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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Momo.05 on February 01, 2010, 11:08:11 pm

Title: Momo's Spech Hw help trend !
Post by: Momo.05 on February 01, 2010, 11:08:11 pm: Hey , there this question that is really bothering me can someone please help me out there!

PROVE:
cot(a+b) = cota.cotb -1/cot(a+b)

Thank you in advance for the help ! :)
Title: Re: Momo's Spech Hw help trend !
Post by: hyperblade01 on February 01, 2010, 11:11:16 pm: Is the meant to be

$cot(a+b) = \frac{cot(a)cot(b)-1}{cot(a+b)}$

OR

$cot(a+b) = cot(a)cot(b) - \frac{1}{cot(a+b)}$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 01, 2010, 11:11:40 pm: Its the first one :)
Title: Re: Momo's Spech Hw help trend !
Post by: kamil9876 on February 01, 2010, 11:14:38 pm: $cot(a+b)=\frac{1}{tan(a+b)}$

$=\frac{1-tan(a)tan(b)}{tan(a)+tan(b)}$

$=\frac{1-tan(a)tan(b)}{tan(a)+tan(b)} \times \frac{cot(a)cot(b)}{cot(a)cot(b)}$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 01, 2010, 11:23:30 pm: thank you
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 08:13:46 pm: This heat is getting to me, can someone please solve this question for me;

Solve for x, -pie <(equal to) x <(equal to) pie.

(a) sin(x) = (root2)/(2)
(b) sin(2x) = -(root3)/(2)
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 08:14:31 pm: Sorry in advance that i cant use the latex thing, since my net is capped and its goes funny when i type and stuff =="
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 08:18:41 pm: Nevermind i got it :)
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 08:36:07 pm: Wait i need some help in the question above but for (b) =="
Thank you for the help :]
Title: Re: Momo's Spech Hw help trend !
Post by: superflya on February 02, 2010, 08:56:05 pm: ur restricted domain wood now become -2pi<2x<2pi

sinx=-root3/2......x= -pi/3

sin is negative in the 3rd and 4th quadrants so ur other solutions wood be -pi/3 + pi, 2pi + pi/3 ..... Find all values within ur restricted domain
Title: Re: Momo's Spech Hw help trend !
Post by: brightsky on February 02, 2010, 09:05:42 pm: $\sin(2x) = -\frac{\sqrt{3}}{2}$ , $x \in [-\pi, \pi]$

Because $\sin(2x)$ is a negative number, we are looking at quadrants 3 and 4.

Draw a unit circle like attached.

The basic angle, that is, the "magnitude" of the angle 2x when drawn on a graph would be $\csc\left(\left|-\frac{\sqrt{3}}{2}\right|\right) = \frac{\pi}{3}$

But we are in quadrants 3 and 4 because that is where sine is negative.

By symmetry, we get $2x = -\frac{2\pi}{3}$

But because it is a $2x$ , we must multiply the domain by 2 as well.

So we need to spin the angle around a whole $2\pi$ , and so we get the angles $-\frac{\pi}{3} + 2\pi = \frac{4\pi}{3}$ and $-\frac{2\pi}{3} + 2\pi = \frac{5\pi}{3}$

Dividing through by 2 we get the angles:

$x = -\frac{\pi}{3}, -\frac{\pi}{6}, \frac{2\pi}{3}, \frac{5\pi}{6}$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 09:26:38 pm: Sorry my computer was abit laggy and i couldnt say thank you in time >< :)
yup, i understand it muchmuchmuch better now :)
Thank you !
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 10:22:24 pm: Oh i have another question i dont quite understand..

In this question i was asked to prove that cos(2A) = (1 - tan^2(A))/(1 + tan^2(A))

Which i proved. Then the question asked me to find the exact value of ;

a. tan(pie/12) -> which i used to the tan(A+B) to solve

BUT for part b. the question is

b. tan (pie/8) -> which is 22.5? there isnt any exact values for it.. and i dont know how that related to the cos(2A) thing i proved .. in other words im CONFUSED !!

=.=
Title: Re: Momo's Spech Hw help trend !
Post by: brightsky on February 02, 2010, 10:39:21 pm: Hmmm...probably a damn question, but is there something you missed whilst reading it? The only exact value I can think of for $\tan\left(\frac{\pi}{8}\right)$ is just $\tan\left(\frac{\pi}{8}\right)$ lol. What does the BoB say?
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 10:41:42 pm: Root2 - 1 ?
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 10:44:25 pm: Here is the exact question from the sheet

Prove that cos(2A) = (1 - tan^2(A))/(1 + tan^2(A)) , and hence find exact values of:

a. tan(pie/12)
b. tan (pie/8)
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 11:00:48 pm: OHH smart, yup i shall try that right after i finish this other question of mine :)
Thanks!
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 11:07:00 pm: Hey.. wait..isnt tan(pie/2) undef?
Title: Re: Momo's Spech Hw help trend !
Post by: hyperblade01 on February 02, 2010, 11:17:43 pm: Not too sure about the hence part, but what I would do is start off with the double angle formula

$tan(\frac{\pi}{4}) = \frac{2tan(\frac{\pi}{8})}{1-tan^2(\frac{\pi}{8})}$

We know $tan(\frac{\pi}{4}) = 1$ - transpose, use quadratic formula etc

EDIT: Actually..i just realised its similar to what i've just done

Start off with:

$cos(\frac{\pi}{4}) = \frac{1-tan^2(\frac{\pi}{8})}{1+tan^2(\frac{\pi}{8})}$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 11:20:56 pm: Umm, i know its late.. But i have an issue where i cant sleep until i fully get a problem i started on.. so i was hoping you can show me the full working out to this question please? If you dont mind that is.
Title: Re: Momo's Spech Hw help trend !
Post by: hyperblade01 on February 02, 2010, 11:48:38 pm: $cos(\frac{\pi}{4}) = \frac{1-tan^2(\frac{\pi}{8})}{1+tan^2(\frac{\pi}{8})}$

$\frac{\sqrt{2}}{2} = \frac{1-tan^2(\frac{\pi}{8})}{1+tan^2(\frac{\pi}{8})}$

$(\sqrt{2}+2)tan^2(\frac{\pi}{8}) + (\sqrt{2}-2) = 0$

Use the quadratic formula, taking the positive solution - b = 0 so i'll omit it

$tan(\frac{\pi}{8}) = \frac{\sqrt{-4(\sqrt{2}-2)(\sqrt{2}+2)}}{2(\sqrt{2}+2)}$

$tan(\frac{\pi}{8}) = \frac{\sqrt{8}}{2(\sqrt{2}+2)}$

$tan(\frac{\pi}{8}) = \frac{\sqrt{2}}{(\sqrt{2}+2)}$

Rationalise denominator..

$tan(\frac{\pi}{8}) = \frac{2-2\sqrt{2}}{(-2)}$

$tan(\frac{\pi}{8}) = \sqrt{2} - 1$

EDIT: I know you did ask for full working but I'm also tired :buck2: gotta wake up early tomorrow >.>
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 11:53:25 pm: wait... why cos?
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 02, 2010, 11:54:28 pm: nevermind
Im a bit lala ish... i missed read it ahaha
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 03, 2010, 09:27:34 pm: Ahh, i got a feeling this forum is likely to be my second home now ...

Any how, im back, but this time with another question which i hope i can get help on lol :)

Find the value of a for which there are infinitely many solutions to the equations

2x + ay - z = 0
3x + 4y - (a+1)z = 13
10x + 8y + (a-4)z = 26

-> can someone here explain to me the meaning of "infinitely many solutions" please?

Also, after finding the value of a from above, solve the equations.

Thank you very much in advance for the help, just in case i forget to thank you :)
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 03, 2010, 09:31:20 pm: Ohh this topic is matrices if that helps with the solution :)
Title: Re: Momo's Spech Hw help trend !
Post by: the.watchman on February 03, 2010, 10:27:32 pm: For infinite many solutions graphically, the 'lines' or whatever have to be the same (think of y1=x and y2=x, if simultaneous, they have infinite sols)

(Lame answer, I know...)
Title: Re: Momo's Spech Hw help trend !
Post by: Mao on February 03, 2010, 10:46:22 pm: Quote from: Momo.05 on February 03, 2010, 09:27:34 pm
Ahh, i got a feeling this forum is likely to be my second home now ...

Any how, im back, but this time with another question which i hope i can get help on lol :)

Find the value of a for which there are infinitely many solutions to the equations

2x + ay - z = 0
3x + 4y - (a+1)z = 13
10x + 8y + (a-4)z = 26

-> can someone here explain to me the meaning of "infinitely many solutions" please?

Also, after finding the value of a from above, solve the equations.

Thank you very much in advance for the help, just in case i forget to thank you :)

I'm pretty sure this is not part of the specialist syllabus [because it's a 3x3 system, the interpretation is a little more complicated]. So here's some reading materials if you are interested:

Visual representation of unique, none and infinite case for a 3x3 system:
http://www.mathwarehouse.com/algebra/planes/systems/three-variable-equations.php

Systematic technique of solving linear equations:
http://www.sosmath.com/matrix/system1/system1.html

Infinite solutions using Gaussian Elimination:
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/gauss/gauss.html

But I'd recommend not really worrying about it that much.
Title: Re: Momo's Spech Hw help trend !
Post by: the.watchman on February 03, 2010, 10:52:09 pm: Just wondering, is it possible to use Laplace expansion to work out when the system of equations has none or infinite solutions in this situation?
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 03, 2010, 11:00:34 pm: OHH right, this is methods.. !
Oopsssssssssssssssss ahaha my bad !
Title: Re: Momo's Spech Hw help trend !
Post by: Mao on February 03, 2010, 11:17:05 pm: Quote from: the.watchman on February 03, 2010, 10:52:09 pm
Just wondering, is it possible to use Laplace expansion to work out when the system of equations has none or infinite solutions in this situation?

Yes, you can. In SM, you are welcome to use methods beyond the syllabus, so long as you get to the right answer via a correct method.

However, Laplace expansion [i.e. determinant] does not give information about whether the system has no solutions or infinite solutions. It simply states 'unique solution' or 'no unique solutions'. Gaussian elimination is still a far better choice.
Title: Re: Momo's Spech Hw help trend !
Post by: the.watchman on February 04, 2010, 06:43:07 am: Quote from: Mao on February 03, 2010, 11:17:05 pm
Quote from: the.watchman on February 03, 2010, 10:52:09 pm
Just wondering, is it possible to use Laplace expansion to work out when the system of equations has none or infinite solutions in this situation?

Yes, you can. In SM, you are welcome to use methods beyond the syllabus, so long as you get to the right answer via a correct method.

However, Laplace expansion [i.e. determinant] does not give information about whether the system has no solutions or infinite solutions. It simply states 'unique solution' or 'no unique solutions'. Gaussian elimination is still a far better choice.

Ok, thanks for that Mao!
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 20, 2010, 11:12:22 am: Hey once again, i need some help :)

Im doing complex numbers instead and atm im quite confuse on how to solve this question

If z = 1 + i.root3

Hence then find conjugate of z^-5 in Cartesian form ..?

THANK YOU! :)
Title: Re: Momo's Spech Hw help trend !
Post by: fady_22 on February 20, 2010, 11:34:42 am: Quote from: Momo.05 on February 20, 2010, 11:12:22 am
Hey once again, i need some help :)

Im doing complex numbers instead and atm im quite confuse on how to solve this question

If z = 1 + i.root3

Hence then find conjugate of z^-5 in Cartesian form ..?

THANK YOU! :)

Change the complex number z into polar form: z=2cis(pi/3)
Using De Moivre's theorem, z^-5=(2cis(pi/3))^-5= 1/32 cis(pi/3)
The conjugate is 1/32cis(-pi/3)
Change this back into cartesian: z^-5=rcosx+rsinx.i=1/32.1/2+1/32.-sqrt(3)/2
= $\frac{1}{64}-\frac{\sqrt{3}}{64}i$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 20, 2010, 11:38:30 am: OHH thank you :)
I got it now ! ~
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 20, 2010, 11:57:20 am: Ohh another question, can someone check if w=root2(cis(pi/4)) equals to 1 + i in Cartesian form? Since thats what i got .. is it right? O.O
Title: Re: Momo's Spech Hw help trend !
Post by: GerrySly on February 20, 2010, 12:28:19 pm: $\begin{align*}<br />\sqrt{2} \text{cis}(\frac{\pi}{4})&=\sqrt{2}\left (\cos{(\frac{\pi}{4})}+i\sin{(\frac{\pi}{4})}\right )\\<br />&=\sqrt{2}\left (\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right )\\<br />&=1+i<br />\end{align*}$
Title: Re: Momo's Spech Hw help trend !
Post by: Stroodle on February 20, 2010, 12:29:36 pm: Quote from: Momo.05 on February 20, 2010, 11:57:20 am
Ohh another question, can someone check if w=root2(cis(pi/4)) equals to 1 + i in Cartesian form? Since thats what i got .. is it right? O.O

Yeah, it's correct:

$\sqrt{2}cis\left (\frac{\pi}{4}\right )$

$\sqrt{2}cos\left (\frac{\pi}{4}\right ) + \sqrt{2}sin\left (\frac{\pi}{4}\right ) i$

$\sqrt{2}\times \frac{1}{\sqrt{2}}+\sqrt{2}\times \frac{1}{\sqrt{2}}i$

$1+i$
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 20, 2010, 01:56:01 pm: Phew, thanks ! :)
Title: Re: Momo's Spech Hw help trend !
Post by: Momo.05 on February 20, 2010, 04:17:05 pm: Okay, ive been stuck on this question for so long, can someone out there help me with it ? PRETTY PLEASE !? ITS DRIVING ME NUTS ATM!

By considering z = cis(x) and z - z^-1 find the expression for sin3x in terms of sinx

I tired using (z -z^-1)^3 but i dont understand where im going wrong... =="

THANK YOU FOR THE HELP!!
Title: Re: Momo's Spech Hw help trend !
Post by: Mao on February 20, 2010, 11:59:08 pm: Expanding: $z - z^{-1} = \cos(x) + i \sin(x) - (\cos (-x) + i \sin(-x)) = 2i \sin(x)$

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

Substituting polar forms: $(2i \sin (x))^3 = cis(3x) - 3(2i \sin(x)) - cis(-3x)$

Simplifying: $-8i \sin^3 (x) + 6i \sin(x) = \cos (3x) + i \sin (3x) - \cos(-3x) - i \sin(-3x)$

$\implies 2i \sin (3x) = -8i \sin^3(x) + 6i \sin (x)$

$\implies \sin(3x) = -4 \sin^3(x) + 3 \sin(x)$