4400 Higher November 2008 Paper 4 Q22
22
Diagram NOT accurately drawn
\(ABC\) is an isosceles triangle.
\(AB = AC = 1\)
\(M\) is the midpoint of \(BC\).
(a)
(i) Use trigonometry to find an expression, in terms of \(x\), for \(BM\).
(ii) Hence write down an expression, in terms of \(x\), for \(BC\).
(2)(b) Use the cosine rule to find an expression, in terms of \(\cos(2x)\), for \(BC^2\). (1)
(c) Hence show that \(\cos(2x) = 1 - 2(\sin x)^2\) (2)
| Scheme | Marks |
|---|---|
| (i) Answer: \(\sin x\) | B1 |
| (ii) Answer: \(2\sin x\) | B1 |
| (2) |
Notes
B1: not \(\sin x = \frac{BM}{1}\) not \(x = \sin^{-1} BM\)
| Scheme | Marks |
|---|---|
| \(2 - 2\cos 2x\) | B1 |
| (1) |
Notes
B1: oe eg \(1^2 + 1^2 - 2 \times 1 \times 1 \times \cos 2x\) not ISW
| Scheme | Marks |
|---|---|
| \((2\sin x)^2 = 2 - 2\cos(2x)\) oe | M1 |
| \(2(\sin x)^2 = 1 - \cos(2x)\) \((\cos(2x) = 1 - 2(\sin x)^2)\) | A1 |
| (2) | |
| (5 marks) |
Notes
A1: or \(2\cos(2x) = 2 - 4(\sin x)^2\) not ISW