The Student Room Group

Edexcel C3,C4 June 2013 Thread

Scroll to see replies

Original post by masryboy94
thank you both !!! now i get it :biggrin:



Original post by gaffer dean
Are you sure? It's important to Remember these identities:
sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1), cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos2x1cos4x=2cos^2x-1


Yeah definitely remember those identities, or at least remember how to derive them from sin2x+cos2x=1 sin^2x + cos^2x = 1 and cos2x=cos2xsin2x cos2x = cos^2x - sin^2x
I'd recommend writing them down at the side of any working out so that you are sure of the trig identities that you use, and to possibly remind yourself that you can use them if you get stuck in a question
Original post by gaffer dean
It's important to Remember these identities:
sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1), cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos2x1cos4x=2cos^2x-1


how about when it is 2cos22x12cos^2 2x-1 because i've seen it in a couple of questions and i didn't know which identity matched with that?
Original post by gaffer dean
It's important to Remember these identities:
sin2x=12(cos2x1)sin^2x=\frac{-1}{2}(cos2x-1), cos2x=12(cos2x+1)cos^2x= \frac{1}{2}(cos2x+1), cos4x=2cos2x1cos4x=2cos^2x-1


btw isn't cos2x=2cos2x1cos2x=2cos^2x-1 because you wrote cos4x=2cos2x1cos4x=2cos^2x-1 ??
Original post by masryboy94
how about when it is 2cos22x12cos^2 2x-1 because i've seen it in a couple of questions and i didn't know which identity matched with that?


Original post by masryboy94
btw isn't cos2x=2cos2x1cos2x=2cos^2x-1 because you wrote cos4x=2cos2x1cos4x=2cos^2x-1 ??

I've edited my post now, Sorry I was supposed to say: 2cos22x1=cos4x2cos^2 2x-1=cos4x
Original post by gaffer dean
I've edited my post now, Sorry I was supposed to say: 2cos22x1=cos4x2cos^2 2x-1=cos4x


ooo okay, yep that answers my question. THANK YOU ! :biggrin:
Original post by gaffer dean
I've edited my post now, Sorry I was supposed to say: 2cos22x1=cos4x2cos^2 2x-1=cos4x


is that the same for when 12sin22x=cos4x1-2sin^2 2x=cos4x ?
Original post by masryboy94
is that the same for when 12sin22x=cos4x1-2sin^2 2x=cos4x ?

yes because 2cos22x1=cos4x2cos^2 2x-1=cos4x can be written as 2(1sin22x)12(1-sin^22x)-1 (Since cos22x=1sin22xcos^22x=1-sin^22x) = 22sin22x1=12sin22x2-2sin^22x-1=1-2sin^22x
(edited 10 years ago)
Original post by gaffer dean
yes because 2cos22x1=cos4x2cos^2 2x-1=cos4x can be written as 2(1sin22x)12(1-sin^22x)-1 (Since cos22x=1sin22xcos^22x=1-sin^22x) = 22sin22x1=12sin22x2-2sin^22x-1=1-2sin^22x


ahh brilliant thank you very much !
6x + 9y -2pie = 0 this is mark scheme answer

i got

-6x - 9y + 2pie =0

I mean the only difference is that they transferred everything to the other side, to make positive but I did not? will I still get the mark
Original post by masryboy94
ahh brilliant thank you very much !

No Problem. :smile:
Original post by gaffer dean
yeah you would still get the marks.


but when I sub x=5 and y=9 for example
I get postitive value and the other one negatvie value
Original post by otrivine
but when I sub x=5 and y=9 for example
I get postitive value and the other one negatvie value

wait what's the question?
Original post by gaffer dean
wait what's the question?


the question was on jan 2010 question 3)c)
Original post by otrivine
6x + 9y -2pie = 0 this is mark scheme answer

i got

-6x - 9y + 2pie =0

I mean the only difference is that they transferred everything to the other side, to make positive but I did not? will I still get the mark


Yes, it's the same thing.
Original post by justinawe
Yes, it's the same thing.


so why when I sub x=5 and y=9 on to either equation I get one being 104.something and the other being -104.something
Original post by otrivine
so why when I sub x=5 and y=9 on to either equation I get one being 104.something and the other being -104.something


What?

They're both supposed to equal 0... they can't equal 104 or -104. Why did you sub in x=5 and y=9?

6x+9y2π=(6x9y+2π)6x + 9y -2\pi = -(-6x - 9y + 2\pi), this is why you get the positive for one and negative for the other.

However, the equation of the line is 6x+9y2π=06x + 9y -2\pi = 0 . This means it can only equal 0, it can't equal 104 or -104 or whatever.

see here,

6x+9y2π=06x + 9y -2\pi = 0

(6x9y+2π)=0-(-6x - 9y + 2\pi) = 0

6x9y+2π=0-6x - 9y + 2\pi = -0

but "negative zero" is still zero, so:

6x9y+2π=0-6x - 9y + 2\pi = 0
Original post by justinawe
What?

They're both supposed to equal 0... they can't equal 104 or -104. Why did you sub in x=5 and y=9?

6x+9y2π=(6x9y+2π)6x + 9y -2\pi = -(-6x - 9y + 2\pi), this is why you get the positive for one and negative for the other.

However, the equation of the line is 6x+9y2π=06x + 9y -2\pi = 0 . This means it can only equal 0, it can't equal 104 or -104 or whatever.

see here,

6x+9y2π=06x + 9y -2\pi = 0


(6x9y+2π)=0-(-6x - 9y + 2\pi) = 0

6x9y+2π=0-6x - 9y + 2\pi = -0


but "negative zero" is still zero, so:

6x9y+2π=0-6x - 9y + 2\pi = 0

So would I get full marks
Original post by otrivine
So would I get full marks


yes
For C3 and C4, Here's a good way of remembering differential and integral of sinx, cosx, -sinx and -cosx (these will not be given in the formula booklet):
(edited 10 years ago)
anyone know where I can get some differential questions in context questions? done the solomon ones, but would like some more -hate them aha!

Quick Reply