1.05o Trigonometric equations: solve in given intervals

1022 questions

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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]
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 Q1
6 marks Moderate -0.3
Given that \(\cosec^2 \theta - \cot \theta = 3\), show that \(\cot^2 \theta - \cot \theta - 2 = 0\). Hence solve the equation \(\cosec^2 \theta - \cot \theta = 3\) for \(0° \leqslant \theta \leqslant 180°\). [6]
OCR MEI C4 Q3
7 marks Moderate -0.3
Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin (\theta - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and \(0° < \alpha < 90°\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0° \leqslant \theta \leqslant 360°\). [7]
OCR MEI C4 Q5
7 marks Standard +0.3
Show that \(\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}\). Hence solve the equation $$\cot 2\theta = 1 + \tan \theta \quad \text{for } 0° < \theta < 360°.$$ [7]
OCR MEI M1 Q1
18 marks Moderate -0.3
Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is \(40\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal. \includegraphics{figure_1}
  1. Show that, according to the standard projectile model in which air resistance is neglected, the flight time, \(T\) s, and the range, \(R\) m, are given by $$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6] A company is designing a new type of ball and wants to model its flight.
  2. Initially the company uses the standard projectile model. Use this model to show that when \(\alpha = 30°\) and the initial speed is \(40\text{ms}^{-1}\), \(T\) is approximately \(4.08\) and \(R\) is approximately \(141.4\). Find the values of \(T\) and \(R\) when \(\alpha = 45°\). [3] The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at \(40\text{ms}^{-1}\). When the angle of projection is set at \(30°\), the range is found to be \(125\) m.
  3. Comment briefly on the accuracy of the standard projectile model in this situation. [1] The company refines the model by assuming that the ball has a constant deceleration of \(2\text{ms}^{-2}\) in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still \(4.08\) s when the angle of projection is \(30°\).
  4. Using the new model, with \(\alpha = 30°\), show that the horizontal displacement from the point of projection, \(x\) m at time \(t\) s, is given by $$x = 40t\cos 30° - t^2.$$ Find the range and hence show that this new model is reasonably accurate in this case. [4] The company then sets the angle of projection to \(45°\) while retaining a projection speed of \(40\text{ms}^{-1}\). With this setting the range of the ball is found to be \(135\) m.
  5. Investigate whether the new model is also accurate for this angle of projection. [3]
  6. Make one suggestion as to how the model could be further refined. [1]
OCR MEI M1 Q2
19 marks Moderate -0.3
\includegraphics{figure_2} Fig. 7 shows a platform \(10\) m long and \(2\) m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O, just misses the other end, P. The ball is projected at \(68.5°\) to the horizontal with a speed of \(U\text{ms}^{-1}\). Air resistance may be neglected. At time \(t\) seconds after projection, the horizontal and vertical displacements of the ball from O are \(x\) m and \(y\) m.
  1. Obtain expressions, in terms of \(U\) and \(t\), for
    1. \(x\),
    2. \(y\). [3]
  2. The ball takes \(T\) s to travel from O to P. Show that \(T = \frac{U\sin 68.5°}{4.9}\) and write down a second equation connecting \(U\) and \(T\). [4]
  3. Hence show that \(U = 12.0\) (correct to three significant figures). [3]
  4. Calculate the horizontal distance of the ball from the platform when the ball lands on the ground. [5]
  5. Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is $$y = x\tan 68.5° - \frac{4.9x^2}{U^2(\cos 68.5°)^2}.$$ Use this equation to show again that \(U = 12.0\) (correct to three significant figures). [4]
AQA FP1 2014 June Q8
9 marks Standard +0.3
  1. Find the general solution of the equation $$\cos\left(\frac{5}{4}x - \frac{\pi}{3}\right) = \frac{\sqrt{2}}{2}$$ giving your answer for \(x\) in terms of \(\pi\). [5 marks]
  2. Use your general solution to find the sum of all the solutions of the equation $$\cos\left(\frac{5}{4}x - \frac{\pi}{3}\right) = \frac{\sqrt{2}}{2}$$ that lie in the interval \(0 \leqslant x \leqslant 20\pi\). Give your answer in the form \(k\pi\), stating the exact value of \(k\). [4 marks]
AQA FP1 2016 June Q4
7 marks Moderate -0.3
  1. Given that \(\sin \frac{\pi}{3} = \cos \frac{\pi}{k}\), state the value of the integer \(k\). [1 mark]
  2. Hence, or otherwise, find the general solution of the equation $$\cos \left( 2x - \frac{5\pi}{6} \right) = \sin \frac{\pi}{3}$$ giving your answer, in its simplest form, in terms of \(\pi\). [4 marks]
  3. Hence, given that \(\cos \left( 2x - \frac{5\pi}{6} \right) = \sin \frac{\pi}{3}\), show that there is only one finite value for \(\tan x\) and state its exact value. [2 marks]
OCR FP3 Q7
11 marks Challenging +1.2
    1. Verify, without using a calculator, that \(\theta = \frac{1}{8}\pi\) is a solution of the equation \(\sin 6\theta = \sin 2\theta\). [1]
    2. By sketching the graphs of \(y = \sin 6\theta\) and \(y = \sin 2\theta\) for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\), or otherwise, find the other solution of the equation \(\sin 6\theta = \sin 2\theta\) in the interval \(0 < \frac{1}{2}\pi\). [2]
  2. Use de Moivre's theorem to prove that $$\sin 6\theta \equiv \sin 2\theta(16\cos^4 \theta - 16\cos^2 \theta + 3).$$ [5]
  3. Hence show that one of the solutions obtained in part (i) satisfies \(\cos^2 \theta = \frac{1}{4}(2 - \sqrt{2})\), and justify which solution it is. [3]
OCR FP3 Q4
9 marks Standard +0.8
  1. By expressing \(\cos \theta\) in terms of \(e^{i\theta}\) and \(e^{-i\theta}\), show that $$\cos^5 \theta \equiv \frac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos \theta).$$ [5]
  2. Hence solve the equation \(\cos 5\theta + 5\cos 3\theta + 9\cos \theta = 0\) for \(0 \leqslant \theta \leqslant \pi\). [4]
OCR FP3 2008 January Q7
11 marks Challenging +1.3
    1. Verify, without using a calculator, that \(\theta = \frac{1}{8}\pi\) is a solution of the equation \(\sin 6\theta = \sin 2\theta\). [1]
    2. By sketching the graphs of \(y = \sin 6\theta\) and \(y = \sin 2\theta\) for \(0 < \theta < \frac{1}{2}\pi\), or otherwise, find the other solution of the equation \(\sin 6\theta = \sin 2\theta\) in the interval \(0 < \theta < \frac{1}{2}\pi\). [2]
  2. Use de Moivre's theorem to prove that $$\sin 6\theta = \sin 2\theta (16 \cos^4 \theta - 16 \cos^2 \theta + 3).$$ [5]
  3. Hence show that one of the solutions obtained in part (i) satisfies \(\cos^2 \theta = \frac{1}{4}(2 - \sqrt{2})\), and justify which solution it is. [3]
OCR FP3 2010 January Q7
13 marks Challenging +1.3
  1. Solve the equation \(\cos 6\theta = 0\), for \(0 < \theta < \pi\). [3]
  2. By using de Moivre's theorem, show that $$\cos 6\theta \equiv (2\cos^2\theta - 1)(16\cos^4\theta - 16\cos^2\theta + 1).$$ [5]
  3. Hence find the exact value of $$\cos\left(\frac{1}{12}\pi\right)\cos\left(\frac{5}{12}\pi\right)\cos\left(\frac{7}{12}\pi\right)\cos\left(\frac{11}{12}\pi\right),$$ justifying your answer. [5]
Edexcel AEA 2002 June Q1
8 marks Challenging +1.8
Solve the following equation, for \(0 \leq x \leq \pi\), giving your answers in terms of \(\pi\). $$\sin 5x - \cos 5x = \cos x - \sin x.$$ [8]
Edexcel AEA 2004 June Q1
9 marks Challenging +1.8
Solve the equation \(\cos x + \sqrt{(1 - \frac{1}{2} \sin 2x)} = 0\), in the interval \(0° \leq x < 360°\). [9]
Edexcel AEA 2008 June Q3
12 marks Challenging +1.8
  1. Prove that \(\tan 15° = 2 - \sqrt{3}\) [4]
  2. Solve, for \(0 < \theta < 360°\), $$\sin(\theta + 60°) \sin(\theta - 60°) = (1 - \sqrt{3}) \cos^2 \theta$$ [8]
OCR H240/02 2020 November Q4
5 marks Standard +0.3
In this question you must show detailed reasoning. Solve the equation \(3\sin^4 \phi + \sin^2 \phi = 4\), for \(0 \leq \phi < 2\pi\), where \(\phi\) is measured in radians. [5]
OCR H240/03 2019 June Q4
14 marks Standard +0.3
\includegraphics{figure_4} The diagram shows the part of the curve \(y = 3x \sin 2x\) for which \(0 \leqslant x \leqslant \frac{1}{2}\pi\). The maximum point on the curve is denoted by \(P\).
  1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(\tan 2x + 2x = 0\). [3]
  2. Use the Newton-Raphson method, with a suitable initial value, to find the \(x\)-coordinate of \(P\), giving your answer correct to 4 decimal places. Show the result of each iteration. [4]
  3. The trapezium rule, with four strips of equal width, is used to find an approximation to $$\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx.$$ Show that the result can be expressed as \(k\pi^2(\sqrt{2} + 1)\), where \(k\) is a rational number to be determined. [4]
    1. Evaluate \(\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx\). [1]
    2. Hence determine whether using the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region enclosed by the curve \(y = 3x \sin 2x\) and the \(x\)-axis for \(0 \leqslant x \leqslant \frac{1}{2}\pi\). [1]
    3. Explain briefly why it is not easy to tell from the diagram alone whether the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region in this case. [1]
OCR H240/03 2022 June Q7
8 marks Standard +0.8
In this question you must show detailed reasoning.
  1. Show that the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\) can be expressed in the form $$m \tan^2 \theta - 4 \tan \theta + (m + 3) = 0.$$ [3]
  2. It is given that there is only one value of \(\theta\), for \(0 < \theta < \pi\), satisfying the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\). Given also that \(m\) is a negative integer, find this value of \(\theta\), correct to 3 significant figures. [5]
OCR H240/03 2023 June Q2
5 marks Moderate -0.3
  1. Express \(3 \sin x - 4 \cos x\) in the form \(R \sin(x - \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). Give the value of \(\alpha\) correct to 4 significant figures. [3]
  2. Hence solve the equation \(3 \sin x - 4 \cos x = 2\) for \(0° < x < 90°\), giving your answer correct to 3 significant figures. [2]
AQA AS Paper 1 2019 June Q1
1 marks Easy -1.8
State the number of solutions to the equation \(\tan 4\theta = 1\) for \(0° < \theta < 180°\) Circle your answer. [1 mark] 1 2 4 8
AQA AS Paper 1 2019 June Q6
6 marks Moderate -0.3
    1. Show that \(\cos \theta = \frac{1}{2}\) is one solution of the equation $$6\sin^2 \theta + 5\cos \theta = 7$$ [2 marks]
    2. Find all the values of \(\theta\) that solve the equation $$6\sin^2 \theta + 5\cos \theta = 7$$ for \(0° \leq \theta \leq 360°\) Give your answers to the nearest degree. [2 marks]
  2. Hence, find all the solutions of the equation $$6\sin^2 2\theta + 5\cos 2\theta = 7$$ for \(0° \leq \theta \leq 360°\) Give your answers to the nearest degree. [2 marks]