C3 January 2013 25/01/2013

Original post by suncake

That seems vaguely familiar but I can't remember it. :frown:

January 2006, question 4b. Here you go.

ahhh thank you :smile:

usually I get up to 1/something and then cant do it D:

Reply 781

Original post by suncake

That seems vaguely familiar but I can't remember it. :frown:

January 2006, question 4b. Here you go.

is there a mark scheme?

Reply 782

Original post by MedMed12

is there a mark scheme?

https://docs.google.com/viewer?a=v&q=cache:TteG7qF0OLkJ:www.edexcel.com/migrationdocuments/QP%2520GCE%2520Curriculum%25202000/Maths/C3_January_2006_Mark_Scheme.doc+&hl=en&gl=uk&pid=bl&srcid=ADGEESj9v4ZnfMPullGnFiGL0lFLWf5jNbtHjUFdTxKemoOMGaCEP1phDIKTBnH7tGFPjebmVQZ7u28kXUqnGOMYF0XCakYaDgFpE2XvTc_9Pok_fLJjF8mWOzJQOXPahWO3HezuWE0R&sig=AHIEtbTQyyC5tOpReur1l4rh_GB9eT30dA
Click "download original" so you can actually read it ahah
Did you manage 4b?

Reply 783

Original post by suncake

https://docs.google.com/viewer?a=v&q=cache:TteG7qF0OLkJ:www.edexcel.com/migrationdocuments/QP%2520GCE%2520Curriculum%25202000/Maths/C3_January_2006_Mark_Scheme.doc+&hl=en&gl=uk&pid=bl&srcid=ADGEESj9v4ZnfMPullGnFiGL0lFLWf5jNbtHjUFdTxKemoOMGaCEP1phDIKTBnH7tGFPjebmVQZ7u28kXUqnGOMYF0XCakYaDgFpE2XvTc_9Pok_fLJjF8mWOzJQOXPahWO3HezuWE0R&sig=AHIEtbTQyyC5tOpReur1l4rh_GB9eT30dA
Click "download original" so you can actually read it ahah
Did you manage 4b?

oh i am doing the whole paper :tongue:

Reply 784

posthumus

20

Original post by suncake

That seems vaguely familiar but I can't remember it. :frown:

January 2006, question 4b. Here you go.

You can use the expression

cos(x) = root( 1-sin^2(x))

[rearranging]

once you have done that, you will get something like :

1/8 x root[1-sin^2(2y+6)]

Now sub in: 2y+6 = arcsin(x/4)

I believe when you do that... the arcsin and sin cancel out :smile:

Then you can simplify from there

Reply 785

Original post by posthumus

You can use the expression

cos(x) = root( 1-sin^2(x))

[rearranging]

once you have done that, you will get something like :

1/8 x root[1-sin^2(2y+6)]

Now sub in: 2y+6 = arcsin(x/4)

I believe when you do that... the arcsin and sin cancel out :smile:

Then you can simplify from there

Ahh I see! Thanks :smile:

I read the examiner's report, and they accepted the step before that for full marks, but it was just really bugging me :tongue:

Reply 786

Original post by suncake

Ahh I see! Thanks :smile:

I read the examiner's report, and they accepted the step before that for full marks, but it was just really bugging me :tongue:

want to explain the bit after 1/8cos(2y+6) to me :tongue:

Reply 787

Original post by MedMed12

want to explain the bit after 1/8cos(2y+6) to me :tongue:

Rearrange x = 4sin(2y+6), so...
x/4 = sin(2y+6)
arcsin(x/4) = 2y+6
Then just sub that in :biggrin:

Those questions are so fiddly aha, I bet they throw one in this time, just to be difficult :unimpressed:

Reply 788

gaffer dean

hi,
can someone please explain to me how to do question 8c from jan 2012 paper?
I don't have a clue how to go about solving it. :s-smilie:

Thanks in advance,

(edited 11 years ago)

Reply 789

Original post by suncake

Rearrange x = 4sin(2y+6), so...
x/4 = sin(2y+6)
arcsin(x/4) = 2y+6
Then just sub that in :biggrin:

Those questions are so fiddly aha, I bet they throw one in this time, just to be difficult :unimpressed:

thank you

how did you find the june 2012 paper? :smile:

Reply 790

posthumus

20

Original post by suncake

Ahh I see! Thanks :smile:

I read the examiner's report, and they accepted the step before that for full marks, but it was just really bugging me :tongue:

Ah that's good !

I can't imagine anyone being able to think of that and work it out in an actual exam :tongue:

Reply 791

franzk

OP

11

Original post by gaffer dean

hi,
can someone please explain to me how to do question 8c on january 2012 paper?
I don't have a clue how to go about solving it. :s-smilie:

Thanks in advance,

its pretty much solving this

Reply 792

Original post by suncake

Rearrange x = 4sin(2y+6), so...
x/4 = sin(2y+6)
arcsin(x/4) = 2y+6
Then just sub that in :biggrin:

Those questions are so fiddly aha, I bet they throw one in this time, just to be difficult :unimpressed:

hey suncake,
are there any other questions like this in any past papers that youve come across? x

Reply 793

gaffer dean

Original post by franzk

its pretty much solving this

that is the same as b which was 1+root3 tan theta/ root 3 - tan theta
so now what do i do?

Reply 794

amarjit700

Original post by tr12

how difficult is solomon trig compared to normal edexcel c3 trig? I'm finding some of the solomon trig quite difficult :s-smilie:

I found solomon trig more simpler than Edexcel :s-smilie:

?

Reply 795

username994051

1

Original post by suncake

That seems vaguely familiar but I can't remember it. :frown:

January 2006, question 4b. Here you go.

hey for the same paper jan 06 how do you get the range for question 8 part c??

Reply 796

gaffer dean

can someone please explain to me how to do question 8c from jan 2012 paper?

(edited 11 years ago)

Reply 797

Original post by MedMed12

thank you

how did you find the june 2012 paper? :smile:

Original post by MedMed12

hey suncake,
are there any other questions like this in any past papers that youve come across? x

I'm sure there was actually one like that in the June 2012 one, if you can't find it online I'll take a photo? I actually sat that paper last year and got an A, aiming for a A* though and want more leeway with C4, so I'm resitting :redface:

Original post by parishah

hey for the same paper jan 06 how do you get the range for question 8 part c??

Did you draw the graph in 8b? Range is just which y values the graph can take, so see where there's an asymptote on the graph. gf(x) has to be above that, so gf(x) > 0 :smile:

Reply 798

Original post by suncake

I'm sure there was actually one like that in the June 2012 one, if you can't find it online I'll take a photo? I actually sat that paper last year and got an A, aiming for a A* though and want more leeway with C4, so I'm resitting :redface:

Did you draw the graph in 8b? Range is just which y values the graph can take, so see where there's an asymptote on the graph. gf(x) has to be above that, so gf(x) > 0 :smile:

ahh i thought it was a nice paper :smile:

Reply 799