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2003 - paper 1 - question 10 - HSN forum

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2003 - paper 1 - question 10


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#1 xClairex

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Posted 13 May 2007 - 02:31 PM

i don't have a clue on how to even start part (a) unsure.gif

any help would be much appreciated


#2 Azie

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Posted 13 May 2007 - 03:41 PM

Ok, for 10.a.(i). ...draw a line down from point A to the x axis an call it B so u make a triangle. This means u have 3 points, O(0,0), A (8,4) and B(8,0) ...now the distance from O to A is 8, and from A to B is 4, so u use pythagoras's thereom to work out the distance from O to A which turns about to be the square root of 80 ...then u use SOHCAHTOA to work out sin(p) which = (8/square root of 80) an cos(p) which = (4/square root of 80) ...then u use the double angle formula so that sin(2p) = 2sin(p)cos(p) ...then substitute in ur answers an u shud get 2((8/square root of 80)(4/square root of 80) which = 64/80, an then if u divide both by 16, u get 4/5

As For 10.b.(ii), im still tryin 2 work that 1 out but i'll let u no when i get it smile.gif


#3 Pete

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Posted 13 May 2007 - 03:50 PM

LaTeX-ing the above to make it a bit more readable tongue.gif :
OA^2 = 8^2 + 4^2 \\= 64 + 16 \\= 80 \\OA = \surd 80 \\ \\\sin(2p) = 2\sin(p)\cos(p) \\= 2 \times \frac{4}{\surd 80} \times \frac{8}{\surd 80} \\= 2 \times \frac{32}{80} \\= \frac{64}{80}  \\ \\= \frac{4}{5}

(Azie: you might want to read up on formatting your posts with LaTeX to make posting in here a bit easier smile.gif )

#4 Azie

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Posted 13 May 2007 - 03:54 PM

i managed 2 work out 10.a.(ii) but i got -3/5 an the answer is 3/5 but i'll tell u how i worked it out an if u spot where i've went wrong, let me no ...use the double angle formula again so that cos(2p) = 1-2sin(squared)p ...then substitute the sin(p) value from part (i) so u get 1-2(8/square root of 80)squared ...this then becums 1-2(64/80) which equals 1-2(4/5) and then 1-8/5 which after u take 5 as the common denominater, gives u -3/5 ...hope thats helped a bit.

#5 xClairex

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Posted 13 May 2007 - 04:01 PM

yeah thanks this has helped alot

i now see how to get (ii) as -3/5 but i cant see how it would be positive :S


#6 Pete

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Posted 13 May 2007 - 04:08 PM

10 a ii) should work like this:
\cos(2p) = 1 - 2\sin^2(p) \\= 1 - 2(\frac{4}{\surd 80} \times \frac{4}{\surd 80}) \\= 1 - 2(\frac{16}{80}) \\= 1 - \frac{32}{80} \\= 1 - \frac{2}{5} \\= \frac{3}{5}

Azie: your error arises with your substitution of 2\sin^2(p): your squaring is slightly off. It's a simple error, but it's what's causing all the problems smile.gif

#7 Azie

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Posted 13 May 2007 - 04:13 PM

ah, i c where i went wrong now, thanks smile.gif

#8 xClairex

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Posted 13 May 2007 - 04:19 PM

thank you pete and azie smile.gif






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