(a)     Prove the trigonometric identity $\sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y$.

(b)     Given $f(x) = \sin({x + \frac{\pi }{6}})\sin({x - \frac{\pi }{6}}),{\text{ }}x \in \left[ {0,{\text{ }}\pi } \right]$, find the range of $f$.

(c)     Given $g(x) = \csc( {x + \frac{\pi }{6}})\csc( {x - \frac{\pi }{6}}),{\text{ }}x \in \left[ {0,{\text{ }}\pi } \right],{\text{ }}x \ne \frac{\pi }{6},{\text{ }}x \ne \frac{{5\pi }}{6}$, find the range of $g$.

(a)     $\sin (x + y)\sin (x - y)$

$= (\sin x\cos y + \cos x\sin y)(\sin x\cos y - \cos x\sin y)$     M1A1

$= {\sin ^2}x{\cos ^2}y + \sin x\sin y\cos x\cos y - \sin x\sin y\cos x\cos y - {\cos ^2}x{\sin ^2}y$

$= {\sin ^2}x{\cos ^2}y - {\cos ^2}x{\sin ^2}y$     A1

$= {\sin ^2}x(1 - {\sin ^2}y) - {\sin ^2}y(1 - {\sin ^2}x)$     A1

$= {\sin ^2}x - {\sin ^2}x{\sin ^2}y - {\sin ^2}y + {\sin ^2}x{\sin ^2}y$

$= {\sin ^2}x - {\sin ^2}y$     AG

[4 marks]

(b)     $f(x) = {\sin ^2}x - \frac{1}{4}$

range is $f \in \left[ { - \frac{1}{4},{\text{ }}\frac{3}{4}} \right]$     A1A1

Note:     Award A1 for each end point. Condone incorrect brackets.

[2 marks]

(c)     $g(x) = \frac{1}{{{{\sin }^2}x - \frac{1}{4}}}$

range is $g \in \left] { - \infty ,{\text{ }} - 4} \right] \cup \left[ {\frac{4}{3},{\text{ }}\infty } \right[$     A1A1

Note:     Award A1 for each part of range. Condone incorrect brackets.

[2 marks]

Total [8 marks]

Part a) often proved to be an easy 4 marks for candidates. A number were surprisingly content to gain the first 3 marks but were unable to make the final step by substituting $1 - {\sin ^2}y$ for ${\cos ^2}y$.

Parts b) and c) were more often than not, problematic. Some puzzling ‘working’ was often seen, with candidates making little headway. Otherwise good candidates were able to answer part b), though correct solutions for c) were a rarity. The range $g \in \left[ { - 4,{\text{ }}\frac{4}{3}} \right]$ was sometimes seen, but gained no marks.