Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

QC · « **Reply #780 on:** January 29, 2017, 12:56:23 pm »

Idk what to do, are you allowed to cross multiply since it isn't a proof?

jakesilove · « **Reply #781 on:** January 29, 2017, 12:59:34 pm »

Quote from: aroon217 on January 29, 2017, 12:56:23 pm

Idk what to do, are you allowed to cross multiply since it isn't a proof?

You definitely can cross multiply! If you didn't want to do that, I would rationalise the denominator on the LHS, and see what pops out. Let me know if you want a full proof!

However, I think that if you multiply the LHS (both denom and numer) by (1+sin(theta)+icos(theta)), things should work out.

Shadowxo · « **Reply #782 on:** January 29, 2017, 01:13:46 pm »

Here's my solution if you're struggling

RuiAce · « **Reply #783 on:** January 29, 2017, 01:46:05 pm »

Quote from: aroon217 on January 29, 2017, 12:56:23 pm

Idk what to do, are you allowed to cross multiply since it isn't a proof?

$\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&=\frac{1+i(\cos \theta-i\sin \theta)}{1-i(\cos \theta+i\sin \theta)}\\ &= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\frac{-\theta}{2}\left( \text{ cis }\frac{\theta}{2}+i\text{ cis }\frac{-\theta}{2} \right)}{\text{ cis }\frac{\theta}{2}\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}\\ &= \text{ cis }(-\theta) \times \frac{i\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}{\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}}\\ &= i \text{ cis }{-\theta}\\ &= \sin \theta + i \cos \theta\end{align*}$
This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.

QC · « **Reply #784 on:** January 29, 2017, 03:10:44 pm »

Quote from: RuiAce on January 29, 2017, 01:46:05 pm

$\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&=\frac{1+i(\cos \theta-i\sin \theta)}{1-i(\cos \theta+i\sin \theta)}\\ &= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\frac{-\theta}{2}\left( \text{ cis }\frac{\theta}{2}+i\text{ cis }\frac{-\theta}{2} \right)}{\text{ cis }\frac{\theta}{2}\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}\\ &= \text{ cis }(-\theta) \times \frac{i\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}{\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}}\\ &= i \text{ cis }{-\theta}\\ &= \sin \theta + i \cos \theta\end{align*}$
This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.

WOW that sol is so nice, that's a pretty good trick factorising the i out at the start, i'll keep it in mind. Thanks !

beau77bro · « **Reply #785 on:** January 29, 2017, 09:08:16 pm »

Quote from: RuiAce on January 29, 2017, 01:46:05 pm

$\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&=\frac{1+i(\cos \theta-i\sin \theta)}{1-i(\cos \theta+i\sin \theta)}\\ &= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\frac{-\theta}{2}\left( \text{ cis }\frac{\theta}{2}+i\text{ cis }\frac{-\theta}{2} \right)}{\text{ cis }\frac{\theta}{2}\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}\\ &= \text{ cis }(-\theta) \times \frac{i\left(\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}\right)}{\text{ cis }\frac{-\theta}{2} -i\text{ cis }\frac{\theta}{2}}\\ &= i \text{ cis }{-\theta}\\ &= \sin \theta + i \cos \theta\end{align*}$
This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.

this is such a nice proof, but i was just wondering what did you do in the middle step where u got half angles for cis? what exactly were the steps/ method you used

RuiAce · « **Reply #786 on:** January 29, 2017, 09:15:25 pm »

Quote from: beau77bro on January 29, 2017, 09:08:16 pm

this is such a nice proof, but i was just wondering what did you do in the middle step where u got half angles for cis? what exactly were the steps/ method you used

$\text{ cis }x \times \text{ cis }y = \text{ cis }(x+y)\\ \text{Factorised out the }\text{ cis }\frac{-\theta}{2}\text{ and }\text{ cis }\frac{\theta}{2}\\ \text{If it's unclear, you should expand it back in}$

michaelalt · « **Reply #787 on:** January 31, 2017, 10:33:18 pm »

Hi! I've been struggling with complex numbers alot and was wondering if i could get some help with the questions below. I have tried to make sense of them, but just end up confused. Any help on any of them would be greatly appreciated!!

Shadowxo · « **Reply #788 on:** January 31, 2017, 10:45:11 pm »

I'll just do the first question
For this question, it's helpful to graph the points.
If point A represents alpha, alpha = cis(theta)
If B is A rotated anticlockwise 2pi/3 and halved, it equals 1/2(cis(theta + 3pi/2))
So it equals 1/2(cis(theta)*cis(3pi/2)) = 1/2(A*cis(3pi/2)) = A/2cis(3pi/2) = D

RuiAce · « **Reply #789 on:** January 31, 2017, 11:24:39 pm »

Quote from: michaelalt on January 31, 2017, 10:33:18 pm

Hi! I've been struggling with complex numbers alot and was wondering if i could get some help with the questions below. I have tried to make sense of them, but just end up confused. Any help on any of them would be greatly appreciated!!

With these questions though, you're not going to get anywhere without a diagram. For any question, post up a diagram with any progress, and we can finish it off.

RuiAce · « **Reply #790 on:** February 01, 2017, 09:46:42 am »

Quote from: michaelalt on February 01, 2017, 09:19:34 am

Thanks for your help on 1, it does make sense. But I'm also stuck on 3. This is the diagram I have drawn so far.

$\text{Note that }ABCD\text{ is a }\textbf{SQUARE}\\ \text{Your diagram misleadingly shows it as any rectangle that's not a square.}\\ \textit{Whilst your diagram need not be to scale, try to make it as less misleading as possible}$
$\text{We will utilise the geometric properties of a square.}\\ \text{Begin by noting that the diagonals of a square bisect each other.}$
$E\text{ must be the midpoint of }AC\\ \text{So if }C\text{ represents the complex number }z\\ \text{We know that }\frac{z}{2}+\frac{1+5i}{2}=-2+i\\ \therefore z=-5-3i$
___________________________
$\text{There are now multiple ways of approaching this}\\ \text{One way is to somehow fiddle around with other properties of a square}\\ \text{and potentially Pythagoras' theorem and trigonometry.}\\ \textit{I mean, squares are convenient. We have a right angled isosceles triangle.}$
$\textbf{I like to think of vectors now.}\\ \text{The property I want to exploit is that the diagonals of a square}\\ \text{bisect each other }\textbf{at right angles}.$

$\text{The vector }\overrightarrow{EA}\text{ represents the complex number }(1+5i)-(-2+i)\\ \text{i.e. }3+4i\\ \text{So using the above theorem, the vector }overrightarrow{EB}\text{ represents the complex number}\\ i(3+4i) = -4+3i\\ \text{Recall: Multiplication by }i\text{ rotates a vector by }90^\circ \text{ anti-clockwise.}$
$\text{So then }B\text{ must simply be }(1+5i)+(-4+3i) = -6+4i\\ \text{And it can then be checked that }D\text{ represents }2-2i$

armtistic · « **Reply #791 on:** February 01, 2017, 05:13:27 pm »

Hey guys

I've been doing some revision questions on conics and I came across this one that I'm struggling with.

Any help would be appreciated

RuiAce · « **Reply #792 on:** February 01, 2017, 05:54:22 pm »

Quote from: armtistic on February 01, 2017, 05:13:27 pm

Hey guys

I've been doing some revision questions on conics and I came across this one that I'm struggling with.

Any help would be appreciated

$\textbf{You cannot possibly do this question without a diagram.}$

$\text{It is easy to prove that the tangent at }P\text{ is given}\\ x+p^2y=2cp\\ \text{and the gradient of this line is }-\frac{1}{p^2}$
$\text{Hence it can be easily checked that}\\ A\text{ is the point }(2cp,0)\\ B\text{ is the point }\left(0, \frac{2c}{p}\right)$
$\text{Now, supposedly }Q\text{ is the point }\left(cq, \frac{c}{q}\right)\\ \text{But we know that the tangent at }Q\text{ is }\textbf{parallel}\text{ to the tangent at }p\\ \text{It is easy to check that the tangent at }Q\text{ has gradient }-\frac{1}{q^2}$
$\begin{align*}\therefore -\frac{1}{p^2}&=-\frac{1}{q^2}\\ \therefore q&=-p\end{align*}\\ \text{Note that }q\neq p\text{ or }P\text{ and }Q\text{ coincide.}$
$\text{So in fact }Q\text{ is the point }\left(-2cp, -\frac{2c}{p}\right)\\ \text{and the tangent at }Q\text{ has equation }x+p^2y=-2cp$
$\text{Hence }C\text{ is }\left(2cp,0\right)\\ D\text{ is }\left(0,-\frac{2c}{p}\right)$
______________________________
$\textbf{The diagram is extremely useful as clearly, by symmetry}\\ ABCD\text{ is a }\textbf{rhombus}\\ \text{It should be easy to calculate the area now by formula.}$
______________________________
$\text{Note that, the diagram also shows us that }\\ Q\text{ and }P\text{ are points obtained by rotational symmetry about the origin}\\ \text{This is because the coordinates of }Q\text{ are the exact NEGATIVES}\, \text{of that of }P\\ \text{So verily, }O\text{ is just the MIDPOINT of }QP$
$\text{This leads to something important:}\\ QOP\text{ is a straight line AND is the DIAMETER of the circle.}\\ \text{Hence }\angle QRP=\angle QSP=\frac{\pi}{2}\text{ as they are the angles in a semi-circle}$
$\text{So using }m_1m_2=-1\text{ for perpendicularity}\\ \begin{align*}m_{PR}m_{QR}&=-1\\ \frac{\frac{c}{p}-\frac{c}{r}}{cp-cr}\cdot \frac{-\frac{c}{p}-\frac{c}{r}}{-cp-cr}&=-1\\ \frac{\frac{r-p}{pr}}{p-r}\cdot \frac{\frac{r+p}{pr}}{r+p}&=-1\\ \frac{-1}{p^2r^2}&=-1\\ r^2&=\frac{1}{p^2}\\ \therefore p=\frac{1}{r}\end{align*}\\ \text{And similarly}\\ s=-\frac{1}{p}$
$\therefore pqrs = p \times -p \times \frac{1}{p}\times -\frac{1}{p}=1\text{ as required.}$
$\text{Note, my diagram was drawn so that }p\text{ and }r\text{ are both positive}\\ \text{Hence }p\neq -r,\text{ rather }p=r\\ \text{If }R\text{ and S}\text{ were }\textit{swapped}\text{, then }r=-\frac{1}{p}$

QC · « **Reply #793 on:** February 02, 2017, 06:11:16 pm »

How do you systematically make sure you can answer any circle geo question or is it not possible. Also, do you have a checklist or something you run through when u see a circle geo question?

RuiAce · « **Reply #794 on:** February 02, 2017, 07:20:56 pm »

Quote from: aroon217 on February 02, 2017, 06:11:16 pm

How do you systematically make sure you can answer any circle geo question or is it not possible. Also, do you have a checklist or something you run through when u see a circle geo question?

$\text{In the HSC, they will not give an impossible question.}$
$\text{Checklist feels too confined. As mentioned previously, I look for key details (e.g. tangents)}\\ \text{to determine which theorem to use.}$
$\text{If they tell you to copy it out, then it means copy and out}\\ \text{Because it means you have to put your own lines everywhere on it,}\\ \text{It is also worth 4 marks, so something simple is out of question.}$

$\text{Working backwards helped me here.}\\ \text{I utilised the fact that if }BGFC\text{ was a cyclic quadrilateral}\\ \text{Then by the exterior angle theorem, }\angle AGF = \angle BCF$
$\text{Notes: Since I knew I had a diameter, I may or may not have needed the angle in a semicircle.}\\\text{Joining }GF\text{ just makes sense, because }BGFC\text{ is what we're interested in.}$
Apologies for the somewhat messy handwriting

Edit: First line should say join GF

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