FP2 MEI OCR Official Thread - 23rd June 2014

Reply 80

username1256732

1

Yes since i screwed up STEP :frown:

Reply 81

pujah0ntas

Original post by mathematigeek

I am so confused :frown:

I get to that part then they just jump to the conclusion and im so confused :frown:

You basically divide everything by cos^5(theta) and then you should get what they asked for

Reply 82

pujah0ntas

If a matrix has an infinite amount of solutions, how do you know whether it forms a sheaf of planes or a triangular prism cause I just usually guess lol

Reply 83

Political Cake

6

Original post by theCreator

Anyone's university offers riding on this exam? :tongue:

Haha frankly I don't know. Anywhere from 55 UMS for an A to 80 UMS for an A*... And If I haven't gotten that in normal maths, it might have to be!

Reply 84

username1107833

2

Original post by pujah0ntas

If a matrix has an infinite amount of solutions, how do you know whether it forms a sheaf of planes or a triangular prism cause I just usually guess lol

I think it's if you can express it under one parameter then it has to be a sheaf of planes. But if you have to express it in terms of two parameters then there are 3 options which I forgot how to differentiate between lol.

Reply 85

username1107833

2

edit: Made a mistake in my question, essentially can anyone explain how you get the modulus of w in Q 2)b)ii) June 2010. Thanks.

(edited 9 years ago)

Reply 86

Biologist0

Original post by theCreator

edit: Made a mistake in my question, essentially can anyone explain how you get the modulus of w in Q 2)b)ii) June 2010. Thanks.

Best thing to do is look at it as an isosceles triangle with the tip at the midpoint. The lengths to P and Q will be root(3), and then draw the line from the origin to the midpoint. Now the length of this line is the modulus of a fourth root of w, and we can find it by simple trig. You now have two right-angled triangles, and the angle at the centre between your bisector and one of your hypotenuses (say OP) is pi/4, because this is a square and we have cut a right angle at the centre of the square in two. Now you can find your length to the midpoint with simple trig - |OM|=root(3)*cos(pi/4). Now simply raise that to the power of four (as it was a fourth root) and you have |w|. Hope this helps!

Reply 87

username1107833

2

Original post by Biologist0

Best thing to do is look at it as an isosceles triangle with the tip at the midpoint. The lengths to P and Q will be root(3), and then draw the line from the origin to the midpoint. Now the length of this line is the modulus of a fourth root of w, and we can find it by simple trig. You now have two right-angled triangles, and the angle at the centre between your bisector and one of your hypotenuses (say OP) is pi/4, because this is a square and we have cut a right angle at the centre of the square in two. Now you can find your length to the midpoint with simple trig - |OM|=root(3)*cos(pi/4). Now simply raise that to the power of four (as it was a fourth root) and you have |w|. Hope this helps!

Damn, would of missed all those marks in the exam, thank you!

Reply 88

wizardmunchahoy

Hi there! Hope everyone's preparation is going well for tomorrow!

Just re-did the June 2013 paper, and I still haven't got a satisfactory answer to question 2 (a) (ii). Obviously the mark scheme's not online, and the examiner's report isn't very helpful, so I'd appreciate an explanation!

It wants you to show that

\cos 18^{\circ} = (\frac{5 + \sqrt 5}{8})^{1/2}

Q.2(a)(i) asks you to show that

\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta

, which is easy enough to show using de Moivre's theorem.

Q.2(a)(ii) wants you to first of all find two possible values for

\cos^2 \theta

. The question tells you that

\cos 5 \theta = 0

but

\cos \theta \not= 0

. That's fine, because you just factorise the equation for

\cos 5 \theta

above to get a quadratic in

\cos^2 \theta

, and you use the quadratic formula to get

\cos^2 \theta = \frac{5 \pm \sqrt 5}{8}

.

Then of course,

\cos \theta = \pm (\frac{5 \pm \sqrt 5}{8})^{1/2}

. But how do you show that both those plus/minus signs are positive to give

\cos 18^{\circ}

, without resorting to a calculator?

Our teacher's explanation was that, out of the four possible roots, using both plus signs gives the value that is closest to 1 (0.951...) and

\cos 18^{\circ} = 0.951...

. Of course,

\cos 5*18^{\circ} = \cos 90^{\circ} = 0

which satisfies the original equation, but so does

\theta = 54^{\circ}

. (Using my calculator,

\cos 54^{\circ} = (\frac{5 - \sqrt 5}{8})^{1/2}

, i.e. the negative root.)

So are there any good explanations on how to prove that

\cos 18^{\circ} = (\frac{5 + \sqrt 5}{8})^{1/2}

and not one of the other three roots?

Sorry about the length of the post, just wanted to make sure that all the necessary details were included! :smile:

Reply 89

WinOrDie

1

Original post by pujah0ntas

If a matrix has an infinite amount of solutions, how do you know whether it forms a sheaf of planes or a triangular prism cause I just usually guess lol

Loool u joker

Reply 90

username1107833

2

Original post by Law-Hopeful

For 2ii when you find C+jS you find that it equals the binomial expansion (1+e^(2jtheta))^n, however on the mark they go from this to something completely different (attached below).

Can anyone help explain how they got that? :confused:

Oh damn, I only just understood it now. You have to use your answer to part i) and substitute that answer in. Hence (1+e^j2theta)^n becomes (2cos(theta)*(cos(theta)+jsin(theta)))^n and then you just expand.

Reply 91

Illiberal Liberal

2

Original post by theCreator

Oh damn, I only just understood it now. You have to use your answer to part i) and substitute that answer in. Hence (1+e^j2theta)^n becomes (2cos(theta)*(cos(theta)+jsin(theta)))^n and then you just expand.

Yeah I just realised what an idiot I was! Thank you! :smile:

Reply 92

piguy

8

Original post by wizardmunchahoy

Then of course,

\cos \theta = \pm (\frac{5 \pm \sqrt 5}{8})^{1/2}

....)

So are there any good explanations on how to prove that

\cos 18^{\circ} = (\frac{5 + \sqrt 5}{8})^{1/2}

and not one of the other three roots?

Sorry about the length of the post, just wanted to make sure that all the necessary details were included! :smile:

Do a quick sketch of cos(x)
the four solutions are stars in diagram
Clearly cos(18) is going to be positive, so the external plus/minus is a plus
Also cos(18) is more than cos(54) (since between 0 and 90 cos(x) decreases), so cos(18) has the bigger magnitude and you add the internal square root.

Spoiler

This exam is going to kill me. :sigh:

Oh well, if things have gone as I think they have then I only need 50-60 UMS for an A in Further Maths.

Reply 94

Illiberal Liberal

2

Can anyone help with this? I get that each line from the origin will be 2pi/3 apart from each other and I can therefore work out the arguments on each point, but I don't have the first clue on how to get the moduli or the x/y points :confused:

Screen Shot 2014-06-22 at 23.19.46.png32.8KB

Reply 95

Kool_Panda

12

Original post by Law-Hopeful

Can anyone help with this? I get that each line from the origin will be 2pi/3 apart from each other and I can therefore work out the arguments on each point, but I don't have the first clue on how to get the moduli or the x/y points :confused:

Well, the moduli of the other points will be the same as the modulus of the original point (2+4j).

Reply 96

wizardmunchahoy

Original post by piguy

Do a quick sketch of cos(x)
the four solutions are stars in diagram
Clearly cos(18) is going to be positive, so the external plus/minus is a plus
Also cos(18) is more than cos(54) (since between 0 and 90 cos(x) decreases), so cos(18) has the bigger magnitude and you add the internal square root.

Spoiler

Ahhhh, thank you so much! That is literally so simple, can't believe I didn't spot it! :tongue:

Such a neat solution! Once again, thank you!

Reply 97

piguy

8

Original post by Law-Hopeful

Can anyone help with this? I get that each line from the origin will be 2pi/3 apart from each other and I can therefore work out the arguments on each point, but I don't have the first clue on how to get the moduli or the x/y points :confused:

The only feasible way to get exact solutions for the x, y co-ordinates is by applying a transformation:

Each point is rotated by 2pi/3 but not increased by any factor, so the transformation is multiplying a point by e^(2pi/3)j to get to the next point on the triangle (i.e. rotation by 120 deg.)

e^(2pi/3)j = cos(2pi/3) + jsin(2pi/3) = -1/2 + sqrt(3)j/2

You just multiply that by (2 + 4j) to get final co-ordinates for x,y.

For the third point, it's twice the rotation, so you multiply (2+4j) by e^(2pi/3)j twice = (e^(2pi/3)j)^2 = e^(4pi/3)j = e^(-2pi/3)j = -1/2 -sqrt(3)j/2