OCR C4 (Core Mathematics 4)

1 Express \(2 \sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Hence write down the greatest and least possible values of \(1 + 2 \sin \theta - 3 \cos \theta\).

2 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < { } _ { 2 } ^ { 1 } \pi\).-
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).

3 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).

1. Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_457_422_457_936} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
   \end{figure} (A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
   (B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
   (C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\).
2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure} (A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
   (B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\). Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
   (C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\).
3. Use the results in parts (i)( \(C\) ) and (ii)( \(C\) ) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form.

4 Solve the equation \(\cos 2 \theta = \sin \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\), giving your answers in terms of \(\pi\).

5 Express \(\sqrt { 3 } \sin x - \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Express \(\alpha\) in the form \(k \pi\). Find the exact coordinates of the maximum point of the curve \(y = \sqrt { 3 } \sin x - \cos x\) for which \(0 < x < 2 \pi\).

6 Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7 Fig. 1 shows part of the graph of \(y = \sin x \quad \sqrt { 3 } \cos x\). \begin{figure}[h] \end{figure} Express \(\quad \sqrt { } \quad\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 \leqslant \alpha \leqslant \frac { 1 } { 2 } \pi\).
Hence write down the exact coordinates of the turning point P .