Given that $\frac{\pi }{2} < \alpha  < \pi$ and $\cos \alpha  = - \frac{3}{4}$, find the value of sin 2α .

$\sin \alpha  = \sqrt {1 - {{\left( { - \frac{3}{4}} \right)}^2}}  = \frac{{\sqrt 7 }}{4}$     (M1)A1

attempt to use double angle formula     M1

$\sin 2\alpha  = 2\frac{{\sqrt 7 }}{4}\left( { - \frac{3}{4}} \right) = - \frac{{3\sqrt 7 }}{8}$     A1

Note: $\frac{{\sqrt 7 }}{4}$ seen would normally be awarded M1A1.

[4 marks]

Many candidates scored full marks on this question, though their explanations for part a) often lacked clarity. Most preferred to use some kind of right-angled triangle rather than (perhaps in this case) the more sensible identity ${\sin ^2}\alpha  + {\cos ^2}\alpha  = 1$.