hatsuko
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could somebody please help with this
http://www.ocr.org.uk/Images/61382-q...hematics-3.pdf

questions 6iii), i dont have a clue where to start
and the whole of question 9
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TenOfThem
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(Original post by hatsuko)
could somebody please help with this
http://www.ocr.org.uk/Images/61382-q...hematics-3.pdf

questions 6iii), i dont have a clue where to start
Use the graph that you drew in 6(ii)
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Ranibizumab
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It's a matter of rearranging the equation. Do you know how to 'undo' the modulus function?
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TenOfThem
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(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)
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TenOfThem
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(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
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Ranibizumab
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(Original post by TenOfThem)
There is no need to "undo" the modulus function
True, but that's how I would have approached it.
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TenOfThem
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(Original post by Ranibizumab)
True, but that's how I would have approached it.

The previous part of the question suggests the method expected by the board
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hatsuko
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(Original post by TenOfThem)
Use the graph that you drew in 6(ii)
what do you mean use the graph?
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Ranibizumab
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(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.
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TenOfThem
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(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
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hatsuko
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(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)?
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TenOfThem
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(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
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hatsuko
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(Original post by TenOfThem)
yes
in that case i dont know how to simplify the mess ive got
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TenOfThem
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(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
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hatsuko
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(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
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TenOfThem
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(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
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hatsuko
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(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) ?
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TenOfThem
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(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)
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hatsuko
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(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?
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TenOfThem
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(Original post by hatsuko)
why are they in front?
I am going to guess what you mean by that

Because it is convention to put the numbers in front of the function

You would write 2sin(x) rather than sin(x)2 wouldn't you
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