19N.1.AHL.TZ0.H_4

pestleMathematicsAAHLPaper 119N· ahl-3-10-compound-angle-identitiessource ↗

$A$ and $B$ are acute angles such that ${\text{cos}}\,A = \frac{2}{3}$ and ${\text{sin}}\,B = \frac{1}{3}$ .

Show that ${\text{cos}}\left( {2A + B} \right) = - \frac{{2\sqrt 2 }}{{27}} - \frac{{4\sqrt 5 }}{{27}}$ .

Markscheme / solution

attempt to use ${\text{cos}}\left( {2A + B} \right) = {\text{cos}}\,2A\,{\text{cos}}\,B - {\text{sin}}\,2A\,{\text{sin}}\,B$ (may be seen later) M1

attempt to use any double angle formulae (seen anywhere) M1

attempt to find either ${\text{sin}}\,A$ or ${\text{cos}}\,B$ (seen anywhere) M1

${\text{cos}}\,A = \frac{2}{3} \Rightarrow {\text{sin}}\,A\left( { = \sqrt {1 - \frac{4}{9}} } \right) = \frac{{\sqrt 5 }}{3}$ (A1)

${\text{sin}}\,B = \frac{1}{3} \Rightarrow {\text{cos}}\,B\left( { = \sqrt {1 - \frac{1}{9}} = \frac{{\sqrt 8 }}{3}} \right) = \frac{{2\sqrt 2 }}{3}$ A1

${\text{cos}}\,2A\left( { = 2\,{\text{co}}{{\text{s}}^2}\,A - 1} \right) = - \frac{1}{9}$ A1

${\text{sin}}\,2A\left( { = 2\,{\text{sin}}\,A\,{\text{cos}}\,A} \right) = \frac{{4\sqrt 5 }}{9}$ A1

So ${\text{cos}}\left( {2A + B} \right) = \left( { - \frac{1}{9}} \right)\left( {\frac{{2\sqrt 2 }}{3}} \right) - \left( {\frac{{4\sqrt 5 }}{9}} \right)\left( {\frac{1}{3}} \right)$

$= - \frac{{2\sqrt 2 }}{{27}} - \frac{{4\sqrt 5 }}{{27}}$ AG

[7 marks]

Examiners’ report

[N/A]