Author Topic: 4U Maths Question Thread (Read 665195 times) Tweet Share

Wales · « **Reply #720 on:** December 03, 2016, 11:47:29 pm »

http://puu.sh/sCJ9I/789660a744.png

I keep struggling with (ii), I get i but ii gets me nowhere

It's some drawn out proof...

RuiAce · « **Reply #721 on:** December 03, 2016, 11:54:35 pm »

Quote from: Wales on December 03, 2016, 11:47:29 pm

http://puu.sh/sCJ9I/789660a744.png

I keep struggling with (ii), I get i but ii gets me nowhere It's some drawn out proof...

$\text{Using the result of (i) multiple times}\\ \text{that is, }\cos n\theta = \frac12 (z^n+z^{-n})$
$\begin{align*}\cos \theta \cos 2\theta&= \frac12(z+z^{-1})\times \frac12(z^2+z^{-2})\\ &= \frac14(z^3+z+z^{-1}+z^{-3})\\ &= \frac12\left(\frac{1}{2}(z+z^{-1})+\frac12(z^3+z^{-3})\right)\\ &= \frac12(\cos \theta+\cos 3\theta) \end{align*}$

Yasminpotts1105 · « **Reply #722 on:** December 11, 2016, 04:22:06 pm »

What is the difference when graphing f(x^{2^{)) to [f(x)]² ?}}

RuiAce · « **Reply #723 on:** December 11, 2016, 04:40:18 pm »

Quote from: Yasminpotts1105 on December 11, 2016, 04:22:06 pm

What is the difference when graphing f(x^{2^{)) to [f(x)]² ?}}

Careful with the typos in your BBC code there; you forgot to [/sup] the first one because you forgot the forward slash

The short answer is that the difference is what you tamper with. It comes back down to bracketing - remember, you always do what's INSIDE the brackets first. Not what's OUTSIDE.

For y=[f(x)]², you're finding the function, and then squaring it AFTER you've done that computation.
For y=f(x²), you're squaring the function BEFORE you apply the function to it.
_____________________

For the longer answer, it is easier to illustrate with an example. Open up Desmos and consider the function f(x)=sin(x)
(Just because it's a bit clearer do illustrate what's going on, and isn't TOO laggy.)
$\text{First, consider }y=[f(x)]^2\\ \text{When }f(x)=\sin x\text{, we have }[f(x)]^2=\sin^2 x\\ \text{Sketch both }y=\sin x\text{ and }y=\sin^2 x\text{ on Desmos.}$
$\text{Now, compare the two curves.}\\ \text{Once again, I will establish the obvious fact that }\boxed{(\sin x)^2=\sin^2x}$
$\text{The reason this is important, is that in the first curve it is }y_1=\sin x\\ \text{but in the second, we have }y_2=\sin^2x\\ \text{Effectively, we are saying that }\boxed{y_1^2=y_2}$
$\text{In words, that is to say}\\ \text{The }\textbf{y-coordinates}\text{ are what's been tampered with.}\\ \text{Note how the }y\text{-coordinates in }y=\sin^2x\\ \text{are always the SQUARE of that of }y=\sin x$
________________________________
$\text{Now delete }y=\sin^2x\\ \text{When }f(x)=\sin x\text{, we know that }f(x^2)=\sin(x^2)\\ \text{Keep }y=\sin x\text{ and sketch }y=\sin(x^2)\text{ on Desmos.}$
$\text{At first glance, it is pretty hard to see how they're related.}\\ \text{But we must think.}\\ \textit{What did we do?}$
$\text{Before we put }\sin\text{ around it, we squared }x\\ \text{We have the sine, of }\boxed{x^2}$
$\text{So we SQUARED the }\boxed{x\text{-coordinates first.}}\\ \text{Here, the }\textbf{x-coordinates}\text{ are what's been tampered with}\\ \text{That is why the curve looks extremely stretched and squashed compared to the first}$

kiwiberry · « **Reply #724 on:** December 12, 2016, 10:03:36 pm »

Points P and Q are the endpoints of focal chord of the ellipse x^2/a^2 + y^2/b^2 = 1. if the parameters at P and Q are θ and α, show that the eccentricity is given by e = sin(θ-α)/(sinθ-sinα)

help please $:-\$

RuiAce · « **Reply #725 on:** December 12, 2016, 10:16:39 pm »

Quote from: kiwiberry on December 12, 2016, 10:03:36 pm

Points P and Q are the endpoints of focal chord of the ellipse x^2/a^2 + y^2/b^2 = 1. if the parameters at P and Q are θ and α, show that the eccentricity is given by e = sin(θ-α)/(sinθ-sinα)

help please $:-\$

$\text{Note that the gradient of }PQ\text{ is}\\ m_{PQ}=\frac{b(\sin \theta-\sin \alpha)}{a(\cos \theta-\cos \alpha)}\\ \text{This implies that the equation of }PQ\text{ is}\\ y-b\sin \theta=\frac{b(\sin \theta-\sin \alpha)}{a(\cos \theta-\cos \alpha)}(x-a\cos \theta)$
$\text{If }PQ\text{ is a focal chord, it passes through }(ae,0)\\ \text{where }e\text{ is the eccentricity of the ellipse.}\\ \text{Hence we can substitute this point in.}\\ \begin{align*}-b\sin \theta&=\frac{b(\sin \theta-\sin \alpha)}{a(\cos \theta-\cos \alpha)}(ae-a\cos \theta)\\ -\sin \theta(\cos \theta-\cos \alpha)&=(\sin \theta-\sin \alpha)(e-\cos \theta)\end{align*}$
$\text{Remember that we want to find the eccentricity.}\\ \text{Hence we must rearrange to make }e\text{ the subject.}\\ \begin{align*}-\sin\theta\cos\theta+\sin\theta\cos\alpha&=e(\sin\theta-\sin\alpha)-\cos\theta(\sin\theta-\sin\alpha)\\ -\sin\theta\cos\theta+\sin\theta\cos\alpha&=e(\sin\theta-\sin\alpha)-\cos\theta\sin\theta+\cos\theta\sin\alpha\\ \sin\theta\cos\alpha-\cos\theta\sin\alpha&=e(\sin\theta-\sin\alpha)\\ \sin(\theta-\alpha)&=e(\sin\theta-\sin\alpha)\\ e&=\frac{\sin(\theta-\alpha)}{\sin\theta-\sin\alpha}\end{align*}\\ \text{as required.}$

Yasminpotts1105 · « **Reply #726 on:** December 13, 2016, 06:53:55 pm »

Sketch |w+3i| / |w-4i| = 1

I tried to let w = x + iy and realize the denominator but I don't understand how to do it with the three terms I end up with.

Wales · « **Reply #727 on:** December 13, 2016, 06:58:59 pm »

http://puu.sh/sNOEr/240e0f639a.jpg

I get part i but for part ii I don't quite get the arg(z1+z2) expansion. Do I throw in Z1 and Z2 and simplify the expression?

RuiAce · « **Reply #728 on:** December 13, 2016, 07:06:18 pm »

Quote from: Yasminpotts1105 on December 13, 2016, 06:53:55 pm

Sketch |w+3i| / |w-4i| = 1

I tried to let w = x + iy and realize the denominator but I don't understand how to do it with the three terms I end up with.

$\text{Multiply the expression out first:}\\ |w+3i|=|w-4i|$
$\text{It should now be clear that the locus is the perpendicular}\\ \text{bisector of the line joining }(0,-3)\text{ and }(0,4)\\ \text{This is the line }y=\frac{1}{2}$
$\text{Alternatively, let }w=x+iy\text{ and square both sides}\\ \begin{align*}|x+i(y+3)|^2&=|x+i(y-4)|^2\\ x^2+(y+3)^2&=x^2+(y-4)^2\\ 6y+9&=-8y+16\\ y&=\frac12\end{align*}$

RuiAce · « **Reply #729 on:** December 13, 2016, 07:26:31 pm »

Quote from: Wales on December 13, 2016, 06:58:59 pm

http://puu.sh/sNOEr/240e0f639a.jpg

I get part i but for part ii I don't quite get the arg(z1+z2) expansion. Do I throw in Z1 and Z2 and simplify the expression?

$\text{In part (i), we noted that}\\ \begin{align*}z_1=-\sqrt{2}+\sqrt{2}i&=2\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)\\z_2=\sqrt{3}+i&=2\left(\cos \frac{\pi}{6}+i\sin\frac{\pi}6\right) \end{align*}$
$\text{So clearly from }\arg\left(\frac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)\text{ we get}\\ \arg\left(\frac{z_1}{z_2}\right)=\frac{7\pi}{12}\\ \text{To show that }\arg(z_1+z_2)=\frac{11\pi}{24}\text{ is harder and cannot be easily done with algebra.}$
$\text{Consider the following Argand diagram.}$

$\text{Observe that }z_1\text{ and }z_2\text{ have the same modulus, so }OA=OB\\ \text{Hence }OACB\text{ is a rhombus.}\\ \text{In a rhombus, we know that }\textit{the diagonals bisect the angles of the quadrilateral.}$
$\text{Note that }\angle AOB = \frac{3\pi}{4}-\frac{\pi}{6}=\frac{7pi}{12}\\ \therefore \angle COB=\frac{7\pi}{24}$
$\text{This gives }\arg(z_1+z_2)=\angle COB + \arg(z_2) = \frac{11\pi}{24}\text{ (check diagram)}$
$\text{And plug everything back in.}$

Yasminpotts1105 · « **Reply #730 on:** December 15, 2016, 08:37:40 pm »

If z^3 = 1, z cannot equal 1, show that (1+z)^5 = - z

I have tried using demoivre's theorem and mathematical induction or just substituting a value for z but I just can't get an answer. Any helpful hints as to how to approach this?

RuiAce · « **Reply #731 on:** December 15, 2016, 08:43:05 pm »

Quote from: Yasminpotts1105 on December 15, 2016, 08:37:40 pm

If z^3 = 1, z cannot equal 1, show that (1+z)^5 = - z

I have tried using demoivre's theorem and mathematical induction or just substituting a value for z but I just can't get an answer. Any helpful hints as to how to approach this?

$\text{By factorising, we obtain a useful result}\\ \begin{align*}z^3&=1\\ z^3-1&=0\\ (z-1)(z^2+z+1)&=0\\ 1+z+z^2&=0\\ 1+z&=-z^2\end{align*}\\ \text{Noting that }z-1\text{ was divided since }z\neq 1$
$\text{So if we plug it in}\\ (1+z)^5=(-z^2)^5=-z^{10}$
$\text{However, we know that }z^3=1\\ \text{This means, cubing both sides, }z^9=1\\ \text{Hence }z^{10}=z\\ \text{Puzzling everything back shows our original thing equals }-z$
The question can be done by explicitly finding the two values for z, but that is a bit tedious and algebraic tricks save more time here.

Yasminpotts1105 · « **Reply #732 on:** December 15, 2016, 09:09:03 pm »

THANK YOU SO MUCH, you've just allowed me to actually sleep tonight.

jamonwindeyer · « **Reply #733 on:** December 15, 2016, 09:34:40 pm »

That math above was a thing of beauty

bluecookie · « **Reply #734 on:** December 23, 2016, 11:58:49 am »

Find a and b.
2+i=(1+31)/(a+bi)

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