You're probably right, it was a tough question and I thought I'd cracked it with that but your solution sounds like it works pretty nicely.(Original post by M^2012)
But surely it's tan(theta) = sin(theta)/cos(theta), not tan=sin/cos, so you'd need the sin part to equal the cos part for that to work, as sin's angle wasn't equal to cos's angle?
Sorry for double posting :P
Well, I was confident about the paper :P
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

OFFICIAL AQA FP1 18th MAY 2012 Thread

 Follow
 41
 18052012 11:52

 Follow
 42
 18052012 11:57
(Original post by CalumE5)
You're probably right, it was a tough question and I thought I'd cracked it with that but your solution sounds like it works pretty nicely.
Well, I was confident about the paper :P
Just out interest, did anybody get the matrix transformation for part (iii) to be scale factor 2 and 90 degrees clockwise (or 270 degrees anticlockwise)?Post rating:3 
 Follow
 43
 18052012 11:57
The general solution question, you had to do sin1 of (cos20) to get 70. Then just work from there like normal.
You get x=540n
x=540n 60
our teacher went through this so im pretty sure its rightPost rating:1 
 Follow
 44
 18052012 11:57
(Original post by M^2012)
You'll have done fine :P The grade boundaries won't be 67/75 for an A like they were in January
Just out interest, did anybody get the matrix transformation for part (iii) to be scale factor 2 and 90 degrees clockwise (or 270 degrees anticlockwise)? 
 Follow
 45
 18052012 11:59
(Original post by M^2012)
In the trig question, I said that:
cos(theta) = sin(theta + 90), therefore, sin(whatever it was) = sin(20 + 90)
Since, I've realised that it should have been cos(theta) = sin(theta90), but both seem to give me the same answer when I try it now...
In the complex numbers I think I got 0.4+1.2i, although I've seen somebody else on here saying it was something else?
Another one that somebody disagreed with me on was the roots of equations one... I got the new product of the roots to be 65/5, whereas they got 64/5?
Edit: Also, I got "n" to be 1006 in the matrices question?
I got 1/2 + 3/2 i
I agree with you on the roots product.
And also for n being 1006. I actually worked it out my calculator and it got a very large number but the first few digits were the same as 2^1006 so I assume it was right. 
 Follow
 46
 18052012 12:13
For the last answer, were the coordinates (8/3,4/3) and (16/3,4/3) right? I think I got those although I may have the signs of the 4/3 part wrong as i cant really remember.
I also got Z=1/2 + 3/2i,
540n, 540n 60, for the general solution
part iii) of the matrix question to be SF 2 and clockwise 90 degrees,
i think part ii) was SF root2 and rotation 135 degrees anticlockwise
i think n of the next question was 1006
new sum was 13 and new product 64/something i think the new equation was 25x^2  64x +325 or something like thatLast edited by eddieb189; 18052012 at 12:44. 
 Follow
 47
 18052012 13:00
(Original post by eddieb189)
For the last answer, were the coordinates (8/3,4/3) and (16/3,4/3) right? I think I got those although I may have the signs of the 4/3 part wrong as i cant really remember.
I also got Z=1/2 + 3/2i,
540n, 540n 60, for the general solution
part iii) of the matrix question to be SF 2 and clockwise 90 degrees,
i think part ii) was SF root2 and rotation 135 degrees anticlockwise
i think n of the next question was 1006
new sum was 13 and new product 64/something i think the new equation was 25x^2  64x +325 or something like that 
 Follow
 48
 18052012 13:03
(Original post by eddieb189)
For the last answer, were the coordinates (8/3,4/3) and (16/3,4/3) right? I think I got those although I may have the signs of the 4/3 part wrong as i cant really remember.
I also got Z=1/2 + 3/2i,
540n, 540n 60, for the general solution
part iii) of the matrix question to be SF 2 and clockwise 90 degrees,
i think part ii) was SF root2 and rotation 135 degrees anticlockwise
i think n of the next question was 1006
new sum was 13 and new product 64/something i think the new equation was 25x^2  64x +325 or something like that
But why is it 60 for the general solution? The basic thing book states is Sin X = Sin Alpha. and Alpha was 20, because it says CosAlpha (cos20), because it doesn't matter if its cos or sin because angle is still same, either its CosAngle or SinAngle, Angle is still Angle. And then you just use x = 360n + alpha and x = 360n + (180  alpha). And x was 70  2/3 x. so 2/3 x = 360n  20  70 which gives you x = 360n 90 then divide by 2/3 to get x = 540n + 135 and then do 70  2/3 x = 360n + (180  20) to get 2/3 x = 360n + 160 and then divide by 2/3 to get x = 540n  240Last edited by Miyata; 18052012 at 13:12. 
 Follow
 49
 18052012 13:25
wait what.. For the general solution question I thought sin(theta)'s general solution was "180n + (1)^n x (alpha).
So doing sin1 of cos(20) you get 70, after some stuff I got something like (3(70  (180n + (1)^n x70))) all divide by 2.
I didn't expand out the brackets. 
 Follow
 50
 18052012 13:30
(Original post by bratwast)
wait what.. For the general solution question I thought sin(theta)'s general solution was "180n + (1)^n x (alpha).
So doing sin1 of cos(20) you get 70, after some stuff I got something like (3(70  (180n + (1)^n x70))) all divide by 2.
I didn't expand out the brackets. 
 Follow
 51
 18052012 13:32
I got my complex number z to be z= 3 2i. So I got that wrong:/ Also forgot to put my p values back into the equation to find the x value. Think I got the matrices, except the M^2004 question. What did everyone get for the Newton Raphson method?

 Follow
 52
 18052012 13:33
(Original post by Miyata)
Not sure about the method you learned, but in our book for Sin general solution is x = 360n + alpha and x = 360n + (180  alpha). Where Sinx = SinAlpha. so in this case x was (70  2/3 x) and alpha was 20, because alpha is still alpha, imo. 
 Follow
 53
 18052012 13:33
guys, x^2 + 42x +65 = 0 ? is that what you guys got?
might of been a 42 can't rememberLast edited by mathslover1; 18052012 at 13:38. 
 Follow
 54
 18052012 13:43
(Original post by Miyata)
Not sure about the method you learned, but in our book for Sin general solution is x = 360n + alpha and x = 360n + (180  alpha). Where Sinx = SinAlpha. so in this case x was (70  2/3 x) and alpha was 20, because alpha is still alpha, imo.
was the question something like sin(70  2/3x) = cos20 then?
because I think I worked out sin1 of cos(20) = 70, so (alpha) is 70.
therefore, 702/3X = 180n + (1)^n x 70 [180n + (1)^n x (alpha) is the general solution)
so, 2/3X = 70(180n + (1)^n x 70)
x = (3(70180n(1)^n x 70))/2
x = (210540n(210 x (1)^n))/2
so, x = (105270n(105x(1)^n))
But, I've obviously gone horribly wrong somewhere... 
 Follow
 55
 18052012 13:46
(Original post by bratwast)
was the question something like sin(70  2/3x) = cos20 then?
because I think I worked out sin1 of cos(20) = 70, so (alpha) is 70.
therefore, 702/3X = 180n + (1)^n x 70 [180n + (1)^n x (alpha) is the general solution)
so, 2/3X = 70(180n + (1)^n x 70)
x = (3(70180n(1)^n x 70))/2
x = (210540n(210 x (1)^n))/2
so, x = (105270n(105x(1)^n))
But, I've obviously gone horribly wrong somewhere... 
 Follow
 56
 18052012 13:47
(Original post by mathslover1)
I see what youve done, we did it this way at first too.. but our teacher showed us the way the Mark Scheme does it. 
 Follow
 57
 18052012 13:50
(Original post by bratwast)
Ahh, okay so would this be wrong? :/
Sorry 
 Follow
 58
 18052012 13:53
I still don't see what I did wrong..

 Follow
 59
 18052012 13:54
(Original post by bratwast)
I still don't see what I did wrong.. 
 Follow
 60
 18052012 13:55
(Original post by bratwast)
Ahh, okay so would this be wrong? :/
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: June 15, 2012
Share this discussion:
Tweet
Related discussions:
 Edexcel AS Mathematics C1/C2  18th/25th May 2016  Official ...
 AQA Physics PHYA5  Thursday 18th June 2015 [Exam ...
 Edexcel C1 13th May 2015 *official thread*
 2015 AS/A2 Results Day (Thursday 13th August)  Official ...
 2014 AS/A2 Results Day (Thursday 14th August) Official ...
 Ask a Current UCL Student: The Official Thread
 2014 AS and A2 Results Day (Thursday 14th August) Official ...
 The Official Cambridge Applicants Thread 2016 Entry MK II
 The Official Cambridge Offer Holders Thread 2016 Entry MK II
 Ask a Current UCL Student: The Official Thread
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by: