Expand and simplify \({\left( {1 - \sqrt 3 } \right)^2}\).

By writing \(15^\circ \) as \(60^\circ  - 45^\circ \) find the value of \(\cos (15^\circ )\).

Find BC in the form \(a + \sqrt b \) where \(a,{\text{ }}b \in \mathbb{Z}\).

\({\left( {1 - \sqrt 3 } \right)^2} = 4 - 2\sqrt 3 \)    A1

Note:     Award A0 for \(1 - 2\sqrt 3  + 3\).

[1 mark]

\(\cos (60^\circ  - 45^\circ ) = \cos (60^\circ )\cos (45^\circ ) + \sin (60^\circ )\sin (45^\circ )\)    M1

\( = \frac{1}{2} \times \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2}{\text{ }}\left( {{\text{or }}\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}} \right)\)    (A1)

\( = \frac{{\sqrt 2  + \sqrt 6 }}{4}{\text{ }}\left( {{\text{or }}\frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}} \right)\)    A1

[3 marks]

\(B{C^2} = 2 + 4 - 2 \times \sqrt 2  \times 2\cos (15^\circ )\)    M1

\( = 6 - \sqrt 2 \left( {\sqrt 2  + \sqrt 6 } \right)\)

\( = 4 - \sqrt {12} {\text{ }}\left( { = 4 - 2\sqrt 3 } \right)\)    A1

\(BC =  \pm \left( {1 - \sqrt 3 } \right)\)    (M1)

\(BC =  - 1 + \sqrt 3 \)    A1

Note:     Accept \(BC = \sqrt 3  - 1\).

Note:     Award M1A0 for \(1 - \sqrt 3 \).

Note:     Valid geometrical methods may be seen.

[4 marks]

The main error here was to fail to note the word ‘simplify’ in the question and some candidates wrote \(1 + 3\) in their final answer rather than 4.

This was well done by the majority of candidates, though a few wrote \(\cos (60 - 45) = \cos 60 - \cos 45\).

Candidates were able to use the cosine rule correctly but then failed to notice the result obtained was the same as that obtained in part (a).