In the diagram below, AD is perpendicular to BC.

CD  =  4, BD  =  2 and AD  =  3. ${\rm{C}}\hat {\rm{A}}{\rm{D}} = \alpha$ and ${\rm{B}}\hat {\rm{A}}{\rm{D}} = \beta$ .

Find the exact value of $\cos (\alpha  - \beta )$ .

METHOD 1

${\text{AC}} = 5$ and $\text{AB} = \sqrt {13}$   (may be seen on diagram)     (A1)

$\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$     (A1)

$\cos \beta = \frac{3}{{\sqrt {13} }}$ and $\sin \beta = \frac{2}{{\sqrt {13} }}$     (A1) 

Note: If only the two cosines are correctly given award (A1)(A1)(A0).

Use of $\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$     (M1)

$= \frac{3}{5} \times \frac{3}{{\sqrt {13} }} + \frac{4}{5} \times \frac{2}{{\sqrt {13} }}$   (substituting)     M1

$= \frac{{17}}{{5\sqrt {13} }}$     $\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)$     A1     N1

[6 marks]

METHOD 2

${\text{AC}} = 5$ amd ${\text{AB}} = \sqrt {13}$   (may be seen on diagram)     (A1)

Use of $\cos (\alpha + \beta ) = \frac{{{\text{A}}{{\text{C}}^2} + {\text{A}}{{\text{B}}^2} - {\text{B}}{{\text{C}}^2}}}{{{\text{2(AC)(AB)}}}}$     (M1)

$= \frac{{25 + 13 - 36}}{{2 \times 5 \times \sqrt {13} }}\,\,\,\,\,\left( { = \frac{1}{{5\sqrt {13} }}} \right)$     A1

Use of $\cos (\alpha + \beta ) + \cos (\alpha - \beta ) = 2\cos \alpha \cos \beta$     (M1)

$\cos \alpha = \frac{3}{5}$ and $\cos \beta = \frac{3}{{\sqrt {13} }}$     (A1)

$\cos (\alpha - \beta ) = \frac{{17}}{{5\sqrt {13} }}\,\,\,\,\,\left( { = 2 \times \frac{3}{5} \times \frac{3}{{\sqrt {13} }} - \frac{1}{{5\sqrt {13} }}} \right){\text{ }}\left( { = \frac{{17\sqrt {13} }}{{65}}} \right)$     A1     N1

[6 marks]

Many candidates used a lot of space answering this question, but were generally successful. A few candidates incorrectly used the formula for the cosine of the difference of angles. An interesting alternative solution was noted, in which the side AB is reflected in AD and the required result follows from the use of the cosine rule.