4U Maths Question Thread

[3.14159265359]

also should I take a math set, like protractor, compass or nah

[RuiAce]

also should I take a math set, like protractor, compass or nah

Like, sure if you want. It helps to draw circles for circle geometry problems where you need to copy the diagram I guess

[Maths4life]

From,
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)

Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)

Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!

[RuiAce]

From,
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)

Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)

Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!

You should only assume stuff from the first equation. You can work with one side at a time for what you wish to prove, but that probably was not the intended approach of the question since it says "deduce".

For this question, as a hint the first step is to identify that:
\begin{align*}\frac{z^6+1}{z^3} &= \frac{z^2-2z\cos \frac\pi6+1}{z} \times \frac{z^2-2z\cos\frac\pi2 + 1}{z} \times \frac{z^2-2z\cos\frac{5\pi}6 + 1}{z} \\ \therefore z^3 + z^{-3} &= \left(z - 2\cos \frac\pi6 + z^{-1} \right)\left(z - 2 \cos \frac\pi2 + z^{-1} \right)\left( z - 2\cos \frac{5\pi}6 + z^{-1}\right)\end{align*}
Now think about what you can do to throw away all the \(z\)'s and \(z^{-1}\)'s for a bunch of cosines. And yes, it is a de Moivre's question

[hassrax]

Thank you so much!!! Just got the answer by letting z=cisthtea.  I just had some other questions. How did you know to divide by z^3, did it have soething to do with the cos3theta? Also, is there a general approach to these questions or does it just take experience?

[RuiAce]

Thank you so much!!! Just got the answer by letting z=cisthtea.  I just had some other questions. How did you know to divide by z^3, did it have soething to do with the cos3theta? Also, is there a general approach to these questions or does it just take experience?

Yeah. I saw that \( \cos \frac\pi6\) and the other three were already there so I wanted to leave them untouched. My initial idea was to somehow kill off the \(z^2+1\) with \(z=-i\sin\theta\) or something, but then I decided nope there's gonna be too many sine's floating about.

Then it hit me that \(\operatorname{cis}\theta + \operatorname{cis}(-\theta) = 2\cos\theta\), so I thought about if that was helpful. That's when I realised that division by \(z^3\) would probably lead somewhere, because as mentioned it forces the \(\cos 3\theta\) to appear when combined with de Moivre's. What was left was to figure out how I would fix up the RHS.

[hassrax]

Thank you for the detailed thought process and answer!! Legend!

[24221]

Hey! can someone pls help me with this question
thank you!!

[RuiAce]

Hey! can someone pls help me with this question
thank you!!

______________________________________
\begin{align*}|z+w|^2 &= (z+w) \overline{(z+w)} \\ &= (z+w)(\overline{z} + \overline{w})\\ &= z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w}\\ &= |z|^2 + z\overline{w} + \overline{z\overline{w}} + |w|^2\\ &= |z|^2 + 2 \operatorname{Re}(z\overline{w}) + |w|^2.\end{align*}

[24221]

thank you!

[terassy]

Hello, could someone help me clarify something.

This is the question:
Show that the following equations represent hyperbolas and find their centre, focu and the equation of directrices and asymptotes.

1) (x-3)^2/64   -    (y+1)^2/36  = 1

For the first part of the question, in ‘showing that the equation is a hyperbola’ do I show that e > 1 or something else like |PS-PS’|=2a

Thanks 

[RuiAce]

Hello, could someone help me clarify something.

This is the question:
Show that the following equations represent hyperbolas and find their centre, focu and the equation of directrices and asymptotes.

1) (x-3)^2/64   -    (y+1)^2/36  = 1

For the first part of the question, in ‘showing that the equation is a hyperbola’ do I show that e > 1 or something else like |PS-PS’|=2a

Thanks 

Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?

[terassy]

Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?

From the old Fitzpatrick 4U textbook. Thanks

[3.14159265359]

Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?

wouldn't you show that it is in the form

xx       yy
---  -   ----  = 1
aa       bb

because that is how my teacher told me to do it

[RuiAce]

wouldn't you show that it is in the form

xx       yy
---  -   ----  = 1
aa       bb

because that is how my teacher told me to do it

It's not of that form because it's not centred at the origin.