C3 paper

Original post by hatsuko

could somebody please help with this
http://www.ocr.org.uk/Images/61382-question-paper-unit-4723-core-mathematics-3.pdf

questions 6iii), i dont have a clue where to start

Use the graph that you drew in 6(ii)

Reply 2

Ranibizumab

9

It's a matter of rearranging the equation. Do you know how to 'undo' the modulus function?

Reply 3

Original post by hatsuko

and the whole of question 9

I am struggling to see any difficulty with 9

(i)
Do you know the formula for Cos(A+B)

Do you know what the Trig ratios for 30 and 60 are

(ii)
Use (i) is it not obvious what theta =

(iii) and (iv)

Use (i)

(edited 10 years ago)

Reply 4

Original post by Ranibizumab

It's a matter of rearranging the equation. Do you know how to 'undo' the modulus function?

There is no need to "undo" the modulus function

Reply 5

Ranibizumab

9

Original post by TenOfThem

There is no need to "undo" the modulus function

True, but that's how I would have approached it. :smile:

Reply 6

Original post by Ranibizumab

True, but that's how I would have approached it. :smile:

The previous part of the question suggests the method expected by the board

Reply 7

OP

Original post by TenOfThem

Use the graph that you drew in 6(ii)

what do you mean use the graph?

Reply 8

Ranibizumab

9

Original post by TenOfThem

The previous part of the question suggests the method expected by the board

Fair enough! Didn't think about the additional solutions.

Reply 9

Original post by hatsuko

what do you mean use the graph?

Draw a line going across at pi/3

One solution can be calculated and the others found using the symmetry of the graph

Reply 10

OP

Original post by TenOfThem

I am struggling to see any difficulty with 9

(i)
Do you know the formula for Cos(A+B)

Do you know what the Trig ratios for 30 and 60 are

(ii)
Use (i) is it not obvious what theta =

(iii) and (iv)

Use (i)

yes I know the formula and the trig ratio
am I just using the formula on 4cos(o+60) and cos(o+30)?

Reply 11

Original post by hatsuko

yes I know the formula and the trig ratio
am I just using the formula on 4cos(o+60) and cos(o+30)?

yes

Reply 12

OP

Original post by TenOfThem

yes

in that case i dont know how to simplify the mess ive got

Reply 13

Original post by hatsuko

in that case i dont know how to simplify the mess ive got

Can you show where you have got too

Reply 14

OP

Original post by TenOfThem

Can you show where you have got too

I got 4cos(o)cos(1/2) - sin(o)sin(1/2) + cos(o)cos((root3)/2) - sin(o)sin((root3)/2)

o = theta because i dont know how to type it

(edited 10 years ago)

Reply 15

Original post by hatsuko

I got 4cos(o)cos(1/2) - sin(o)sin(1/2) + cos(o)cos((root3)/2) - sin(o)sin((root3)/2)

You are adding instead of multiplying

And you have written cos(1/2) etc where you meant 1/2 etc

Reply 16

OP

Original post by TenOfThem

You are adding instead of multiplying

And you have written cos(1/2) etc where you meant 1/2 etc

can you simplify

4cos^2cos(1/2)cos((root3)/2) ?

Reply 17

Original post by hatsuko

can you simplify

4cos^2cos(1/2)cos((root3)/2) ?

At no point should you have cos(1/2)

You have

4(\frac{1}{2}\cos \theta - \frac{\sqrt3}{2}\sin \theta)(\frac{\sqrt3}{2}\cos \theta - \frac{1}{2}\sin \theta)

Reply 18

OP

Original post by TenOfThem

At no point should you have cos(1/2)

You have

4(\frac{1}{2}\cos \theta - \frac{\sqrt3}{2}\sin \theta)(\frac{\sqrt3}{2}\cos \theta - \frac{1}{2}\sin \theta)

why are they in front?

Reply 19