6 The function \(\mathrm { f } ( x ) = \frac { \sin x } { 2 - \cos x }\) has domain \(- \pi \leqslant x \leqslant \pi\).
Fig. 8 shows the graph of \(y = \mathrm { f } ( x )\) for \(0 \leqslant x \leqslant \pi\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bce065bf-a56c-4686-8fa7-cb18cb95012e-3_557_844_602_646}
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\caption{Fig. 8}
\end{figure}
- Find \(\mathrm { f } ( - x )\) in terms of \(\mathrm { f } ( x )\). Hence sketch the graph of \(y = \mathrm { f } ( x )\) for the complete domain \(- \pi \leqslant x \leqslant \pi\).
- Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 \cos x - 1 } { ( 2 - \cos x ) ^ { 2 } }\). Hence find the exact coordinates of the turning point P .
State the range of the function \(\mathrm { f } ( x )\), giving your answer exactly.
- Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int _ { 0 } ^ { \pi } \frac { \sin x } { 2 \cos x } \mathrm {~d} x\).
- Sketch the graph of \(y = \mathrm { f } ( 2 x )\).
- Using your answers to parts (iii) and (iv), write down the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 2 \cos 2 x } \mathrm {~d} x\).