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# AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

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• View Poll Results: How did you find the exam?
Pshh I could do it in my sleep!
14
5.56%
Went pretty well
29
11.51%
It was okay... (maybe made silly mistakes?)
82
32.54%
62
24.60%
I died.
61
24.21%
Fine I guess, couldn't careless.
4
1.59%

1. And I don't remember the triple angle formulae. I still need to learn them.
2. Can anyone help me with this vectors question (part c)?

I need to find the coordinates of D. Since AD and BC are equal, I found D simply by adding the difference from B to C if you know what I mean. But I want to know how to do the method they used in the mark scheme (OD = OC + BA). How do I do this? How do I know where the origin is?

http://store.aqa.org.uk/qual/gceasa/...W-MS-JAN07.PDF
3. (Original post by mojopin1)
Can anyone help me with this vectors question (part c)?

I need to find the coordinates of D. Since AD and BC are equal, I found D simply by adding the difference from B to C if you know what I mean. But I want to know how to do the method they used in the mark scheme (OD = OC + BA). How do I do this? How do I know where the origin is?

http://store.aqa.org.uk/qual/gceasa/...W-MS-JAN07.PDF
If you look at the attachment, using the parallel sides, Vector OC (which is given) + Vector BA (which you have already worked out) will give the co-ordinates of "D"
Attached Images

4. (Original post by JulietheCat)
And I don't remember the triple angle formulae. I still need to learn them.
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD
5. (Original post by Quexx)
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD
Re(cosx+isinx)^3= Re(cos^3x + 3(cos^2x(isinx)) + 3(cosx(isinx)^2)+ (isinx)^3)

cos(3x) = cos^3(x)+ 3cosx(-sin^2(x))
= cos^3(x) -3cosx(1-cos^2x)

= 4cos^3(x)-3cosx

6. (Original post by Quexx)
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD
What unit are the triple angle formulae used in?
7. (Original post by 1platinum)
What unit are the triple angle formulae used in?
Might come up in the FP3 exam.
It could have come up during the FP2 exam, but luckily it didn't... =P
8. (Original post by Quexx)
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD
Thank you so much! I guess it means that just like linear interpolation in S1 and standard integrals in C4, just because it's in the textbook doesn't mean it would come up.
9. Useful in understanding the binomial expansion. How to rearrange equations in the form of (1+ax)^n
and also describes how to work out the range of values for which the binomial expansion is valid.

11. these vectors are killing me!
12. (Original post by snoopydoops)
these vectors are killing me!
13. Is anyone else struggling with cartesian equations? I know it's simply rearranging but I find it difficult getting it into the form they want.
14. (Original post by mojopin1)
Is anyone else struggling with cartesian equations? I know it's simply rearranging but I find it difficult getting it into the form they want.
For example...? :/
And do you mean the form they state in the question or in the mark scheme?
15. Yes, when they ask you to 'show' that the cartesian equation can be written in the form. What is the best stratergy to this?
16. (Original post by mojopin1)
Yes, when they ask you to 'show' that the cartesian equation can be written in the form. What is the best stratergy to this?
It depends on the question... I'd say to attempt to pick out parts of the form which you can manipulate the x or y equation for.. Like if the form they ask for has x^2 somewhere, try squaring the x equation, maybe rearranging it a bit and substituting that in. Or if it's trig, possibly mess around with trig identities until it looks like it might fit the form. There can be quite a bit of trial and error involved. I find that the mark scheme provides some insight on how to do it too!
17. Can anyone help me with jan 2012 question 1c? I don't understand how ln has been rearranged to get the answer! Thanks.
Can anyone help me with jan 2012 question 1c? I don't understand how ln has been rearranged to get the answer! Thanks.
after you've intergrated and inserted the limits, you need to use the log rules which you learnt from C2 to help you rearrange (ln is the natural logarithm)
-3ln3=-ln3^3=-ln27
ln5 - ln27 = ln(5/27)
19. (Original post by Cath-ay)
after you've intergrated and inserted the limits, you need to use the log rules which you learnt from C2 to help you rearrange (ln is the natural logarithm)
-3ln3=-ln3^3=-ln27
ln5 - ln27 = ln(5/27)
Got it, thanks very much!
20. Hi Cathay
I only left one type of question don't know how to do it...
Can anyone help me the JUN 10 question 7c and 8?

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Updated: December 4, 2012
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