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2003 - Paper 1 - Q.10 - HSN forum # 2003 - Paper 1 - Q.10

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### #1ice_illusion

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Posted 10 February 2006 - 03:58 PM

I'm having problems with question 10 in the 2003 Non calc paper. I had a look at some of the sites that give answers and can see where they are coming from in manipulating the sin(2p) etc but I can't work out how they actually got p in the first place.
So how can you find p?

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Posted 12 February 2006 - 09:48 AM

A is the point (8.4), so if you make a triangle between the origin, A and the x axis (call this point B) you'll have the points O(0,0), A(8,4) and B(8,0).

Now go for the distance formula to get the lengths of the side OA ( 80). The side OB will be 8 units and the side AB will be 4 units.

Looking at the diagram shows what angle p is, now you should know that sine of an angle is opp/hyp and cosine is adj/hyp. So sin(p) = 4/ 80 and cos(p) = 8/ 80.

Using double angle formula, sin(2p) = 2sin(p)cos(p)

Substituting in the values obtained from the triangle, sin(2p) = 2 * 4/ 80 * 8/ 80. Mutliply all the numerators and all the denominators together and cancel down and you should get the correct answer.
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### #3ice_illusion

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Posted 12 February 2006 - 02:27 PM

Aah, that makes sense. Thank you.

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