OCR C4 (Core Mathematics 4)

1 Express \(3 \cos \theta + 4 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence find the range of the function \(\mathbf { f } ( \theta )\), where $$f ( \theta ) = 7 + 3 \cos \theta + 4 \sin \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$ Write down the greatest possible value of \(\frac { 1 } { 7 + 3 \cos \theta + 4 \sin \theta }\).

2 Express \(3 \sin x + 2 \cos x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\).
Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = 3 \sin x + 2 \cos x , 0 \leqslant x \leqslant \pi$$

3 Show that \(\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta\).

5 Solve the equation \(2 \sin 2 \theta + \cos 2 \theta = 1\), for \(0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }\).

6 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7

1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

8 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h] \end{figure}

1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
5. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
6. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).