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
OFFICIAL AQA FP1 18th MAY 2012 Thread
Announcements  Posted on  

Score your uni out of 10 and help thousands of students chose the right uni  21092015  
Applying to uni? Take our 60second survey  you could bag an Amazon voucher.  19092015 


(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)? 
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 right 
(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)? 
(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. 
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 
(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 
(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  240 
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. 
(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. 
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?

(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. 
guys, x^2 + 42x +65 = 0 ? is that what you guys got?
might of been a 42 can't remember 
(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... 
(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... 
(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. 
(Original post by bratwast)
Ahh, okay so would this be wrong? :/
Sorry 
I still don't see what I did wrong..

(Original post by bratwast)
I still don't see what I did wrong.. 
(Original post by bratwast)
Ahh, okay so would this be wrong? :/
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: