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C3 January 2013 25/01/2013

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Original post by AS01
1/2 for that?


as their answer
Reply 1581
Original post by Y_123456
I work out the points before plotting the graph.
For example, is P is at (4, 2), I will plot the point at (7, 4).
Because the horizontal translation is +3 to the right (X + 3), and vertical stretch is by the factor 2 (Y x 2).


Sorry, I asked the question incorrectly.

It was a different question. Thanks anyway. :smile:
Original post by TheNoobyPotato
hey guys, say we had 2f(x) + 2, since they both affect y values, which is applied first, and why?

you have to do y-coordinate * 2 first and then y-co-ordinate +4. But, i'm not sure why.
Reply 1583
Original post by otrivine
as their answer


so its not even root3/root2?
0=Theta

When doing RSin(0-Alpha) for exampe, for 2sin0-1.5cos0

How do you know which number is A and which is B?!
Original post by AS01
so its not even root3/root2?


well its not either of our answers, so i was not sure of that
Original post by TheNoobyPotato
hey guys,
say we had 2f(x) + 2, since they both affect y
values, which is applied first, and why?



Original function was f(x)

You multiplied it by 2

2f(x)

And then added 2 to the product

2f(x) + 2
Original post by Hadley112
0=Theta

When doing RSin(0-Alpha) for exampe, for 2sin0-1.5cos0

How do you know which number is A and which is B?!


What do you mean by A & B ?

Are you talking about factor formulae ? :smile:
Reply 1588
do you think they would ever ask you to draw the graphs of arcsin or arccos etc?? just noticed that section in the book...... :/
Need some help :frown: I think i'm being really silly, or my calculator is just playing up.
i'm doing Jan 2012,Q4. So im subbing in the y coordinate into the equation however i get 0.03655...
It should be 2 root 3. like on here http://www.examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=30&solution=4.1
what am I dong wrong?
I'm in degrees, but I don't know if any other set up is wrong on my calc.

Thank you in advance!
Original post by JenniS
do you think they would ever ask you to draw the graphs of arcsin or arccos etc?? just noticed that section in the book...... :/


I hope not.... >.<
Reply 1591
Original post by otrivine
well its not either of our answers, so i was not sure of that


I don't think we can get 1/2 as the answer.
Original post by AS01
I don't think we can get 1/2 as the answer.


but the mark scheme got 1/2 root 6 ?
Original post by posthumus
What do you mean by A & B ?

Are you talking about factor formulae ? :smile:


Yeah, writing it in the form RCos(0-Alpha) !
Best of luck to all of you. I'm going to keep quiet now :smile:
Reply 1595
Original post by otrivine
but the mark scheme got 1/2 root 6 ?


is root6 in the denominatior or numerator?
if root 6 is in numerator then you have to just rationalise root3/root2
Reply 1596
Original post by Nehal:)
Need some help :frown: I think i'm being really silly, or my calculator is just playing up.
i'm doing Jan 2012,Q4. So im subbing in the y coordinate into the equation however i get 0.03655...
It should be 2 root 3. like on here http://www.examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=30&solution=4.1
what am I dong wrong?
I'm in degrees, but I don't know if any other set up is wrong on my calc.

Thank you in advance!
switch to radians
Original post by Olive123
Can you explain it to me step by step I dont understand it please.


Factor formulae states:

sinA+sinB=2sin(A+B2)cos(AB2)\sin{A} + \sin{B} = 2\sin{(\frac{A + B}{2})} \cos{(\frac{A - B}{2})}

cosA+cosB=2cos(A+B2)cos(AB2)\cos{A} + \cos{B} = 2\cos{(\frac{A + B}{2})} \cos{(\frac{A - B}{2})}

sinAsinB=2cos(A+B2)sin(AB2)\sin{A} - \sin{B} = 2\cos{(\frac{A + B}{2})} \sin{(\frac{A - B}{2})}

cosAcosB=2sin(A+B2)sin(AB2)\cos{A} - \cos{B} = -2\sin{(\frac{A + B}{2})} \sin{(\frac{A - B}{2})}

so an example:

sin5x+sin3x\sin{5x} + \sin{3x}

A=5xA = 5x, B=3xB = 3x

Substiting those values in to sinA+sinB\sin{A} + \sin{B}

=2sin(5x+3x2)cos(5x3x2)= 2\sin{(\frac{5x + 3x}{2})} \cos{(\frac{5x - 3x}{2})}

=2sin4xcosx= 2\sin{4x}\cos{x}


Some Questions may in fact ask the reverse:

Example:

Express 2sin4xcosx2\sin{4x}\cos{x} in terms of sin only.

Well we know A+B2=4x\frac{A + B}{2} = 4x, AB2=x\frac{A - B}{2} = x

So A+B=8xA + B = 8x, AB=2xA - B = 2x

Adding:

2A=10x2A = 10x

A=5xA = 5x

B=8xA=8x5x=3xB = 8x - A = 8x - 5x = 3x

=sin5x+sin3x = \sin{5x} + \sin{3x}
(edited 11 years ago)
Original post by Qmakz
switch to radians


-_- so it was just me being super stupid!
THANK YOU!
Can't believe that never occurred to me :/
Original post by AS01
is root6 in the denominatior or numerator?
if root 6 is in numerator then you have to just rationalise root3/root2


No they got

1/2root6, oh so that means ours are the same

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