HSC Mathematics Marathon (1 Viewer)

UnrealAnchovies · Feb 7, 2011

Riproot said:
Fuck it. I'll just do my English essay... ewwwwwwwww...

+1

Ugh.

roryclifford · Feb 7, 2011

Riproot said:
OMFG That's the book we use for 4u at my school! Just sayin'
I havent actually opened it yet though lol

Haha yewwww
I went sweet in my first 4 unit assessment for the year, then did the summary exersise for the topic out of that textbook and mmmm... Struggle is the word that comes to mind.

Worried now hahah

Riproot · Feb 7, 2011

roryclifford said:
Haha yewwww
I went sweet in my first 4 unit assessment for the year, then did the summary exersise for the topic out of that textbook and mmmm... Struggle is the word that comes to mind.

Worried now hahah

Shit... How's the multiplicity stuff? I have to do that tomorrow night :/

ohexploitable · Feb 7, 2011

Riproot said:
Shit... How's the multiplicity stuff? I have to do that tomorrow night :/

Multiplication is easy peasy, it gets a bit harder when you're multiplying by double digits though :S

Riproot · Feb 7, 2011

ohexploitable said:
Multiplication is easy peasy, it gets a bit harder when you're multiplying by double digits though :S

You seriously just made me like rofl. like fuckin' shit.
That's the best! LOL
I need to spread rep before giving you more :-(

ohexploitable · Feb 7, 2011

Riproot said:
You seriously just made me like rofl. like fuckin' shit.
That's the best! LOL
I need to spread rep before giving you more

There was some good stuff in the "Hardest general maths topic?" thread but it doesn't work anymore.

Riproot · Feb 7, 2011

ohexploitable said:
There was some good stuff in the "Hardest general maths topic?" thread but it doesn't work anymore.

:-(
I hope they get it working soon lol

ohexploitable · Feb 7, 2011

Hmm... lets get another question going...

$\\\text{Using }z^n+z^{-n}=2\cos n\theta\text{ and }z^n-z^{-n}=2i\sin n\theta,\\\\\text{Prove }\sin^2\theta\cos^3\theta=\frac{1}{16}(2\cos\theta-\cos3\theta-\cos5\theta)$

Riproot · Feb 7, 2011

ohexploitable said:
Hmm... lets get another question going...

$\\\text{Using }z^n+z^{-n}=2\cos n\theta\text{ and }z^n-z^{-n}=2i\sin n\theta,\\\\\text{Prove }\sin^2\theta\cos^3\theta=\frac{1}{16}(2\cos\theta-\cos3\theta-\cos5\theta)$

I was gonna answer this, but then I thought whether anyone has seen me do maths on here before. I don't think they have. I don't want to ruin it now.
I'm terrible at maths :-D

UnrealAnchovies · Feb 7, 2011

Where are the other 11ers?

I want to keep posting/ answering but I've done topics that exploit/rip haven't. Which sucks.

Drongoski · Feb 7, 2011

(z+z^-1)³ = (2cos@)³ = 8cos³@

(z-z^-1)² = (2isin@)² = -4sin²@

.: multiplying both sides:

-32sin²@cos³@ = (z+z^-1)³ x (z-z^-1)²

$= (z^5+z^{-5}) + (z^3+z^{-3})-2(z+z^{-1})$

$= 2cos5 \theta + 2cos3 \theta -4cos \theta$

$\therefore sin ^{2}\theta cos^{3} \theta = \frac {1}{16} (2cos \theta - cos3 \theta - cos 5 \theta )$

Edit: Sorry - hopeless with TeX

AAEldar · Feb 7, 2011

ohexploitable said:
Hmm... lets get another question going...

$\\\text{Using }z^n+z^{-n}=2\cos n\theta\text{ and }z^n-z^{-n}=2i\sin n\theta,\\\\\text{Prove }\sin^2\theta\cos^3\theta=\frac{1}{16}(2\cos\theta-\cos3\theta-\cos5\theta)$

If I was bothered I would, however I loathe questions like that D:

cutemouse · Feb 7, 2011

(i) If [maths]P(x)=3x^4-11x^3+14x^2-11x+3[/maths] show that

[maths]P(x)=x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8)[/maths]

(ii) Hence solve P(x)=0 over C.

Bored Of Fail 3 · Feb 7, 2011

jm01 said:
(i) If [maths]P(x)=3x^4-11x^3+14x^2-11x+3[/maths] show that

[maths]P(x)=x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8)[/maths]

(ii) Hence solve P(x)=0 over C.

thats just tons of algebraic manipulation

and lol u trying to copy ohexploitable with the sig :|

ohexploitable · Feb 7, 2011

jm01 said:
(i) If [maths]P(x)=3x^4-11x^3+14x^2-11x+3[/maths] show that

[maths]P(x)=x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8)[/maths]

(ii) Hence solve P(x)=0 over C.

$\\(i)\,3x^2(x+\frac{1}{x})^2-11x^2(x+\frac{1}{x})+8x^2\\=3(x^2+1)^2-11x^3-11x+8x^2\\=3x^4+6x^2+3-11x^3-11x+8x^2\\=3x^4-11x^3+14x^2-11x+3\\\\(ii)\,x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8)=0\\x^2(3u^2-11u+8)=0\\x=0,\,u=1,\,\frac{8}{3}\\x=0,\,\pm\sqrt[3]{-1},\,\frac{4\pm\sqrt{7}}{3}$

deterministic · Feb 7, 2011

ohexploitable said:
$\\(i)\,3x^2(x+\frac{1}{x})^2-11x^2(x+\frac{1}{x})+8x^2\\=3(x^2+1)^2-11x^3-11x+8x^2\\=3x^4+6x^2+3-11x^3-11x+8x^2\\=3x^4-11x^3+14x^2-11x+3\\\\(ii)\,x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8)=0\\x^2(3u^2-11u+8)=0\\x=0,\,u=1,\,\frac{8}{3}\\x=0,\,\pm\sqrt[3]{-1},\,\frac{4\pm\sqrt{7}}{3}$

x=0 is not a solution (sub into original equation to see)
and you have 5 solutions for degree 4 polynomial??

Riproot · Feb 7, 2011

deterministic said:
x=0 is not a solution (sub into original equation to see)
and you have 5 solutions for degree 4 polynomial??

If 0 isn't a solution he has 4

Trebla · Feb 7, 2011

Note that for the case where u = 1
x + 1/x = 1
=> x² - x + 1 = 0
=> x = (1 ± i√3)/2

Trebla · Feb 7, 2011

If z = cos θ + i sin θ, then show that (without using induction)

$1+z+z^2+....+z^{n-1}=\dfrac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\left(\cos\dfrac{(n-1)\theta}{2}+i\sin\dfrac{(n-1)\theta}{2}\right)$

NewiJapper · Feb 7, 2011

if i = sqrt -1, find i^2.

HSC Mathematics Marathon (1 Viewer)

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