Find $\cos \theta$ .

Find $\tan 2\theta$ .

METHOD 1

evidence of choosing ${\sin ^2}\theta  + {\cos ^2}\theta  = 1$     (M1)

correct working     (A1)

e.g. ${\cos ^2}\theta  = \frac{9}{{13}}$ , $\cos \theta  =  \pm \frac{3}{{\sqrt {13} }}$ , $\cos \theta  = \sqrt {\frac{9}{{13}}}$

$\cos \theta  = - \frac{3}{{\sqrt {13} }}$    A1     N2

Note: If no working shown, award N1 for $\frac{3}{{\sqrt {13} }}$ .

METHOD 2

approach involving Pythagoras’ theorem     (M1)

e.g. ${2^2} + {x^2} = 13$ ,

finding third side equals 3     (A1)

$\cos \theta  = - \frac{3}{{\sqrt {13} }}$     A1     N2

Note: If no working shown, award N1 for $\frac{3}{{\sqrt {13} }}$ .

[3 marks]

correct substitution into $\sin 2\theta$ (seen anywhere)     (A1)

e.g. $2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)$

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

e.g. ${\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}$ , $2{\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - 1$ , $1 - 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}$

valid attempt to find $\tan 2\theta$     (M1)

e.g. $\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { - \frac{3}{{\sqrt {13} }}} \right)}^2} - {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}$ , $\frac{{2\left( { - \frac{2}{3}} \right)}}{{1 - {{\left( { - \frac{2}{3}} \right)}^2}}}$

correct working     A1

e.g. $\frac{{\frac{{(2)(2)( - 3)}}{{13}}}}{{\frac{9}{{13}} - \frac{4}{{13}}}}$ , $\frac{{ - \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} - 1}}$ , $\frac{{ - \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}$

$\tan 2\theta  = - \frac{{12}}{5}$     A1     N4

Note: If students find answers for $\cos \theta$ which are not in the range $[ - 1{\text{, }}1]$, award full FT in (b) for correct FT working shown.

[5 marks]

While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.

In part (b), many candidates incorrectly tried to calculate $\tan 2\theta$ as $2 \times \tan \theta$ , rather than using the double-angle identities.