The Student Room Group

OFFICIAL AQA FP1 18th MAY 2012 Thread

Scroll to see replies

Reply 40
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


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
Reply 41
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


You'll have done fine :P The grade boundaries won't be 67/75 for an A like they were in January :biggrin:

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)?
Reply 42
The general solution question, you had to do sin-1 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
Reply 43
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 :biggrin:

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)?


I got that :smile:
Reply 44
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(theta-90), 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 took arcsin of both sides, arcsin(cos20) = 70. Then found the general solution in the normal way.

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.
Reply 45
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
(edited 11 years ago)
Reply 46
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


I got the same answer as well:biggrin: But have not idea how to do the matrix transformation (a) and b(i),(IV)
Reply 47
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


got the same Z, and matrix same thing, didn't do n, but other stuff I am pretty sure I got right, I remember getting 13, it was 65/5 I think to get 13 for the product, I think. and don't remember rest.
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
(edited 11 years ago)
Reply 48
wait what.. For the general solution question I thought sin(theta)'s general solution was "180n + (-1)^n x (alpha).

So doing sin-1 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.
Reply 49
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 sin-1 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.


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.
Reply 50
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?
Reply 51
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.


Yes, alpha is still the same angle, but cos(a) doesn't have the same value as sin(a) (usually).
guys, x^2 + 42x +65 = 0 ? is that what you guys got?
might of been a -42 can't remember
(edited 11 years ago)
Reply 53
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 sin-1 of cos(20) = 70, so (alpha) is 70.

therefore, 70-2/3X = 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(70-180n-(-1)^n x 70))/2

x = (210-540n-(210 x (-1)^n))/2

so, x = (105-270n-(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 sin-1 of cos(20) = 70, so (alpha) is 70.

therefore, 70-2/3X = 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(70-180n-(-1)^n x 70))/2

x = (210-540n-(210 x (-1)^n))/2

so, x = (105-270n-(105x(-1)^n))


But, I've obviously gone horribly wrong somewhere...


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.
Reply 55
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.


Ahh, okay so would this be wrong? :/
Original post by bratwast
Ahh, okay so would this be wrong? :/


yeah, well i dont think you will get all the marks? maybe 1 or 2, 3 if your pushing it?

Sorry :frown:
Reply 57
I still don't see what I did wrong..
Original post by bratwast
I still don't see what I did wrong..

check the Mark Scheme for other papers for sin general solution question
Reply 59
Original post by bratwast
Ahh, okay so would this be wrong? :/


Mate you can use either your method or the 360degree + alpha one. Both gives the right answer, just different way of showing it. You won't get mark off for using different general solutions.

Quick Reply

Latest