ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: rainbows. on March 21, 2010, 09:14:09 pm

Title: Homework Help :)
Post by: rainbows. on March 21, 2010, 09:14:09 pm: I was wondering, I dont remember the question really, but i think its

Prove/expand/solve/whatever it is of something cos(x/2).. I know im not really asking a question but i remember it has something about a cos(x/2) i need to remember about it =="
Title: Re: Homework Help :)
Post by: superflya on March 21, 2010, 09:17:03 pm: lol wah?
Title: Re: Homework Help :)
Post by: the.watchman on March 21, 2010, 09:21:52 pm: Quote from: superflya on March 21, 2010, 09:17:03 pm
lol wah?

LOL I don't think we can do much without the question ;D
Title: Re: Homework Help :)
Post by: rainbows. on March 21, 2010, 09:22:40 pm: Umm, i think the question was something double angle formula related to the cos(x/2) like using the double angle formula to prove it ?
Title: Re: Homework Help :)
Post by: superflya on March 21, 2010, 09:24:26 pm: yea umm prove $\cos (\frac{x}{2})$ = wat?
Title: Re: Homework Help :)
Post by: rainbows. on March 21, 2010, 09:27:24 pm: Urgh.. im trying to find the question ==" *runs around room, looking through piles of clothes and books everywhere*

Randomly can someone express this 1 + tan(x)i in modulus-argument form, 0 less than x less than pi/2

i = complex
Title: Re: Homework Help :)
Post by: m@tty on March 21, 2010, 09:35:21 pm: $1+\tan{x}\cdot i=1+\frac{\sin{x}\cdot i}{\cos{x}}=\frac{\cos{x}}{\cos{x}} \cdot \left(1+\frac{\sin{x}\cdot i}{\cos{x}}\right)=\frac{1}{\cos{x}} \cdot \left(\cos{x}+\sin{x}\cdot i\right) = \frac{1}{\cos{x}} \cdot cis{x}$

$\therefore \ |1+\tan{x}\cdot i|=\frac{1}{\cos{x}}=\sec{x}$

I think...

That should be the modulus; but what is 'modulus-argument form'?

EDIT: You could have also just taken the sum of the squares of the imaginary and real parts and square rooted them.

ie. $|1+\tan{x}\cdot i|=\sqrt{Re(1+\tan{x}\cdot i)^2+Im(1+\tan{x}\cdot i)}=\sqrt{1+\tan^2{x}}$

And the Pythagorean identity states:
$\cos^2{x}+\sin^2{x}=1$

$\frac{\cos^2{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}$

$tan^2{x}+1=sec^2{x}$

So $\sqrt{1+\tan^2{x}}=\sqrt{\sec^2{x}}=\sec{x}=\frac{1}{\cos{x}}$
Title: Re: Homework Help :)
Post by: rainbows. on March 21, 2010, 09:38:08 pm: Im not sure, but your right (Y)
Title: Re: Homework Help :)
Post by: superflya on March 21, 2010, 09:39:35 pm: mod-arg form is the same as polar form.
Title: Re: Homework Help :)
Post by: rainbows. on March 21, 2010, 09:41:04 pm: Oh ah..
Title: Re: Homework Help :)
Post by: m@tty on March 21, 2010, 09:44:15 pm: Quote from: superflya on March 21, 2010, 09:39:35 pm
mod-arg form is the same as polar form.
I've never heard of it before... :\
Title: Re: Homework Help :)
Post by: the.watchman on March 21, 2010, 09:49:59 pm: Quote from: m@tty on March 21, 2010, 09:44:15 pm
Quote from: superflya on March 21, 2010, 09:39:35 pm
mod-arg form is the same as polar form.
I've never heard of it before... :\

LOL the modulus is the co-efficient of the 'cis' or |z|
The argument is the angle taken by the 'cis' :P
Title: Re: Homework Help :)
Post by: m@tty on March 21, 2010, 09:53:35 pm: I know what modulus and argument are, I also know what polar form is. I just didn't know that....

Wait, now it makes sense, Magnitude-Angle form -- of course it's polar! :buck2:
Title: Re: Homework Help :)
Post by: samuch on March 21, 2010, 09:54:24 pm: Quote from: the.watchman on March 21, 2010, 09:49:59 pm
Quote from: m@tty on March 21, 2010, 09:44:15 pm
Quote from: superflya on March 21, 2010, 09:39:35 pm
mod-arg form is the same as polar form.
I've never heard of it before... :\

LOL the modulus is the co-efficient of the 'cis' or |z|
The argument is the angle taken by the 'cis' :P
you make me depressed because you are in year 11 and you know all this!!!
*wishes that i was like you last year*
Title: Re: Homework Help :)
Post by: the.watchman on March 21, 2010, 09:55:48 pm: Quote from: m@tty on March 21, 2010, 09:53:35 pm
I know what modulus and argument are, I also know what polar form is. I just didn't know that....

Wait, now it makes sense, Magnitude-Angle form -- of course it's polar :buck2:

Haha, now you get it :D

Quote from: samuch on March 21, 2010, 09:54:24 pm
Quote from: the.watchman on March 21, 2010, 09:49:59 pm
Quote from: m@tty on March 21, 2010, 09:44:15 pm
Quote from: superflya on March 21, 2010, 09:39:35 pm
mod-arg form is the same as polar form.
I've never heard of it before... :\

LOL the modulus is the co-efficient of the 'cis' or |z|
The argument is the angle taken by the 'cis' :P
you make me depressed because you are in year 11 and you know all this!!!
*wishes that i was like you last year*

Well, we don't do anything in GMA, so I learn myself :D
Title: Re: Homework Help :)
Post by: rainbows. on March 21, 2010, 10:05:06 pm: OHHH someone help me factorise Z^6 - 64 please :(
Title: Re: Homework Help :)
Post by: kamil9876 on March 21, 2010, 10:15:14 pm: solve $z^6-64=0$ (Polar form works really sweet for this one)

You should get 6 roots $z_1,z_2...z_6$

Therefore $z^6-64=(z-z_1)(z-z_2)...(z-z_6)$
Title: Re: Homework Help :)
Post by: Mao on March 22, 2010, 02:43:21 am: $z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z+1-\sqrt{3}i)$
Title: Re: Homework Help :)
Post by: moekamo on March 22, 2010, 02:52:11 am: Quote from: Mao on March 22, 2010, 02:43:21 am
$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$
Title: Re: Homework Help :)
Post by: Mao on March 22, 2010, 03:06:40 am: Quote from: moekamo on March 22, 2010, 02:52:11 am
Quote from: Mao on March 22, 2010, 02:43:21 am
$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$

yes, that'll teach me about copy/pasting tex codes.
Title: Re: Homework Help :)
Post by: the.watchman on March 22, 2010, 06:48:49 am: Quote from: Mao on March 22, 2010, 03:06:40 am
Quote from: moekamo on March 22, 2010, 02:52:11 am
Quote from: Mao on March 22, 2010, 02:43:21 am
$z^6-64 = (z^3 + 8)(z^3-8) = (z+2)(z^2-2z+4)(z-2)(z^2+2z+4) = (z-2)(z+2)(z-1+\sqrt{3}i)(z-1-\sqrt{3}i)(z+1+\sqrt{3}i)(z-1-\sqrt{3}i)$

you have two $(z-1-\sqrt{3}i)$ factors in here... shouldnt one of them be $(z+1-\sqrt{3}i)$

yes, that'll teach me about copy/pasting tex codes.

LOL i do that all the time... :P