OCR C4 (Core Mathematics 4)

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 \(f(\theta)\), 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 \(\frac{1}{7 + 3\cos\theta + 4\sin\theta}\). [6]

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 \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 \leqslant x \leqslant \pi.$$ [7]

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

The angle \(\theta\) satisfies the equation \(\sin(\theta + 45°) = \cos\theta\).

1. Using the exact values of \(\sin 45°\) and \(\cos 45°\), show that \(\tan\theta = \sqrt{2} - 1\). [5]
2. Find the values of \(\theta\) for \(0° < \theta < 360°\). [2]

Solve the equation \(2\sin 2\theta + \cos 2\theta = 1\), for \(0° \leqslant \theta < 360°\). [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 \(0° \leqslant \theta \leqslant 360°\). [7]

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

In Fig. 6, OAB is a thin bent rod, with OA = \(a\) metres, AB = \(b\) metres and angle OAB = 120°. 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. \includegraphics{figure_6}

1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a\sin\theta + b\sin(\theta - 60°).$$ [3]
2. Hence show that \(h = (a + \frac{1}{2}b)\sin\theta - \frac{\sqrt{3}}{2}b\cos\theta\). [3]

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.

1. Show that OB is horizontal when \(\tan\theta = \frac{\sqrt{3}b}{2a + b}\). [3]

In the case when \(a = 1\) and \(b = 2\), \(h = 2\sin\theta - \sqrt{3}\cos\theta\).

1. 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 \(\theta\). [7]

1. 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 \(\pi\). [4]
2. Write down the derivative of \(\tan\theta\). Hence show that \(\int_0^{\frac{\pi}{3}} \frac{1}{(\cos\theta + \sqrt{3}\sin\theta)^2} \, d\theta = \frac{\sqrt{3}}{4}\). [4]