Show that $\cos A = \frac{{12}}{{13}}$.

Find $\cos 2A$.

METHOD 1

approach involving Pythagoras’ theorem     (M1)

eg     ${5^2} + {x^2} = {13^2}$, labelling correct sides on triangle

finding third side is 12 (may be seen on diagram)     A1

$\cos A = \frac{{12}}{{13}}$     AG     N0

METHOD 2

approach involving ${\sin ^2}\theta  + {\cos ^2}\theta  = 1$     (M1)

eg     ${\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\theta  = 1,{\text{ }}{x^2} + \frac{{25}}{{169}} = 1$

correct working     A1

eg     ${\cos ^2}\theta  = \frac{{144}}{{169}}$

$\cos A = \frac{{12}}{{13}}$     AG     N0

[2 marks]

correct substitution into $\cos 2\theta$     (A1)

eg     $1 - 2{\left( {\frac{5}{{13}}} \right)^2},{\text{ }}2{\left( {\frac{{12}}{{13}}} \right)^2} - 1,{\text{ }}{\left( {\frac{{12}}{{13}}} \right)^2} - {\left( {\frac{5}{{13}}} \right)^2}$

correct working     (A1)

eg     $1 - \frac{{50}}{{169}},{\text{ }}\frac{{288}}{{169}} - 1,{\text{ }}\frac{{144}}{{169}} - \frac{{25}}{{169}}$

$\cos 2A = \frac{{119}}{{169}}$     A1     N2

[3 marks]