1.05l Double angle formulae: and compound angle formulae

575 questions

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OCR C3 2013 January Q9
10 marks Standard +0.8
  1. Prove that $$\cos^2(\theta + 45°) - \frac{1}{2}(\cos 2\theta - \sin 2\theta) \equiv \sin^2 \theta.$$ [4]
  2. Hence solve the equation $$6\cos^2(\frac{1}{3}\theta + 45°) - 3(\cos \theta - \sin \theta) = 2$$ for \(-90° < \theta < 90°\). [3]
  3. It is given that there are two values of \(\theta\), where \(-90° < \theta < 90°\), satisfying the equation $$6\cos^2(\frac{1}{3}\theta + 45°) - 3(\cos \frac{2}{3}\theta - \sin \frac{2}{3}\theta) = k,$$ where \(k\) is a constant. Find the set of possible values of \(k\). [3]
OCR C3 2009 June Q3
6 marks Standard +0.3
The angles \(\alpha\) and \(\beta\) are such that $$\tan \alpha = m + 2 \quad \text{and} \quad \tan \beta = m,$$ where \(m\) is a constant.
  1. Given that \(\sec^2 \alpha - \sec^2 \beta = 16\), find the value of \(m\). [3]
  2. Hence find the exact value of \(\tan(\alpha + \beta)\). [3]
OCR C3 2010 June Q3
6 marks Standard +0.3
  1. Express the equation \(\cosec \theta(3 \cos 2\theta + 7) + 11 = 0\) in the form \(a \sin^2 \theta + b \sin \theta + c = 0\), where \(a\), \(b\) and \(c\) are constants. [3]
  2. Hence solve, for \(-180° < \theta < 180°\), the equation \(\cosec \theta(3 \cos 2\theta + 7) + 11 = 0\). [3]
Edexcel C3 Q1
8 marks Standard +0.3
  1. Find the exact value of \(x\) such that $$3 \arctan (x - 2) + \pi = 0.$$ [3]
  2. Solve, for \(-\pi < \theta < \pi\), the equation $$\cos 2\theta - \sin \theta - 1 = 0,$$ giving your answers in terms of \(\pi\). [5]
Edexcel C3 Q5
9 marks Challenging +1.2
Find the values of \(x\) in the interval \(-180 < x < 180\) for which $$\tan (x + 45)^{\circ} - \tan x^{\circ} = 4,$$ giving your answers to 1 decimal place. [9]
Edexcel C3 Q7
12 marks Standard +0.3
  1. Use the identity $$\cos (A + B) = \cos A \cos B - \sin A \sin B$$ to prove that $$\cos x \equiv 1 - 2 \sin^2 \frac{x}{2}.$$ [3]
  2. Prove that, for \(\sin x \neq 0\), $$\frac{1 - \cos x}{\sin x} \equiv \tan \frac{x}{2}.$$ [3]
  3. Find the values of \(x\) in the interval \(0 \leq x \leq 360^{\circ}\) for which $$\frac{1 - \cos x}{\sin x} = 2 \sec^2 \frac{x}{2} - 5,$$ giving your answers to 1 decimal place where appropriate. [6]
OCR C3 Q3
7 marks Standard +0.8
  1. Prove the identity $$\sqrt{2} \cos (x + 45)° + 2 \cos (x - 30)° \equiv (1 + \sqrt{3}) \cos x°.$$ [4]
  2. Hence, find the exact value of \(\cos 75°\) in terms of surds. [3]
AQA C4 2010 June Q5
11 marks Standard +0.3
    1. Show that the equation \(3\cos 2x + 2\sin x + 1 = 0\) can be written in the form $$3\sin^2 x - \sin x - 2 = 0$$ [3 marks]
    2. Hence, given that \(3\cos 2x + 2\sin x + 1 = 0\), find the possible values of \(\sin x\). [2 marks]
    1. Express \(3\cos 2x + 2\sin 2x\) in the form \(R\cos(2x - \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\), giving \(\alpha\) to the nearest \(0.1°\). [3 marks]
    2. Hence solve the equation $$3\cos 2x + 2\sin 2x + 1 = 0$$ for all solutions in the interval \(0° < x < 180°\), giving \(x\) to the nearest \(0.1°\). [3 marks]
AQA C4 2016 June Q2
5 marks Standard +0.3
By forming and solving a suitable quadratic equation, find the solutions of the equation $$3 \cos 2\theta - 5 \cos \theta + 2 = 0$$ in the interval \(0° < \theta < 360°\), giving your answers to the nearest \(0.1°\). [5 marks]
AQA C4 2016 June Q5
10 marks Standard +0.3
It is given that \(\sin A = \frac{\sqrt{5}}{3}\) and \(\sin B = \frac{1}{\sqrt{5}}\), where the angles \(A\) and \(B\) are both acute.
    1. Show that the exact value of \(\cos B = \frac{2}{\sqrt{5}}\). [1 mark]
    2. Hence show that the exact value of \(\sin 2B\) is \(\frac{4}{5}\). [2 marks]
    1. Show that the exact value of \(\sin(A - B)\) can be written as \(p(5 - \sqrt{5})\), where \(p\) is a rational number. [4 marks]
    2. Find the exact value of \(\cos(A - B)\) in the form \(r + s\sqrt{5}\), where \(r\) and \(s\) are rational numbers. [3 marks]
Edexcel C4 Q3
14 marks Standard +0.3
  1. Use the identity for \(\cos(A + B)\) to prove that \(\cos 2A = 2\cos^2 A - 1\). [2]
  2. Use the substitution \(x = 2\sqrt{2} \sin \theta\) to prove that $$\int_2^{\sqrt{6}} \sqrt{(8 - x^2)} \, dx = \frac{1}{3}(\pi + 3\sqrt{3} - 6).$$ [7]
A curve is given by the parametric equations $$x = \sec \theta, \quad y = \ln(1 + \cos 2\theta), \quad 0 \leq \theta < \frac{\pi}{2}.$$
  1. Find an equation of the tangent to the curve at the point where \(\theta = \frac{\pi}{3}\). [5]
Edexcel C4 Q1
8 marks Moderate -0.3
  1. Express \(1.5 \sin 2x + 2 \cos 2x\) in the form \(R \sin (2x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{1}{2}\pi\), giving your values of \(R\) and \(\alpha\) to 3 decimal places where appropriate. [4]
  2. Express \(3 \sin x \cos x + 4 \cos^2 x\) in the form \(a \cos 2x + b \sin 2x + c\), where \(a\), \(b\) and \(c\) are constants to be found. [2]
  3. Hence, using your answer to part (a), deduce the maximum value of \(3 \sin x \cos x + 4 \cos^2 x\). [2]
OCR C4 2006 June Q8
9 marks Standard +0.8
  1. Show that \(\int \cos^2 6x dx = \frac{1}{2}x + \frac{1}{24}\sin 12x + c\). [3]
  2. Hence find the exact value of \(\int_0^{\frac{\pi}{12}} x\cos^2 6x dx\). [6]
OCR MEI C4 2009 June Q8
19 marks Standard +0.8
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 O. 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. \includegraphics{figure_8.1}
    1. Show that AB = \(2\sin 15°\). [2]
    2. Use a double angle formula to express \(\cos 30°\) in terms of \(\sin 15°\). Using the exact value of \(\cos 30°\), show that \(\sin 15° = \frac{1}{4}\sqrt{2 - \sqrt{3}}\). [4]
    3. Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6\sqrt{2 - \sqrt{3}}\). [2]
  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. \includegraphics{figure_8.2}
    1. Show that DE = \(2\tan 15°\). [2]
    2. Let \(t = \tan 15°\). Use a double angle formula to express \(\tan 30°\) in terms of \(t\). Hence show that \(t^2 + 2\sqrt{3}t - 1 = 0\). [3]
    3. Solve this equation, and hence show that \(\pi < 12(2 - \sqrt{3})\). [4]
  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. [2]
OCR MEI C4 2012 June Q5
6 marks Moderate -0.3
Given the equation \(\sin(x + 45°) = 2\cos x\), show that \(\sin x + \cos x = 2\sqrt{2}\cos x\). Hence solve, correct to 2 decimal places, the equation for \(0° < x < 360°\). [6]
OCR MEI C4 2013 June Q3
7 marks Moderate -0.8
Using appropriate right-angled triangles, show that \(\tan 45° = 1\) and \(\tan 30° = \frac{1}{\sqrt{3}}\). Hence show that \(\tan 75° = 2 + \sqrt{3}\). [7]
OCR MEI C4 2013 June Q6
18 marks Standard +0.3
The motion of a particle is modelled by the differential equation $$v \frac{dv}{dt} + 4x = 0,$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).
  1. Solve the differential equation to show that \(v^2 = 20 - 4x^2\). [4]
Now consider motion for which \(x = \cos 2t + 2 \sin 2t\), where \(x\) is the displacement from a fixed point at time \(t\).
  1. Verify that, when \(t = 0\), \(x = 1\). Use the fact that \(v = \frac{dx}{dt}\) to verify that when \(t = 0\), \(v = 4\). [4]
  2. Express \(x\) in the form \(R \cos(2t - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v^2 = 20 - 4x^2\). [7]
  3. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value. [3]
OCR MEI C4 2014 June Q4
8 marks Moderate -0.3
  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]
Edexcel C4 Q3
10 marks Moderate -0.3
  1. Use the substitution \(u = 2 - x^2\) to find $$\int \frac{x}{2 - x^2} \, dx.$$ [4]
  2. Evaluate $$\int_0^{\frac{1}{4}} \sin 3x \cos x \, dx.$$ [6]
OCR C4 Q3
3 marks Moderate -0.8
Show that \(\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta\). [3]
OCR C4 Q4
7 marks Moderate -0.3
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]
OCR C4 Q5
6 marks Standard +0.3
Solve the equation \(2\sin 2\theta + \cos 2\theta = 1\), for \(0° \leqslant \theta < 360°\). [6]
OCR C4 Q6
7 marks Standard +0.3
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]
OCR C4 Q7
8 marks Standard +0.3
  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]
OCR MEI C4 Q2
19 marks Standard +0.3
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 O. 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. \includegraphics{figure_1}
    1. Show that \(\text{AB} = 2 \sin 15°\). [2]
    2. Use a double angle formula to express \(\cos 30°\) in terms of \(\sin 15°\). Using the exact value of \(\cos 30°\), show that \(\sin 15° = \frac{1}{2}\sqrt{2 - \sqrt{3}}\). [4]
    3. Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6\sqrt{2 - \sqrt{3}}\). [2]
  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. \includegraphics{figure_2}
    1. Show that \(\text{DE} = 2 \tan 15°\). [2]
    2. Let \(t = \tan 15°\). Use a double angle formula to express \(\tan 30°\) in terms of \(t\). Hence show that \(t^2 + 2\sqrt{3}t - 1 = 0\). [3]
    3. Solve this equation, and hence show that \(\pi < 12(2 - \sqrt{3})\). [4]
  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. [2]