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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: darkphoenix on December 20, 2009, 03:19:52 pm

Title: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 03:19:52 pm: How would you:

Factorise $6z^4 - z^2 -2$ over C
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 20, 2009, 03:23:10 pm: Hint: Let $u = z^2$

$\implies 6u^2-u-2$ .
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 03:43:22 pm: Hm any more help? hahah
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 20, 2009, 03:57:53 pm: From where I left off:

$(2u+1)(3u-2)=0$

$(2z^2+1)(3z^2-2)=0$

$(2z^2-i^2)(3z^2-2)=0$

Try DOPS on the first factor and factorise the second one normally.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 04:06:16 pm: Alrite got it, thanks TT

EDIT: Got another question:

If $2+ 3i$ is a zero of $P(z)= 2z^3-11z^2+38z-39=0$ , find the other two zeroes.
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 20, 2009, 04:42:50 pm: $z-(2+3\mathbf{i})$ is a factor.

$\frac{2z^3-11z^2+38z-39}{z-(2+3\mathbf{i})} = (z-2)(2z-3)+3(2z-3)\mathbf{i}$

$\therefore (z-2)(2z-3)+3(2x-3)\mathbf{i} = 0+0\mathbf{i}$

$\implies (z-2)(2z-3) = 0 \ \mbox{and} \ 3(2z-3) = 0$

Rest is trivial.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 04:50:49 pm: Wait so how did you get the answer after you divided by $z - (2+3i)$
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 20, 2009, 04:51:58 pm: Long division is used. I would use paint to show long division but my paint skills really suck.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 05:28:39 pm: Ah ok fair enough. How about a question on vectors:

A Hiker travels 3km north, then 4km west. How far is the hiker from the starting point? What is the bearing of the resultant displacement?

I've worked out the distance is 5km from starting point already.
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 20, 2009, 05:31:30 pm: Does the question say, the bearing from the starting point or the bearing from the final position of the hiker?

If it doesn't state you can work out either angle (I assume you have drew a graph)
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 20, 2009, 05:41:33 pm: Yeah it doesnt say.

EDIT: How would i approach this question?

Use vectors to show that the diagonals of a square intersect at right angles.

Would i just use the scalar product and show that 2 of the lines equal 0 are are perpendicular or something?
Title: Re: Help with Holiday Homework
Post by: Mao on December 21, 2009, 09:45:29 pm: In a square ABCD, let AB = a, BC = b, then CD = - a, DA = - b, also that a.b = 0, |a| = |b|

The diagonals are AC = AB + BC = a + b, and BD = BC + CD = b - a
Then, AC.BD = a.b - a.a + b.b - a.b = -|a|^2 + |a|^2 = 0

Therefore, diagonals intersect at right angles.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 22, 2009, 12:56:18 am: Alrite cool thanks Mao
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 01:09:26 pm: Hi how would you
use the formula
tan = 2x 2tanx/ 1-tan^2x to find a surd expression for tan pie/8
Title: Re: Help with Holiday Homework
Post by: kyzoo on December 22, 2009, 01:27:37 pm: Let $\frac{\pi}{8} = x$
$2x = \frac{\pi}{4}$

$tan(2x) = \frac{2 tanx}{1-tan^2 x}$
$tan(2x) - (tan(2x))(tan^2 x) -2 tanx = 0$
$tan(2x) = tan(\frac{\pi}{4}) = 1$

So we have $1 - tan^2 x - 2 tanx = 0$
$tan^2 x + 2 tan x - 1 = 0$
$(tan x + 1)^2 - 2 = 0$
$(tan x + 1)^2 = 2$
$tan x + 1 = \pm\sqrt{2}$
$tan x = \pm\sqrt{2} - 1$

But because $tan \frac{\pi}{8}$ is a positive number

Therefore $tan \frac{\pi}{8} = \sqrt{2} - 1$
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 01:35:41 pm: Thanksss ::)
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 02:18:25 pm: Hey just a few other questions

Given that tan(2x) = 4root2 / 7 where x --> [0, pie/4)
find the exact value of sin(x)

and

if cos(A) = sin(A-B)sin(B) prove that tan(A-B)tan(B) = 1/2

thanks
Title: Re: Help with Holiday Homework
Post by: cipherpol on December 22, 2009, 02:19:26 pm: Hey kyzoo, how'd you get $tan(2x) - (tan(2x))(tan^2 x) -2 tanx = 0$ ?
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 02:22:42 pm: Consider $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw a right-angle triangle.

Opposite side is $4\sqrt{2}$ and the adjacent is $7$ .

The hypotenuse is given by $\sqrt{(4\sqrt{2})^2+7^2}$

$\therefore \sin(X) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\implies \sin(2x) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\frac{\sin(2x)}{\cos(2x)} = \frac{4\sqrt{2}}{7}$

$\implies \cos(2x) = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\frac{4\sqrt{2}}{7}}$

I smell some double angle formulas and wishful thinking.

Can you do the rest?
Title: Re: Help with Holiday Homework
Post by: kyzoo on December 22, 2009, 02:33:19 pm: Quote from: cipherpol on December 22, 2009, 02:19:26 pm
Hey kyzoo, how'd you get $tan(2x) - (tan(2x))(tan^2 x) -2 tanx = 0$ ?

$(tan(2x)) = \frac{2 tanx}{1-tan^2 x}$
$(tan(2x))(1-tan^2 x)=2 tan x$
$tan(2x) - (tan(2x))(tan^2 x) - 2 tan x = 0$
Title: Re: Help with Holiday Homework
Post by: cipherpol on December 22, 2009, 02:37:53 pm: i must have been retarded, thanks :]
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 04:31:06 pm: sorry i dont get why you did the last part to find cos(2x)
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 04:33:05 pm: Quote from: TrueTears on December 22, 2009, 02:22:42 pm
Consider $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw a right-angle triangle.

Opposite side is $4\sqrt{2}$ and the adjacent is $7$ .

The hypotenuse is given by $\sqrt{(4\sqrt{2})^2+7^2}$

$\therefore \sin(X) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\implies \sin(2x) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\frac{\sin(2x)}{\cos(2x)} = \frac{4\sqrt{2}}{7}$

$\implies \cos(2x) = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\frac{4\sqrt{2}}{7}}$

I smell some double angle formulas and wishful thinking.

Can you do the rest?
$\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\cos(2x)} = \frac{4\sqrt{2}}{7}$
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 04:40:07 pm: but it says to find the value of sin(x)

sorry
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 04:41:46 pm: I know :)

That is the crux of the question, the rest is just simply playing around with double angle formulas, try it yourself :)

$\cos(2x) = 1-2\sin(x)^2$
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 05:20:42 pm: thanks i got it,
one quick question how do i get started on

evaluate P= sin6sin42sin66sin78
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 05:24:36 pm: $\sin(6) \sin(42) \sin(66) \sin(78) = \left(\frac{1}{8} \left(\sqrt{30 - 6 \sqrt{5}} - 1 - \sqrt{5}\right)\right)\left(\frac{1}{8} \left(1 + \sqrt{30 + 6 \sqrt{5}} - \sqrt{5}\right)\right)\left( \frac{1}{8} \left(1 + \sqrt{30 - 6 \sqrt{5}} + \sqrt{5}\right)\right)\left(\frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}}\right)$ .
Title: Re: Help with Holiday Homework
Post by: ChristineNguyen on December 22, 2009, 05:30:32 pm: LOL

how did you work those out? ???
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 05:49:43 pm: From the Taylor Series

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ where $x$ is in radians.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 22, 2009, 11:32:04 pm: Wow thanks ChristineNguyen for hijacking my topic! Haha jks :)

Um got another question:

Use De Moivre's theorem to solve the equation $z^2= -3-3i$ . (Give your answers in polar form.)

I tried working it out and got a different answer to what it was supposed to be..
Title: Re: Help with Holiday Homework
Post by: zzdfa on December 22, 2009, 11:38:19 pm: polar form: z^2=18^(1/2)cis(-3PI/4)

thus z=18^(1/4)cis(-3pi/8)

and of course -3pi/8 is the same angle as 5pi/8.

was your answerb ''off'' by 2pi?

edit: fail mistake fixd
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 22, 2009, 11:43:35 pm: $-3-3i = 3\sqrt{2}Cis\left(\frac{-3\pi}{4}\right)$

$z^2 = 3\sqrt{2}Cis\left(\frac{-3\pi}{4}\right)$

Let $z = rCis\left(\theta\right)$

$r^2Cis\left(2\theta\right)=3\sqrt{2}Cis\left(\frac{-3\pi}{4}\right)$

$r = \left(3\sqrt{2}\right)^{\frac{1}{2}}$

$2\theta = \frac{-3\pi}{4} + 2k\pi \ \forall \ k \in \mathbb{Z}$

$\theta = \frac{1}{2}\left(\frac{-3\pi}{4} + 2k\pi\right) \ \forall \ k \in \mathbb{Z}$

$k = 0$

$\theta = \frac{-3\pi}{8}$

$k = 1$

$\theta = \frac{5\pi}{8}$

$\therefore z = \left(3\sqrt{2}\right)^{\frac{1}{2}}Cis\left(\frac{-3\pi}{8}\right) , \left(3\sqrt{2}\right)^{\frac{1}{2}}Cis\left(\frac{5\pi}{8}\right)$
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 23, 2009, 12:48:49 am: Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 23, 2009, 12:49:55 am: Quote from: darkphoenix on December 23, 2009, 12:48:49 am
Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.
Are you using Essentials? They sucked for Complex numbers.
Title: Re: Help with Holiday Homework
Post by: kyzoo on December 23, 2009, 01:02:56 am: Quote from: TrueTears on December 23, 2009, 12:49:55 am
Quote from: darkphoenix on December 23, 2009, 12:48:49 am
Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.
Are you using Essentials? They sucked for Complex numbers.

What's with this "The derivation of this formula is left as an exercise for the reader." So annoying, they never had this in the Methods textbook =(
Title: Re: Help with Holiday Homework
Post by: TrueTears on December 23, 2009, 01:11:30 am: Quote from: kyzoo on December 23, 2009, 01:02:56 am
Quote from: TrueTears on December 23, 2009, 12:49:55 am
Quote from: darkphoenix on December 23, 2009, 12:48:49 am
Yeah i forgot to minus the 2pi. Thanks guys, the textbook was harder to understand.
Are you using Essentials? They sucked for Complex numbers.

What's with this "The derivation of this formula is left as an exercise for the reader." So annoying, they never had this in the Methods textbook =(
lol I see so much of that in the book I'm reading atm.
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 23, 2009, 02:19:02 pm: Nah our school is using the new edition of Maths quest, it sucks. :(

EDIT: One more question, this on vectors:

If a = 2i - 2j + k and b = -i + j - 4k, find a unit vector perpendicular to a and b.
Title: Re: Help with Holiday Homework
Post by: Mao on December 23, 2009, 06:46:16 pm: let c = x i + y j + z k

c . a = 0
2x - 2y + z = 0 [1]

c . b = 0
-x + y - 4z = 0 [2]

(simultaneous equations) 2*[2] + [1]
-7z = 0
z = 0

substituting back into [2] (or [1], will yield the same result)
-x + y = 0
y = x

also, since c is a unit vector, $x^2 + y^2 + z^2 = 1$
since z = 0, y = x, substituting yields
$x^2 + x^2 + 0^2 = 1 \implies 2x^2 = 1 \implies x = \pm \frac{\sqrt{2}}{2}$

Hence $\textbf{c} = \frac{\sqrt{2}}{2}\textbf{i} + \frac{\sqrt{2}}{2}\textbf{j}$ or $\textbf{c} = -\frac{\sqrt{2}}{2}\textbf{i} - \frac{\sqrt{2}}{2}\textbf{j}$
Title: Re: Help with Holiday Homework
Post by: darkphoenix on December 23, 2009, 07:06:03 pm: thanks mao for the quick response.
Title: Re: Help with Holiday Homework
Post by: kakar0t on January 07, 2010, 09:25:33 pm: Can someone give me a hand with this question please? :D

(http://i50.tinypic.com/2zydgd5.jpg)
Title: Re: Help with Holiday Homework
Post by: kamil9876 on January 07, 2010, 10:51:25 pm: a.)

$|OA|=|OC|$ because it is a square. Thus: $|u|=|z|$ (1)

$<AOC=\frac{\pi}{2} \implies Arg(z)=Arg(u+\frac{\pi}{2})$

$\therefore z=|z|cis(Arg(z))=|u|cis(Arg(u+\frac{\pi}{2}))=|u|cis(Arg(u)) *i=ui$

b.) $w=u+z=u+ui=u(1+i)$
Title: Re: Help with Holiday Homework
Post by: kakar0t on January 08, 2010, 01:53:52 pm: Thanks man!