Show that \(f(x)\) can be expressed as \(1 + \sin 2x\) .

The graph of f is shown below for \(0 \le x \le 2\pi \) .

Let \(g(x) = 1 + \cos x\) . On the same set of axes, sketch the graph of g for \(0 \le x \le 2\pi \) .

The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
k\\
0
\end{array}} \right)\) .

Write down the value of p and a possible value of k .

attempt to expand     (M1)

e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms

correct expansion     A1

e.g. \({\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\)

\(f(x) = 1 + \sin 2x\)     AG     N0

[2 marks]

     A1A1     N2

Note: Award A1 for correct sinusoidal shape with period \(2\pi \) and range \([0{\text{, }}2]\), A1 for minimum in circle.

\(p = 2\) , \(k = - \frac{\pi }{2}\)     A1A1     N2

[2 marks]

Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates were certainly helped by the fact that it was a "show that" question.

More candidates had difficulty with part (b) with many assuming the first graph was \(1 + \sin (x)\) and hence sketching a horizontal translation of \(\pi /2\) for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor p correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation.

Part (c) certainly served as a discriminator between the grade 6 and 7 candidates.