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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

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Original post by datpr0
No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.


How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I thimk we've only really done it with single 2x using double angle formulae..
Original post by datpr0
Well that's because they're a pain in the a**! ...and everything else is pretty straight forward once you've done a few past papers.


Yeah I know I hate them too, but I mean that it seems everyone commenting thinks that it's just a personal weakness when really they're just awful haha
Can anyone tell me where the ln x comes from question 8 c in june2014?
I don't understand what the markscheme has done for 7c.
I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2
Where does that come from? :/

image.jpg
Original post by Jankie
With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!


I could explain it, but it would take too long, so i wont explain it. Sorry for being useless.
Original post by Haza100
Would also like to know this.

Would something like Cos(4x) be equivalent to
2Cos^2(2x) - 1
1 - 2Sin^2(2x)
cos^2(2x) - sin^2(2x)


I'm pretty sure those are all correct.
I just did a trig question with 4x in them and it worked.
Original post by Becky J
How would you do something like cos(2x +2x)? And do you think that kind of thing might come up - I think we've only really done it with single 2x using double angle formulae..


Well, remember the double angle formulae are derivatives of the formulae given in the formula booklet regarding sin(A+B), cos(A+B) and tan(A+B). That is why you rewrite cos4x as cos(2x + 2x), because it is now in the form cos(A+B) which you know from the formula booklet equals cosAcosB - sinAsinB (in this case cos2xcos2x - sin2xsin2x).

Which simplifies to cos^2(2x) - sin^2(2x). That's interesting, lol :biggrin:!
(edited 8 years ago)
Reply 1267
ImageUploadedByStudent Room1433793777.121609.jpg

Is the value of tanB negative because B is obtuse? How does that work?


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Original post by datpr0
No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.


I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?

Simply using the DA formulae I get the following:
Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)
Original post by emsca
ImageUploadedByStudent Room1433793777.121609.jpg

Is the value of tanB negative because B is obtuse? How does that work?


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Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.

ImageUploadedByStudent Room1433793989.299943.jpg
Reply 1270
Original post by Jimmy20002012
Think about cast circle and where acute angles, obtuse angles and reflex angles are. Here this might help some what. Sine you are given a positive sin value it must be in the second quadrant.

ImageUploadedByStudent Room1433793989.299943.jpg


Thank you so much!


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Original post by Haza100
I'm probably missing something totally obvious (which is a little worrying haha), can you explain in a little more detail how you would identify what cos(4x) is equal to?

Simply using the DA formulae I get the following:
Cos(4x) == Cos(2x + 2x) == Cos(2x) * Cos(2x) - Sin(2x) * Sin(2x) == Cos^2(2x) - Sin^2(2x)


Yes, I reached the same conclusion. I believe it's because you can split 4x into two equal parts. A better example would be cos3x/sin3x/tan3x, which they would be more likely to ask as it doesn't simplify so nicely.
Original post by emsca
ImageUploadedByStudent Room1433793777.121609.jpg

Is the value of tanB negative because B is obtuse? How does that work?


Posted from TSR Mobile


Can someone explain this to me without the use of a CAST diagram?
(My college never taught them)
vector Q.jpg
With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?
Original post by saad97
I don't understand what the markscheme has done for 7c.
I calculated B and then worked out the distance between AB, however they state that AB^2 = 2AP^2
Where does that come from? :/

image.jpg


Oh wow where does that come from?? I got |AP|=|BP|=6sqrt(14)
What paper is this? which year?
Original post by mamacita
vector Q.jpg
With a question like this is it just convention to label the sides of the parallelogram clockwise starting from bottom left? I tried this question and I labelled my diagram in a different way from the markscheme and got a different answer. How are you supposed to know which corner is D?


Is this c4?? Aqa c4?? :O No way...
question - http://prntscr.com/7eopwv
ms - http://prntscr.com/7eoq1x
my workings - http://prntscr.com/7eoyez

would I get the marks?
from june 2014.
Original post by bakedbeans247
Is this c4?? Aqa c4?? :O No way...


Noooo it's from an OCR paper but the one's I was doing have the same content as AQA, I would imagine they'd give us a diagram?
Original post by bakedbeans247
Oh wow where does that come from?? I got |AP|=|BP|=6sqrt(14)
What paper is this? which year?


It's June 2007 Q7.
My answer was sqrt(1085) :/
(edited 8 years ago)
Original post by datpr0


Quick question! Does 2p^t = q^t equal

2 ln p^t = ln q^t

I'm just unsure what to do with the 2!

Thanks

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