one that I'm trying to solve now but I haven't cracked it yet.

$Prove that$

$\tan \theta + \tan \left (\theta + \frac{\pi}{n} \right ) + \tan \left (\theta + \frac{2\pi}{n} \right ) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right ) = -n \cot \left (\frac{n\pi}{2} + n\theta \right )$

I'm still workin on this question homie, tricky question but i'll get it

~peace out homie~

Sy123 · Mar 30, 2013

Re: HSC 2013 4U Marathon

$\int \tan (x) \tan(2x) \tan(3x) \ dx$

(provide proof)

GoldyOrNugget · Mar 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work guys,

$Simplify$

$\sin \theta \sin(2\theta) + \sin (2\theta)\sin(3\theta) + \dots + \sin(n\theta) \sin((n+1)\theta)$

FYI, this same question, but with cos instead of sin, was marked as one of the hardest questions in our algebra exercises book for Higher Maths 1A.

RealiseNothing · Mar 30, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
FYI, this same question, but with cos instead of sin, was marked as one of the hardest questions in our algebra exercises book for Higher Maths 1A.

serious? i thought it was pretty standard

Sy123 · Mar 30, 2013

Re: HSC 2013 4U Marathon

This was a good one (even though I said I won't post more series questions)

$$\sum_{k=1}^{n} \tan^{-1} \left (\frac{1}{1+k(k+1)} \right )$

OMGITzJustin · Mar 30, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
serious? i thought it was pretty standard

questions in that book are relatively easy, so in comparison with the others it was classified as one the hardest ones you can do out of all the questions for that particular exercise

Secant · Mar 31, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\int \tan (x) \tan(2x) \tan(3x) \ dx$

(provide proof)

tan(2x)=2tanx/(1-(tanx)^2) and tan(3x)=(3tanx-(tanx)^3)/(1-3(tanx)^2) therefore the answer is (1/2)[ln(2(cosx)^2-1]+(2/3)ln(cosx)-(1/3)[4(sinx)^2-1] + C

asianese · Mar 31, 2013

Re: HSC 2013 4U Marathon

^ Clearly a proof.

Secant · Mar 31, 2013

Re: HSC 2013 4U Marathon

asianese said:
^ Clearly a proof.

sick cunt dont need to prove

Sy123 · Mar 31, 2013

Re: HSC 2013 4U Marathon

Secant said:
tan(2x)=2tanx/(1-(tanx)^2) and tan(3x)=(3tanx-(tanx)^3)/(1-3(tanx)^2) therefore the answer is (1/2)[ln(2(cosx)^2-1]+(2/3)ln(cosx)-(1/3)[4(sinx)^2-1] + C

So how did you integrate:

$\int \tan x \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \frac{3\tan x - \tan^3 x}{1-3\tan^2 x} \ dx$

?

asianese · Mar 31, 2013

Re: HSC 2013 4U Marathon

Is the antiderivative expressible in terms of elementary functions?

Sy123 · Mar 31, 2013

Re: HSC 2013 4U Marathon

asianese said:
Is the antiderivative expressible in terms of elementary functions?

Very much so.

seanieg89 · Mar 31, 2013

Re: HSC 2013 4U Marathon

asianese said:
Is the antiderivative expressible in terms of elementary functions?

Any rational function of trig functions has elementary primitive.

t substitutions -> partial fractions -> integrating rational functions with at most quadratic denoms, all of which are elementary.

Sy123 · Mar 31, 2013

Re: HSC 2013 4U Marathon

This problem boils down to:

$\tan(3x) = \frac{\tan x + \tan(2x)}{1-\tan x \tan(2x)}$

Rearranging:

$\tan(3x) - \tan x \tan(2x) \tan(3x) = \tan(x) + \tan(2x)$

$\tan(x) \tan(2x) \tan(3x) = \tan(x) + \tan(2x) + \tan(3x)$

Which is easily integratable

RealiseNothing · Mar 31, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
This problem boils down to:

$\tan(3x) = \frac{\tan x + \tan(2x)}{1-\tan x \tan(2x)}$

Rearranging:

$\tan(3x) - \tan x \tan(2x) \tan(3x) = \tan(x) + \tan(2x)$

$\tan(x) \tan(2x) \tan(3x) = \tan(x) + \tan(2x) + \tan(3x)$

Which is easily integratable

$tan(x)tan(2x)tan(3x)=tan(3x)-tan(x)-tan(2x)$

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