OCR MEI C4 (Core Mathematics 4)

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

2

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 \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$

4 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).

5 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\). Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

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