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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# 2003 - Paper 1 - Q.10

Started by ice_illusion, Feb 10 2006 03:58 PM

2 replies to this topic

### #1

Posted 10 February 2006 - 03:58 PM

### #2

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.

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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### #3

Posted 12 February 2006 - 02:27 PM

Aah, that makes sense. Thank you.

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