Author Topic: Maths Questions ^_^ (Read 6480 times) Tweet Share

Odette · « **Reply #15 on:** April 27, 2008, 01:19:36 pm »

Ok i'm a little confused about this, so any help will be appreciated... Oh and an explanation of where they came from would be good..
Deduce the formula in terms of t
$\sin \theta$ = $\frac {2t} {1 + t^2}$
$\cos \theta$ = $\frac {1-t^2} {1+t^2}$
Where t = $\tan \frac{\theta} {2}$

P.S this is my first time using LaTeX

Mao · « **Reply #16 on:** April 27, 2008, 03:54:05 pm »

what is the question asking....? =\

Odette · « **Reply #17 on:** April 27, 2008, 04:32:27 pm »

Oops lol i just realised its already in terms of t.... haha um, my question really should be where did this formula come from?

dcc · « **Reply #18 on:** April 27, 2008, 04:46:12 pm »

$\sin \theta = \dfrac{2t}{1 + t^2}$

where $t = \tan \dfrac{\theta}{2}$

RHS:

$\dfrac{2t}{1+t^2} = \dfrac{2\tan \dfrac{\theta}{2}}{1 + \tan^2 \dfrac{\theta}{2}}$

$\implies \dfrac{2\dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}}}{\mbox{sec}^2 \dfrac{\theta}{2}}$ (since $1 + \tan^2x = sec^2x$ )

$\implies 2\dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}} \cdot \cos^2 \dfrac{\theta}{2}$ (since $\dfrac{1}{\mbox{sec}^2x} = \cos^2 x$ )

$\implies 2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \sin \theta$ (since $\sin 2u = 2\sin u \cos u$ )

Odette · « **Reply #19 on:** April 27, 2008, 04:50:21 pm »

Thanks, but i'm completely lost

Mao · « **Reply #20 on:** April 27, 2008, 04:50:41 pm »

so, you are given that:

$\tan\left(\frac{\theta}{2}\right) = t$

that is, $\tan\left(\frac{\theta}{2}\right) = \frac{Opposite}{Adjacent} = \frac{t}{1}$

if we represent this in a triangle:

the hypotenuse, by pythagoras's theorem, is $\sqrt{1+t^2}$

sine is opposite/hypotenuse, so

$\sin \left( \frac{\theta}{2}\right) = \frac{t}{\sqrt{1+t^2}}$

cosine is then

$\cos \left( \frac{\theta}{2}\right) = \frac{1}{\sqrt{1+t^2}}$

to find $\sin(\theta)$ , we use the double angle formula:

$\sin(\theta) = \sin\left(2\cdot \frac{\theta}{2}\right)$

$\sin(\theta) = 2\cdot \sin\left(\frac{\theta}{2}\right) \cdot \cos\left(\frac{\theta}{2}\right)$

$\sin(\theta) = 2 \cdot \frac{t}{\sqrt{1+t^2}} \cdot \frac{1}{\sqrt{1+t^2}}$

$\sin(\theta) =\frac{2t}{1+t^2}$

$\cos(\theta) = \cos\left(2\cdot \frac{\theta}{2}\right)$

$\cos(\theta) = 2\cdot \cos^2\left(\frac{\theta}{2}\right) -1$

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

$\cos(\theta) = \frac{2-\left(1+t^2\right)}{1+t^2}$

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

dcc · « **Reply #21 on:** April 27, 2008, 05:03:17 pm »

Odette · « **Reply #22 on:** April 27, 2008, 05:10:24 pm »

OH I GET IT!! YAY!! THANK YOU MAO AND DCC

Odette · « **Reply #23 on:** April 28, 2008, 01:14:13 pm »

Ok new set of questions and new topic

COMPLEX NUMBERS!! YAY
Here are some of the questions I need help with...
1.Let z = 4 + 8i and $\zeta$ = 8 + 6i. Evaluate each of the following, giving your answer in the form x + yi.
a) z $\zeta$
b)z + 3 $\overline {\zeta}$
c) $\frac {z} {\zeta}$
d) $\frac {z + \zeta} {\zeta}$

2. Using z from the previous question, evaluate z $^5$ , giving your answer in the form x + yi

Thanks again to whoever helps

Odette · « **Reply #24 on:** April 28, 2008, 01:19:44 pm »

Oh and if anyone could possibly explain the concept of complex numbers that would be greatly appreciated, because the lecturer didn't explain it clearly and this is the first time i've been introduced to complex numbers.

Odette · « **Reply #25 on:** April 28, 2008, 09:40:12 pm »

Anyone? Please ^_^ Dcc? Mao? Anyone? I really need to know this and i just dont get it

Mao · « **Reply #26 on:** April 28, 2008, 09:46:42 pm »

sorry, i would, except i have bloody english SAC 2morrow (and the next two days)

2 english SACs in the same week is seriously not cool =(

AppleXY · « **Reply #27 on:** April 28, 2008, 10:05:00 pm »

Yeah same. Would've helped you but I have other stuphs

.

It's pretty simple, do you have a ref. text?

Odette · « **Reply #28 on:** April 28, 2008, 10:06:25 pm »

No just an exercisebook lolz

Odette · « **Reply #29 on:** April 28, 2008, 10:13:28 pm »

That's ok, um i guess i'll have to figure it out somehow

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