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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Tea.bag on February 24, 2008, 02:06:25 am

Title: Specialist Problem
Post by: Tea.bag on February 24, 2008, 02:06:25 am: Can anyone explain this to me?

1. By finding z⁴ if z = cisØ, show that cos4Ø = cos⁴Ø - 6cos²Ø.sin²Ø + sin⁴Ø and that sin4Ø = 4cos³Ø.sinØ - 4cosØ.sin³Ø
Title: Re: Specialist Problem
Post by: Ahmad on February 24, 2008, 08:14:20 am: $(\cos \theta + i \sin \theta)^4 = \cos 4\theta + i \sin 4\theta$

Expand the left hand side and equate the real and imaginary parts of the expanded left hand side and the right hand side.
Title: Re: Specialist Problem
Post by: Tea.bag on February 24, 2008, 09:11:04 am: Got it.

Thanks.
Title: Re: Specialist Problem
Post by: Tea.bag on February 26, 2008, 07:25:34 pm: Hey i've got another question..

1. Prove that: $cos(\frac{n\pi}{2}-n\theta)+isin(\frac{n\pi}{2}-n\theta)=(\frac{1+sin\theta+icos\theta}{1+sin\theta-icos\theta})^n$

2. If $n$ is a positive integer, prove that $(\sqrt{3}+i)^n+(\sqrt{3}-i)^n=2^{n+1}cos\frac{n\pi}{6}$

3. The centre of a regular hexagon is at the origin and one vertex is given by $\sqrt{3}+i$ on the Argand diagram. Find the complex number represented by the other vertices.
Title: Re: Specialist Problem
Post by: Neobeo on February 26, 2008, 07:33:22 pm: Quote from: Tea.bag on February 26, 2008, 07:25:34 pm
Hey i've got another question..

Prove that: $cos(\frac{n\pi}{2}-nØ)+isin(\frac{n\pi}{2}-nØ)=(\frac{1+sinØ+icosØ}{1+sinØ-icosØ})^n$

Simplify LHS until you get $\text{LHS} = (\sin(Ø)+i\cos(Ø))^n$
Simplify RHS until you get $\text{RHS} = (\sin(Ø)+i\cos(Ø))^n$
Title: Re: Specialist Problem
Post by: Tea.bag on February 26, 2008, 07:36:03 pm: i dont know how to do latex for 2^n+1..can anyone show me?
Title: Re: Specialist Problem
Post by: Neobeo on February 26, 2008, 07:38:24 pm: Quote from: Tea.bag on February 26, 2008, 07:25:34 pm
If $n$ is a positive integer, prove that $(\sqrt{3}+i)^n+(\sqrt{3}-i)^n=2^n+1cos\frac{n\pi}{6}$

$(\sqrt{3}+i)^n+(\sqrt{3}-i)^n$
$=(2cis\frac{\pi}{6})^n+(2cis\frac{-\pi}{6})^n$
$=2^ncis\frac{n\pi}{6}+2^ncis\frac{-n\pi}{6}$
... should provide a starting point

also try 2^{n+1} = $2^{n+1}$
Title: Re: Specialist Problem
Post by: Tea.bag on February 26, 2008, 07:49:29 pm: That really helped thanks.
Title: Re: Specialist Problem
Post by: Tea.bag on February 27, 2008, 05:43:12 pm: Hey guys,

im having my first sac for specialist. I dont know what to revise. Do any of you know. So confusing..
Title: Re: Specialist Problem
Post by: Matt The Rat on February 27, 2008, 06:24:35 pm: Make sure you go thoroughly through the chapter going over and clarifying any questions that you're unsure on. The chapter review has pretty good questions to work on, at least in Essential.

Good luck with it! Mine's on Monday for complex numbers.
Title: Re: Specialist Problem
Post by: Tea.bag on February 27, 2008, 08:14:16 pm: Mines on complex numbers too.
Title: Re: Specialist Problem
Post by: AppleXY on February 27, 2008, 08:25:06 pm: hahaha what a concidence, I had a Complex Numbers test too :P :P :P (well you guys are technically going to have, but still lol)
Title: Re: Specialist Problem
Post by: Tea.bag on February 27, 2008, 10:22:28 pm: Was it a sac or just a practise test?
Title: Re: Specialist Problem
Post by: Tea.bag on March 21, 2008, 09:51:33 pm: yet another question.

1. Show that: $\frac{1×2^2+2×3^2+...+n(n+1)^2}{1^2×2+2^2×3+...+n^2(n+1)}=\frac{3n+5}{3n+1}$
Title: Re: Specialist Problem
Post by: Ahmad on March 21, 2008, 11:48:58 pm: By Force:

Equivalent to,

$(3n+1) \displaystyle \sum_{i=1}^{n} i(i+1)^2 = (3n+5) \sum_{i=1}^n i^2(i+1)$

$3n \displaystyle \sum_{i=1}^{n} i(i+1) = 5 \sum_{i=1}^{n} i^2(i+1) - \sum_{i=1}^{n} i(i+1)^2$

Let $S_j(n) = \displaystyle \sum_{i=1}^n i^j$

Then, we must show, $4 S_3(n) + 3(1-n) S_2(n) - (1+3n)S_1(n) = 0$ , which is trivial.
Title: Re: Specialist Problem
Post by: Tea.bag on March 22, 2008, 05:38:07 pm: A super hard question..

1) Let $a_1,a_2,...$ be positive real numbers in geometric progression. For each $n$ , let $A_n, G_n, H_n$ be respectively, the arithmetic mean, geometric means and harmonic mean of $a_1, a_2, ... , a_n$ . Find an expression for the geometric mean of $G_1, G_2, ... ,G_n$ in terms of $A_1, A_2, ... ,A_n, H_1, H_2, ... ,H_n.$
Title: Re: Specialist Problem
Post by: Ahmad on March 22, 2008, 06:35:01 pm: Let the geometric progression be $a,\ ar,\ ar^2, \ldots ,\ ar^{n-1}$

$A_n = \frac{1}{n} \displaystyle \sum_{i=0}^{n-1} ar^i = \frac{a(r^n - 1)}{n(r-1)}$

$H_n = n \cdot \left( \displaystyle \sum_{i=0}^{n-1} \frac{1}{ar^i} \right) ^{-1} = \frac{an(r^{-1} - 1)}{r^{-n} - 1}$

$G_n = \displaystyle \sqrt[n]{\prod_{i=0}^{n-1} (ar^i)} = ar^{\frac{n-1}{2}}$

$G(G_1,\ G_2,\ \ldots,\ G_n) = \displaystyle \prod_{i=1}^{n} (ar^\frac{i-1}{2}) = ar^\frac{n-1}{4}$

$= \sqrt[4]{A_1^2\cdot A_n\cdot H_n}$
Title: Re: Specialist Problem
Post by: Tea.bag on March 22, 2008, 08:20:18 pm: is this the same to ur solution?

Let $G_m$ be the geometric mean of $G_1, G_2, ... , G_3$

$\implies G_m = (G_1×G_2 .... G_n)^{\frac{1}{n}}$

$=[(a_1)×(a_1×a_1r)^{\frac{1}{2}}×(a_1×a_1r×a_1r^2)^{\frac{1}{3}} ... (a_1×a_1r×a_1r^2 ... a_1r^{n-1})^{\frac{1}{n}}]^{\frac{1}{n}}$

where $r$ is the common ratio of $G.P. a_1, a_2, ... , a_n$

$=[(a_1×a_1 ... n times)(r^{\frac{1}{2}}×r^{\frac{3}{3}}×r^{\frac{6}{4}} ... r^{\frac{n(n-1)}{2n}})]^{\frac{1}{n}}$

$=[a_1^n×r^{\frac{1}{2}+1+\frac{3}{2}+ ... \frac{n-1}{2}}]^{\frac{1}{n}} = a_1×[r^{\frac{1}{2}×[\frac{n(n-1)}{2}]}]^{\frac{1}{n}} = a_1×[r^{\frac{n-1}{4}}]$

Now, $A_n = \frac{a_1+a_2+ ... +a_n}{n} = \frac{a_1(1-r^n)}{n(1-r)}$

And, $H_n = \frac{n}{(\frac{1}{a_1}+\frac{1}{a_2}+ ... +\frac{1}{a_n})} = \frac{n}{\frac{1}{a_1}(1+\frac{1}{r}+ ... + \frac{1}{r^{n-1}})} = \frac{a_1n(1-r)r^{n-1}}{1-r^n}$

Again, $A_n×H_n = \frac{a_1(1-r^n)}{n(1-r)}×\frac{a_1n(1-r)r^{n-1}}{1-r^n} = a_1^2r^{n-1}$

$\implies \prod_{k=1}^{n}A_kH_k = \prod_{k=1}^{n}(a_1^2r^{k-1}) = (a_1^2×a_1^2×a_1^2 ... n times)× r^0×r^1×r^2 ... r^{n-1}$

$=a_1^{2n}×r^{1+2+ ... +(n-1)} = a_1^{2n}r^{\frac{n(n-1)}{2}} = [a_1r^{\frac{n-1}{4}}]^{2n} = [G_m]^{2m}$

$\implies G_m = [\prod_{k=1}^{n}A_kH_k]^{\frac{1}{2n}}$

$\implies G_k = (A_1A_2 ... A_n × H_1H_2 ... H_n)^{\frac{1}{2n}}$
Title: Re: Specialist Problem
Post by: AppleXY on March 22, 2008, 08:39:01 pm: Are you sure this is in the study design of specialist? I have never seen such a question in an any exam. (I mean, you only have 3 hours in total exam time)
Title: Re: Specialist Problem
Post by: cara.mel on March 22, 2008, 08:45:20 pm: Quote from: AppleXY on March 22, 2008, 08:39:01 pm
Are you sure this is in the study design of specialist? I have never seen such a question in an any exam. (I mean, you only have 3 hours in total exam time)

The last two questions aren't. The rest on this thread use spec knowledge but their only place is on like a problem solving SAC, they wouldn't be on the exams.
I do not like my familiarity with VCAA >.<
Title: Re: Specialist Problem
Post by: gfb on March 22, 2008, 09:00:25 pm: Good rittens! .

No way this is in the spesh study design because we are going to do them in semester 2? at Uni lol.
Title: Re: Specialist Problem
Post by: Ahmad on March 22, 2008, 09:03:26 pm: Yup it's the same solution expressed in a different way. :)
Title: Re: Specialist Problem
Post by: Tea.bag on March 22, 2008, 11:35:29 pm: Im pretty sure its not in the spesh course.
i was just letting people have a crack at it. :)
Title: Re: Specialist Problem
Post by: Ahmad on March 22, 2008, 11:43:29 pm: Possibly better place to put them in the future is,

Recreational Problems:
http://vcenotes.com/forum/index.php/topic,11.0.html

:)
Title: Re: Specialist Problem
Post by: Tea.bag on March 23, 2008, 10:05:37 pm: how do u simplify?

Simplify:

1) $tan\left(\displaystyle\frac{\pi}{2}-A\right)$

2) $cot\left(\displaystyle\frac{3\pi}{2}+A\right)$

3) $cot\left(\displaystyle\frac{\pi}{2}+y\right)$

i seem to get undefined answers..
Title: Re: Specialist Problem
Post by: Mao on March 23, 2008, 10:52:00 pm: try looking at the transformations:

1) $tan\left( \frac{\pi}{2} -A\right)=tan\left(-\left(A-\frac{\pi}{2}\right) \right)$

so essentially it was flipped along y after the translation,
and because we have learnt about the relationship between tan(x) and cot(x), we know that this transformation describes cot(A)

2)remember the period for cot is $\pi$ , also looking at the transformations involved here, cot is shifted to the left by one period and $\frac{\pi}{2}$

3)notice the resemblence to the above?
Title: Re: Specialist Problem
Post by: Ahmad on March 24, 2008, 12:48:35 am: Another way,

$\tan (\frac{\pi}{2} - A) = \frac{\sin (\frac{\pi}{2} - A)}{\cos (\frac{\pi}{2} - A)} = \frac{\cos A}{\sin A} = \cot A$
Title: Re: Specialist Problem
Post by: Tea.bag on March 24, 2008, 12:05:59 pm: thnx.

More questions :)

Prove:

1) $secx-tanx=\frac{cosx}{1+sinx}$

2) $\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

3) $(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

4) $sin^6x+cos^6x=1-3sin^2x cos^2x$

5) $(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$
Title: Re: Specialist Problem
Post by: Mao on March 24, 2008, 01:57:23 pm: 1) $secx-tanx=\frac{cosx}{1+sinx}$

$sec(x)-tan(x)$

$\implies \frac{1}{cos(x)}-\frac{sin(x)}{cos(x)}$

$\implies \frac{1-sin(x)}{cos(x)}$

if we multiplied by $1+sin(x)$ on top and bottom:

$\implies \frac{\left( 1-sin(x) \right) \left( 1+sin(x) \right)}{cos(x)\left( 1+sin(x) \right)}$

notice that resembles difference between two squares, which simplifies to $1-sin^2 (x) \implies cos^2(x)$

$\implies \frac{cos^2(x)}{cos(x)\left( 1+sin(x) \right)}$

$\implies \frac{cos(x)}{1+sin(x)}$

$secx-tanx=\frac{cosx}{1+sinx}$

2) $\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

it will be very problematic expanding that square root, so we'll show that RHS is equal to LHS.

$sec(x)+tan(x)$

$\implies \frac{1+sin(x)}{cos(x)}$

using the pythagorean identity, $sin^2(x)+cos^2(x)=1 \implies cos(x)=\sqrt{1-sin^2(x)}$ , so we use this for the denominator:

$\implies \frac{1+sin(x)}{\sqrt{1-sin^2(x)}}$

now we include the numerator inside the square root:

$\implies \sqrt{\frac{\left( 1+ sin(x) \right) ^2}{1-sin^2(x)}$

notice how the denominator is difference between two squares

$\implies \sqrt{\frac{\left( 1+ sin(x) \right) ^2}{\left( 1-sin(x) \right) \left( 1+sin(x) \right)}$

$\implies \sqrt\frac{1+sinx}{1-sinx}$

$\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

3) $(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

from 1):

$secx-tanx=\frac{cosx}{1+sinx}$

$(sec(x)-tan(x))^2=\frac{cos^2(x)}{(1+sin(x))^2}$

$(sec(x)-tan(x))^2=\frac{1-sin^2(x)}{(1+sin(x))^2}$

$(sec(x)-tan(x))^2=\frac{(1+sin(x))(1-sin(x))}{(1+sin(x))^2}$

$(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

4) $sin^6x+cos^6x=1-3sin^2x cos^2x$

given that: $sin^2(x)=1-cos^2(x)$

we can rewrite $sin^6(x)$ as:

$sin^6(x)=\left( sin^2(x) \right) ^3$

$sin^6(x)=\left( 1-cos^2(x) \right) ^3$

$sin^6(x)=1-3cos^2(x)+3cos^4(x)-cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3\cdot cos^2(x) \cdot cos^2(x) -cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3\cdot \left( 1-sin^2(x) \right) \cdot cos^2(x) -cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3cos^2(x)-3sin^2(x)cos^2(x)-cos^6(x)$

$sin^6(x)+cos^6(x)=1-3sin^2(x)cos^2(x)$

5) $(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$

$cosec(\theta)+cot(\theta)=\frac{1+cos(\theta)}{sin(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{sin^2(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{1-cos^2(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{(1+cos(\theta))(1-cos(\theta))}$

$(cosec\theta+cot\theta)^2=\frac{1+cos(\theta}{1-cos(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{1+sec^{-1}(\theta}{1-sec^{-1}(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{sec(\theta) \cdot sec^{-1}(\theta)+sec^{-1}(\theta}{sec(\theta) \cdot sec^{-1}(\theta)-sec^{-1}(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{sec^{-1}(\theta) \cdot (sec(\theta)+1)}{sec^{-1}(\theta)\cdot (sec(\theta)-1)}$

$(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$

:D
Title: Re: Specialist Problem
Post by: Ahmad on March 24, 2008, 02:13:37 pm: 2. Multiply inside the square root of the LHS by $\frac{1 + \sin x}{1 + \sin x}$

4. $(\sin^2 x + \cos^2 x)^3 = 1$
Title: Re: Specialist Problem
Post by: Neobeo on March 24, 2008, 02:59:41 pm: Quote from: Tea.bag on March 24, 2008, 12:05:59 pm
$secx-tanx=\frac{cosx}{1+sinx}$

I claim this page with my awesome proof:

(http://img508.imageshack.us/img508/1438/mathpg1mo3.png)

(http://img177.imageshack.us/img177/7457/mathpg2kf9.png)
Title: Re: Specialist Problem
Post by: ed_saifa on March 24, 2008, 03:04:55 pm: O__O oh wow. i like that alternative!
Title: Re: Specialist Problem
Post by: Ahmad on March 24, 2008, 03:29:18 pm: Nicely done! And hi to you too Neobeo ;) ;D
Title: Re: Specialist Problem
Post by: Mao on March 24, 2008, 03:41:13 pm: Quote from: Ahmad on March 24, 2008, 03:29:18 pm
And hi to you too Neobeo ;)

=O i knew it!

xD jk jk

well done neobeo
Title: Re: Specialist Problem
Post by: Collin Li on March 24, 2008, 04:18:06 pm: I love your handwriting Neobeo.
Title: Re: Specialist Problem
Post by: Neobeo on March 24, 2008, 06:54:28 pm: Quote from: Tea.bag on March 24, 2008, 12:05:59 pm
$sin^6x+cos^6x=1-3sin^2x cos^2x$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
Title: Re: Specialist Problem
Post by: Tea.bag on March 25, 2008, 11:07:20 pm: i got it!
thnx for all the help guys
Title: Re: Specialist Problem
Post by: Tea.bag on April 05, 2008, 12:16:19 am: another problem..

(a) If v = 1 + i and z = x + iy, for any real numbers x and y:
(i) Show that the equation
|z − v| = |vz|
represents a circle, and find its centre and radius.
(ii) Find the intersection of the circle in part (i) with the straight line
|z − v| = |z + v| .

(b) Using the roots of z^5 = 1, or otherwise, write z^4 + z^3 + z^2 + z + 1 as the product of two quadratic
expressions with real coefficients.
Hence find the exact value of the product $cos\frac{2\pi}{5}cos\frac{4\pi}{5}$
Title: Re: Specialist Problem
Post by: ed_saifa on April 05, 2008, 09:42:12 am: $|z-v|=|vz|$
Let z=x+yi
$|x+yi-(1+i)|=|(1+i)(x+yi)|$
Group real and imaginary parts
$|x-1+yi-i|=|x-y+yi+ix|$
$|(x-1)+i(y-1)|=|(x-y)+i(x+y)|$
Find the moduli
$\sqrt{(x-1)^2 +(y-1)^2}=\sqrt{(x-y)^2 + (x+y)^2}$
Square both sides
$(x-1)^2 + (y-1)^2=(x-y)^2 + (x+y)^2$
Expand and collect up terms
$x^2-2x+1+y^2-2y+1=x^2-2xy+y^2+x^2+2xy+y^2$
$x^2+y^2+2x+2y-2=0$
Factorise
$(x^2+2x+1)-1+(y^2+2y+1)-1-2=0$
$(x+1)^2 + (y+1)^2= 4$
centre: (-1,-1) r=2
Title: Re: Specialist Problem
Post by: khalid!!! on April 05, 2008, 09:57:25 am: Quote from: Ahmad on February 24, 2008, 08:14:20 am
$(\cos \theta + i \sin \theta)^4 = \cos 4\theta + i \sin 4\theta$

Expand the left hand side and equate the real and imaginary parts of the expanded left hand side and the right hand side.

-_- i dont get it xD
Title: Re: Specialist Problem
Post by: ice_blockie on April 05, 2008, 10:16:38 am: I want your stationery Neobeo!!!! Where did you get it?
Title: Re: Specialist Problem
Post by: Mao on April 05, 2008, 12:14:18 pm: ice blockie:
Quote from: ice_blockie on April 05, 2008, 10:16:38 am
I want your stationery Neobeo!!!! Where did you get it?
he uses a tablet PC xD

ed_saifa:
Quote from: ed_saifa on April 05, 2008, 09:42:12 am
$\sqrt{(x-1)^2 +(y+1)^2}=\sqrt{(x-y)^2 + (x+y)^2}$
that should be a $y-1$

Tea.Bag:
a) ii)

$|z-v|=|vz|$ : $(x+1)^2 + (y+1)^2= 4$

$|z-v|=|z+v|$

$\implies \sqrt{(x-1)^2+(y-1)^2}=\sqrt{(x+1)^2+(y+1)^2$

$\implies x^2-2x+1+y^2-2y+1=x^2+2x+1+y^2+2y+1$

$\implies -4x-4y=0$

$\implies y=-x$

Hence the intersection between $|z-v|=|vz|$ and $|z-v|=|z+v|$ is when both relationships are satisfied:
substituting $y=-x$ to our circle:

$(x+1)^2 + (-x+1)^2= 4$

$x^2+2x+1+x^2-2x+1=4$

$2x^2=2$

$x^2=1$

$x=\pm 1$

$y=\mp 1$

$z=\pm1 \mp i$

b)
it should be noted that:
$x^n-a^n=(x-a)\left(x^{n-1}+a\cdot x^{n-2}+a^2 \cdot x^{n-3}+ \cdots + a^{n-2}\cdot x + a^{n-1}\right)$

The roots of $z^5=1$ :

$|z|^5cis(5\theta)=cis(0+2k\pi )$

$\implies |z|=1$

$\implies \theta = 0,\frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}$

$\implies Arg(z)=-\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}$

$z=1,cis\left(\pm \frac{2\pi}{5}\right), cis\left(\pm \frac{4\pi}{5}\right)$

This means the expression $z^5-1$ can be expressed as:

$z^5-1=(z-1)\left(z-cis\left(\frac{2\pi}{5}\right)\right)\left(z-cis\left(-\frac{2\pi}{5}\right)\right)\left(z-cis\left(\frac{4\pi}{5}\right)\right)\left(z-cis\left(-\frac{4\pi}{5}\right)\right)$

since the roots above are all conjugates of each other:

$z^5-1=(z-1)\left(z^2-\left(cis\left(\frac{2\pi}{5}\right)+cis\left(-\frac{2\pi}{5}\right)\right)z+|z|\right)\left(z^2-\left(cis\left(\frac{4\pi}{5}\right)+cis\left(-\frac{4\pi}{5}\right)\right)z+|z|\right)$

$z^5-1=(z-1)\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

$(z-1)\left(z^4+z^3+z^2+z+1\right)=(z-1)\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

$z^4+z^3+z^2+z+1=\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

Hence: expanding the RHS:

$\implies z^4-\left(2\cos\left(\frac{4\pi}{5}\right)+2\cos\left(\frac{2\pi}{5}\right)\right)z^3+\left(2+4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)\right)z^2-\left(2\cos\left(\frac{4\pi}{5}\right)+2\cos\left(\frac{2\pi}{5}\right)\right)z+1$

therefore, by equality of polynomials:

$z^2=\left(2+4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)\right)z^2$

$4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)=-1$

$\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)=-\frac{1}{4}$
Title: Re: Specialist Problem
Post by: chid on April 05, 2008, 01:27:39 pm: I've got a question.

Arg(z)= π/4 and Arg(z-3)= π/2. Find Arg(z-6i)

I'm a little confused.

Thanks.
Title: Re: Specialist Problem
Post by: ed_saifa on April 05, 2008, 02:18:39 pm: Thanks Mao I corrected it XD
Title: Re: Specialist Problem
Post by: Tea.bag on April 09, 2008, 05:37:15 pm: another problem...

Find the range of $f(x)=cos(sin^{-1}2x)$
Title: Re: Specialist Problem
Post by: Mao on April 09, 2008, 06:14:45 pm: Quote from: Tea.bag on April 09, 2008, 05:37:15 pm
another problem...

Find the range of $f(x)=cos(sin^{-1}2x)$

this is a composite function, so first we need to find the range of $\sin^{-1}2x$ , which is $[-\frac{\pi}{2},\frac{\pi}{2}]$

that means $\cos$ takes the domain of of $[-\frac{\pi}{2},\frac{\pi}{2}]$ , which would give u a range of $[0,1]$
Title: Re: Specialist Problem
Post by: Tea.bag on April 09, 2008, 06:27:29 pm: i dont get it.. :(

$cos\frac{\pi}{2}=-1?$
Title: Re: Specialist Problem
Post by: Mao on April 09, 2008, 06:29:47 pm: argh ooops that meant to be a +1 =S
Title: Re: Specialist Problem
Post by: Tea.bag on April 09, 2008, 07:04:58 pm: i still dont get it..

can u explain why $cos\frac{\pi}{2}=1$ Isnt it 0??
Title: Re: Specialist Problem
Post by: Mao on April 09, 2008, 07:17:08 pm: in the interval of $u\in [ -\frac{\pi}{2},\frac{\pi}{2}]$ , $\cos(u)$ has its minimum value at $\cos\left(\pm \frac{\pi}{2}\right)=0$ , and the maximum value occurs at $\cos(0)=1$
The range is written as (minimum,maximum), not (left,right).

hence the range is $[0,1]$
Title: Re: Specialist Problem
Post by: Tea.bag on April 17, 2008, 05:00:27 pm: more questions..

1) An aeroplane can travel at a speed of 120km/h when there is no wind. Today there is a wind velocity of $3i+2j$ km/h.

Find:
a) the speed of the wind
b) the position vector required if the aeroplane is to end up due north of its starting point after 1 hour
c) the actual bearing required

2) Another aeroplane can travel at a speed of 240 km/h when there is no wind. Today there is a wind velocity of $-3i-7j$ km/h.

Find:
a) the speed of the wind
b) the position vector required if the aeroplane is to end up due east of its starting point after 2 hours
c) the actual bearing required
Title: Re: Specialist Problem
Post by: Mao on April 17, 2008, 08:28:19 pm: 1)
a)
$v_w = |3i+2j| = \sqrt{13}$ km/h

b)
let $P=ai+bj$ be the position vector of the aeroplane after 1 hour:

for the plane to end up due north, the $i$ component must be 0, hence $a=-3$

$|P|=\sqrt{(-3)^2 + b^2} = 120$

$b=3\cdot \sqrt{1599} \approx 119.96$

$P=-3i + 119.96j$

c) as the word "bearing" is quite ambiguous, i'm going to give it how the weatherman gives it: (N something E or S something W)

from the positive $j$ axis

$tan(\theta) = \frac{-3}{119.96}$

$\theta \approx -1.43^o$

N1.43^oW

2)
a)
$v_w=|-3i-7j|=\sqrt{58}$

b)
east is the positive $i$ direction, hence the displacement in the $j$ direction must be 0:

let $P=ai+bj$ be the position vector of the plane after 2 hrs

$b=-(-7 \times 2) = 14$

$|P|=\sqrt{a^2 + 14^2} = 480$

$a=2 \cdot \sqrt{57551} \approx 479.80$

$P=479.80i + 14j$

c)
$tan(\theta)=\frac{479.80}{14}$

$\theta \approx 88.33^o$

N88.33^oE

hope that helped.
Title: Re: Specialist Problem
Post by: Tea.bag on April 18, 2008, 07:06:50 pm: thnx
Title: Re: Specialist Problem
Post by: Tea.bag on May 01, 2008, 10:50:18 pm: Help!

1) Consider the triangle $ABC$ . Point $D$ is the midpoint of line $AB$ and $E$ is the midpoint of line $BC$ .

Show, using vectors to represent various lines, that line $DE$ is parallel to $AC$ and $DE=\frac{1}{2}AC$

2) If the length of a vector , $\vec{a}$ , is given by $|\vec{a}|$ , show geometrically that for any two vectors $\vec{a}$ and $\vec{b}$ :

$|\vec{a}|+|\vec{b}|\ge|\vec{a}+\vec{b}|$
Title: Re: Specialist Problem
Post by: dcc on May 01, 2008, 11:24:34 pm: Q1: TRIANGLE ABC:

$\mbox{Let }\vec{AB} = \tilde{a}$

$\mbox{Let }\vec{AC} = \tilde{b}$

:. $\vec{BC} = \vec{BA} + \vec{AC} = \tilde{b} - \tilde{a}$

Given $\vec{AD} = \dfrac{1}{2}\tilde{a}$

and: $\vec{BE} = \dfrac{1}{2}\vec{BC} = \dfrac{1}{2}(\tilde{b} - \tilde{a})$

Therefore:

$\vec{DE} = \vec{DA} + \vec{AB} + \vec{BE} = -\dfrac{1}{2}\tilde{a} + \tilde{a} + \dfrac{1}{2}(\tilde{b} - \tilde{a}) = \dfrac{1}{2}\tilde{a} + \dfrac{1}{2}\tilde{b} - \dfrac{1}{2}\tilde{a} = \dfrac{1}{2}\tilde{b} = \dfrac{1}{2}\vec{AC}$
Title: Re: Specialist Problem
Post by: dcc on May 01, 2008, 11:33:06 pm: For the second one, I would suggest drawing a triangle, and note that the shortest distance between two points is a straight line. By drawing two vectors and adding them head to tail, notice that the resultant vector is always the same magnitude (if a & b are parallel) or less. Im not sure how thorough a geometric proof should be, but that seems like a rational explanation.
Title: Re: Specialist Problem
Post by: Tea.bag on May 25, 2008, 06:02:37 pm: Help!!!

A spider builds a web in a garden. The position vectors of the ends A and B of a strand of the web are given by $\vec{OA}=2i+3j+k$ and $\vec{OB}=3i+4j+2k$ respectively.

a) Find:
1) $\vec{AB}$
2) the length of the strand

b) A small insect is at point C where $\vec{OC}=2.5i+4j+1.5k$ . Unlucky it flies in a straight line and hits the strand of web between A and B. Let Q be the point where the insect hits the strand where $\vec{AQ}=\lambda{AB}$ .

1) Find $\vec{CQ}$ in terms of $\lambda$
2) If the insect hits the strand at right angles find the value of $\lambda$ and $\vec{OQ}$

c) Another web strand $\vec{MN}$ has end points M and N with position vectors $\vec{OM}=4i+2j-k$ and $\vec{ON}=6i+10j+9k$ . The spider decides to continue AB and to join MN. Find the position vector of the point of contact.
Title: Re: Specialist Problem
Post by: ed_saifa on May 25, 2008, 06:57:44 pm: a)
1)
$\vec{AB}= \vec{OB}-\vec{OA}$

$\vec{AB}=3i + 4j + 2k -(2i + 3j + k)$

$\vec{AB}= i + j + k$

2)
$|\vec{AB}|=$ length of strand

$|\vec{AB}|= \sqrt{1^2 + 1^2 + 1^2}$

$|\vec{AB}|= \sqrt{3}$
Title: Re: Specialist Problem
Post by: ell on May 25, 2008, 07:20:38 pm: b)
1)
$\vec{CQ}= -\vec{OC}+\vec{OQ}$
$= -\vec{OC}+\vec{OA}+\vec{AQ}$
$= -2.5i-4j-1.5k+2i+3j+k +\vec{AQ}$
$= -2.5i-4j-1.5k+2i+3j+k + \gamma(i+j+k)$ since $\vec{AQ}=\gamma\vec{AB}$

Expand out and simplify:
$= -2.5i-4j-1.5k+2i+3j+k + \gamma i +\gamma j+\gamma k$
$= (\gamma - 0.5)i + (\gamma - 1)j +(\gamma -0.5)k$

2)
"at right angles", i.e. perpendicular.
$\vec{CQ}.\vec{AB}=0$
$[(\gamma-0.5)i+(\gamma-1)j+(\gamma-0.5)k].(i+j+k)=0$
$\gamma-0.5+\gamma-1+\gamma-0.5=0$
$3\gamma-2=0$
$\gamma=\frac{2}{3}$

$\vec{OQ}=\vec{OA}+\vec{AQ}$
$\vec{OQ}=2i+3j+k+\gamma\vec{AB}$
$\vec{OQ}=2i+3j+k+\gamma(i+j+k)$
Sub in the $\gamma$ value:
$\vec{OQ}=2i+3j+k+\frac{2}{3}(i+j+k)$
$\vec{OQ}=2i+3j+k+\frac{2}{3}i+\frac{2}{3}j+\frac{2}{3}k$
$\vec{OQ}=\frac{8}{3}i+\frac{11}{3}j+\frac{5}{3}k$
$\vec{OQ}=\frac{1}{3}(8i+11j+5k)$
Title: Re: Specialist Problem
Post by: Tea.bag on May 25, 2008, 08:24:41 pm: cool thanks..
another question..

VABCD is a square based pyramid and O is the centre of the base. $i, j, k$ are unit vectors in the direction of AB, BC and OV respectively. the point O is to be taken as the origin for position vectors. AB=BC=CD=DA=4cm
OV = 2h cm where h is a positive real number.
P,Q,M and N are the midpoints of AB,BC,VC and VA respectively.

a) Find the position vecots A,B,C and D respectively to O.
b) Find vectors $\vec{PM}$ and $\vec{QN}$ , in terms of h.
c) Find the position vector $\vec{OX}$ , of X the point of intersection of QN and PM.
d) If OX is perpendicular to VB
1) Find the value of h
2)calculate the angle between PM and QN giving your answer correct to the nearest 0.1degrees.
e) 1)Prove that NMQP is a rectangle
2)Find h if NMPQ is a square.
Title: Re: Specialist Problem
Post by: ell on May 25, 2008, 09:05:03 pm: Probably makes it a bit easier if we draw the pyramid.
(http://img81.imageshack.us/img81/6197/vecqqtq6.gif)

a) Position vectors of A,B,C and D:
So $\hat{\vec{AB}}=i$ $\hat{\vec{BC}}=j$ and $\hat{\vec{OV}}=k$ , and the length of one side is 4 cm.
From diagram ,
$\vec{OA}=-2i-2j$
$\vec{OB}=2i-2j$
$\vec{OC}=2i+2j$
$\vec{OD}=-2i+2j$

b)
We have to do a bit of travelling to get from P to M. :P
So $\vec{PM}=\vec{PB}+\vec{BC}+\vec{CM}$

We know $\vec{PB}$ and $\vec{BC}$ :
$\vec{PM}=2i+4j+\vec{CM}$

Now $\vec{CV}=\vec{CQ}+\vec{QO}+\vec{OV}$

$\vec{CV}=-2i-2j+2hk$

$\vec{CM}=\frac{1}{2}\vec{CV}$

Therfore $\vec{PM}=2i+4j+\frac{1}{2}(-2i-2j+2hk)$

$\vec{PM}=2i+4j-i-j+hk$

$\vec{PM}=i+3j+hk$

Similar process for finding $\vec{QN}$ -
$\vec{QN}=\vec{QB}+\vec{BA}+\vec{AN}$

$\vec{QN}=-2j-4i+\vec{AN}$

$\vec{AV}=2i+2j+2hk$

$\vec{AN}=\frac{1}{2}\vec{AV}$

$\vec{QN}=-2j-4i+\frac{1}{2}(2i+2j+2hk)$

$\vec{QN}=-3i-j+hk$
Title: Re: Specialist Problem
Post by: Tea.bag on May 26, 2008, 04:07:40 pm: Solve:

$sin(4x)=cos(2x)$ where $x\in{[0,\pi]}$
Title: Re: Specialist Problem
Post by: Mao on May 26, 2008, 04:16:22 pm: $\begin{align*} \sin(4x)&=\cos(2x) \\ 2\sin(2x)\cos(2x) &= \cos(2x)\\ \end{align*}$

dividing by cos(2x) on both sides, we must apply the constraint $\cos(2x)\neq 0$
hence possible solutions are:
$\begin{align*} \cos(2x)=0 & \mbox{ and } \sin(2x) = \frac{1}{2}\\ x=\frac{\pi}{4} & \mbox{ and } 2x = \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}\\ \end{align*}$

$x = \left\{\frac{\pi}{12},\frac{\pi}{4},\frac{5\pi}{12}\right\}$
Title: Re: Specialist Problem
Post by: AppleXY on May 26, 2008, 04:17:55 pm: yeah, was gonna post but Mao did it lol.

Remember to exhaust all of your trig ids :D ;)
Title: Re: Specialist Problem
Post by: Tea.bag on May 26, 2008, 04:18:54 pm: ahhhh.. so simple :uglystupid2:
thnx
Title: Re: Specialist Problem
Post by: humph on May 27, 2008, 03:05:54 am: Quote from: Mao on May 26, 2008, 04:16:22 pm
$\begin{align*} \sin(4x)&=\cos(2x) \\ 2\sin(2x)\cos(2x) &= \cos(2x)\\ \sin(2x) &= \frac{1}{2}\\ 2x &= \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}\\ x &= \left\{\frac{\pi}{12},\frac{5\pi}{12}\right\}\\ \end{align*}$
you can't just divide through by $\cos(2x)$ as it might be zero, and so you lose possible solutions that way. in fact, the answer should be $x = \left\{\frac{\pi}{12},\frac{\pi}{4},\frac{5\pi}{12}\right\}$
Title: Re: Specialist Problem
Post by: ell on May 27, 2008, 07:41:47 am: You can move the $\cos{(2x)}$ to the LHS to avoid missing out on an answer.

$ \begin{align*} 2\sin{(2x)}\cos{(2x)}-\cos{(2x)}&=0\\ \cos{(2x)}(2\sin{(2x)}-1)&=0\\ \end{align*} $
Title: Re: Specialist Problem
Post by: Mao on May 27, 2008, 05:15:18 pm: thats a very good point~

I forgot about the constraints
Title: Re: Specialist Problem
Post by: Tea.bag on June 29, 2008, 05:15:35 pm: can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$
Title: Re: Specialist Problem
Post by: Mao on June 29, 2008, 05:26:25 pm: Quote from: Tea.bag on June 29, 2008, 05:15:35 pm
can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$

$\int_a^b f(x)-g(x)\; dx$
Title: Re: Specialist Problem
Post by: polky on June 29, 2008, 05:52:04 pm: Quote from: Tea.bag on June 29, 2008, 05:15:35 pm
can u tell me way i should tackle these kind of questions?

Find the area between the curves $y=\sqrt{9-x}$ and $y=x+3$ from $x=-1$ and $x=2$

Sketch both graphs to figure out which one lies on top and which one is at the bottom of the region you want to integrate. Then you can set about integrating it! :)
Title: Re: Specialist Problem
Post by: Tea.bag on June 30, 2008, 09:39:28 pm: Find the antiderivative:

1. $\int\frac{x^2}{x^2+4}$
Title: Re: Specialist Problem
Post by: Mao on June 30, 2008, 09:45:47 pm: Quote from: Tea.bag on June 30, 2008, 09:39:28 pm
Find the antiderivative:

1. $\int\frac{x^2}{x^2+4}$

$\implies \int \frac{x^2+4-4}{x^2+4}\; dx=\int 1-2\frac{2}{x^2+4}\; dx = x-2\tan^{-1} \frac{x}{2}+c$
Title: Re: Specialist Problem
Post by: Tea.bag on June 30, 2008, 09:47:42 pm: thnx. i forgot to that i could divide the numerator by denominator :D