1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

8

1. Show that the equation $$\frac { 1 } { 1 + \cos \theta } + \frac { 1 } { 1 - \cos \theta } = 32$$ can be written in the form $$\operatorname { cosec } ^ { 2 } \theta = 16$$
2. Hence, or otherwise, solve the equation $$\frac { 1 } { 1 + \cos ( 2 x - 0.6 ) } + \frac { 1 } { 1 - \cos ( 2 x - 0.6 ) } = 32$$ giving all values of \(x\) in radians to two decimal places in the interval \(0 < x < \pi\).
   (5 marks)

9

1. Given that \(x = \frac { \sin y } { \cos y }\), use the quotient rule to show that $$\frac { \mathrm { d } x } { \mathrm {~d} y } = \sec ^ { 2 } y$$ (3 marks)
2. Given that \(\tan y = x - 1\), use a trigonometrical identity to show that $$\sec ^ { 2 } y = x ^ { 2 } - 2 x + 2$$
3. Show that, if \(y = \tan ^ { - 1 } ( x - 1 )\), then $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x ^ { 2 } - 2 x + 2 }$$ (l mark)
4. A curve has equation \(y = \tan ^ { - 1 } ( x - 1 ) - \ln x\).
   1. Find the value of the \(x\)-coordinate of each of the stationary points of the curve.
   2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   3. Hence show that the curve has a minimum point which lies on the \(x\)-axis.

4 By forming and solving a quadratic equation, solve the equation $$8 \sec x - 2 \sec ^ { 2 } x = \tan ^ { 2 } x - 2$$ in the interval \(0 < x < 2 \pi\), giving the values of \(x\) in radians to three significant figures.

8

1. Show that the expression \(\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x }\) can be written as \(2 \sec x\).
   [0pt] [4 marks]
2. Hence solve the equation $$\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x } = \tan ^ { 2 } x - 2$$ giving the values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
   [0pt] [6 marks]
3. Hence solve the equation $$\frac { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } { \cos \left( 2 \theta - 30 ^ { \circ } \right) } + \frac { \cos \left( 2 \theta - 30 ^ { \circ } \right) } { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } = \tan ^ { 2 } \left( 2 \theta - 30 ^ { \circ } \right) - 2$$ giving the values of \(\theta\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
   [0pt] [2 marks]
   \includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-16_1517_1709_1190_153}
   \includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-20_2489_1730_221_139}

2. Find, to 2 decimal places, the solutions of the equation $$3 \cot ^ { 2 } x - 4 \operatorname { cosec } x + \operatorname { cosec } ^ { 2 } x = 0$$ in the interval \(0 \leq x \leq 2 \pi\).

2. (a) Prove that, for \(\cos x \neq 0\), $$\sin 2 x - \tan x \equiv \tan x \cos 2 x .$$ (b) Hence, or otherwise, solve the equation $$\sin 2 x - \tan x = 2 \cos 2 x ,$$ for \(x\) in the interval \(0 \leq x \leq 180 ^ { \circ }\).

1 A particle, of mass 4 kg , moves in a horizontal plane under the action of a single force, \(\mathbf { F }\) newtons. The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in the horizontal plane, perpendicular to each other. At time \(t\) seconds, the velocity of the particle, \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), is given by $$\mathbf { v } = 4 \cos 2 t \mathbf { i } + 3 \sin t \mathbf { j }$$

1. 1. Find an expression for the force, \(\mathbf { F }\), acting on the particle at time \(t\) seconds.
   2. Find the magnitude of \(\mathbf { F }\) when \(t = \pi\).
2. When \(t = 0\), the particle is at the point with position vector \(( 2 \mathbf { i } - 14 \mathbf { j } )\) metres. Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\) seconds.
   [0pt] [5 marks]
   \includegraphics[max width=\textwidth, alt={}]{691c50b4-50b2-4e3a-a7e0-60f8ec35ee3c-02_1346_1717_1361_150}

3 The region bounded by the \(y\)-axis and the curves \(y = \sin 2 x\) and \(y = \sqrt { 2 } \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\) is occupied by a uniform lamina. Find the exact value of the \(x\)-coordinate of the centre of mass of the lamina.

3. A particle \(P\) moves in the \(x y\)-plane in such a way that its position vector \(\mathbf { r }\) metres at time \(t\) seconds, where \(0 \leqslant t < \pi\), satisfies the differential equation $$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$ When \(t = 0\), the particle is at the point with position vector \(( - \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).

1. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0 , P\) is \(x\) metres from the origin \(O\) and moving with velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction of \(x\) increasing, where

$$v = 5 \sin 2 t$$ When \(t = 0 , x = 1\) and \(P\) is at rest.

1. Find the magnitude and direction of the acceleration of \(P\) at the instant when \(P\) is next at rest.
2. Show that \(1 \leqslant x \leqslant 6\)
3. Find the total time, in the first \(4 \pi\) seconds of the motion, for which \(P\) is more than 3 metres from \(O\)
   \includegraphics[max width=\textwidth, alt={}]{a7901165-1679-4d30-9444-0c27020e32ea-16_2260_52_309_1982}

1. In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.}

1. Use the substitution \(t = \tan \left( \frac { \theta } { 2 } \right)\) to show that $$\int \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \int \frac { a } { ( t + b ) ^ { 2 } + c } d t$$ where \(a\), \(b\) and \(c\) are constants to be determined.
2. Hence show that $$\int _ { \frac { \pi } { 2 } } ^ { \frac { 2 \pi } { 3 } } \frac { 1 } { 2 \sin \theta + \cos \theta + 2 } d \theta = \ln \left( \frac { 2 \sqrt { 3 } } { 3 } \right)$$

8. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the function \(\mathrm { h } ( x )\) with equation $$h ( x ) = 45 + 15 \sin x + 21 \sin \left( \frac { x } { 2 } \right) + 25 \cos \left( \frac { x } { 2 } \right) \quad x \in [ 0,40 ]$$

1. Show that $$\frac { \mathrm { d } h } { \mathrm {~d} x } = \frac { \left( t ^ { 2 } - 6 t - 17 \right) \left( 9 t ^ { 2 } + 4 t - 3 \right) } { 2 \left( 1 + t ^ { 2 } \right) ^ { 2 } }$$ where \(t = \tan \left( \frac { x } { 4 } \right)\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a52911da-4b69-4d86-975e-d10e3a481e1d-16_581_1403_1263_331} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Source: \({ } ^ { 1 }\) Data taken on 29th December 2016 from \href{http://www.ukho.gov.uk/easytide/EasyTide}{http://www.ukho.gov.uk/easytide/EasyTide} Figure 2 shows a graph of predicted tide heights, in metres, for Portland harbour from 08:00 on the 3rd January 2017 to the end of the 4th January \(2017 { } ^ { 1 }\). The graph of \(k \mathrm {~h} ( x )\), where \(k\) is a constant and \(x\) is the number of hours after 08:00 on 3rd of January, can be used to model the predicted tide heights, in metres, for this period of time.
2. 1. Suggest a value of \(k\) that could be used for the graph of \(k \mathrm {~h} ( x )\) to form a suitable model.
   2. Why may such a model be suitable to predict the times when the tide heights are at their peaks, but not to predict the heights of these peaks?
3. Use Figure 2 and the result of part (a) to estimate, to the nearest minute, the time of the highest tide height on the 4th January 2017.

3 In this question you must show detailed reasoning.

1. Given that \(\sin \alpha = \frac { 2 } { 3 }\), find the exact values of \(\cos \alpha\).
2. Given that \(2 \tan ^ { 2 } \beta - 7 \sec \beta + 5 = 0\), find the exact value of \(\sec \beta\).

1. (a) Given that \(\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1\), show that \(1 + \tan ^ { 2 } \theta \equiv \sec ^ { 2 } \theta\).
   (b) Solve, for \(0 \leq \theta < 360 ^ { \circ }\), the equation

$$2 \tan ^ { 2 } \theta + \sec \theta = 1 ,$$ giving your answers to 1 decimal place.

4 It is given that \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\).

1. Show that the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\) can be written in the form $$2 \cot ^ { 2 } x + 5 \cot x - 3 = 0$$
2. Hence show that \(\tan x = 2\) or \(\tan x = - \frac { 1 } { 3 }\).
3. Hence, or otherwise, solve the equation \(2 \operatorname { cosec } ^ { 2 } x = 5 - 5 \cot x\), giving all values of \(x\) in radians to one decimal place in the interval \(- \pi < x \leqslant \pi\).

4

1. Solve the equation \(\sec x = \frac { 3 } { 2 }\), giving all values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).
2. By using a suitable trigonometrical identity, solve the equation $$2 \tan ^ { 2 } x = 10 - 5 \sec x$$ giving all values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).

3

1. Solve the equation $$\operatorname { cosec } x = 3$$ giving all values of \(x\) in radians to two decimal places, in the interval \(0 \leqslant x \leqslant 2 \pi\).
   (2 marks)
2. By using a suitable trigonometric identity, solve the equation $$\cot ^ { 2 } x = 11 - \operatorname { cosec } x$$ giving all values of \(x\) in radians to two decimal places, in the interval \(0 \leqslant x \leqslant 2 \pi\).
   (6 marks)

8

1. Show that the equation \(4 \operatorname { cosec } ^ { 2 } \theta - \cot ^ { 2 } \theta = k\), where \(k \neq 4\), can be written in the form $$\sec ^ { 2 } \theta = \frac { k - 1 } { k - 4 }$$
2. Hence, or otherwise, solve the equation $$4 \operatorname { cosec } ^ { 2 } \left( 2 x + 75 ^ { \circ } \right) - \cot ^ { 2 } \left( 2 x + 75 ^ { \circ } \right) = 5$$ giving all values of \(x\) in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
   [0pt] [5 marks] \includegraphics[max width=\textwidth, alt={}, center]{2df59047-3bfe-4b9c-a77f-142bc7506cbc-18_72_113_1055_159}
   
   \includegraphics[max width=\textwidth, alt={}]{2df59047-3bfe-4b9c-a77f-142bc7506cbc-20_2288_1707_221_153}

4 Given that \(x = 2 \sec \theta\) and \(y = 3 \tan \theta\), show that \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1\).

8

1. Given that $$9 \sin ^ { 2 } \theta + \sin 2 \theta = 8$$ show that $$8 \cot ^ { 2 } \theta - 2 \cot \theta - 1 = 0$$ 8
2. Hence, solve $$9 \sin ^ { 2 } \theta + \sin 2 \theta = 8$$ in the interval \(0 < \theta < 2 \pi\) Give your answers to two decimal places.
   8
3. Solve $$9 \sin ^ { 2 } \left( 2 x - \frac { \pi } { 4 } \right) + \sin \left( 4 x - \frac { \pi } { 2 } \right) = 8$$ in the interval \(0 < x < \frac { \pi } { 2 }\) Give your answers to one decimal place.

10

1. Given that $$y = \tan x$$ use the quotient rule to show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec ^ { 2 } x$$ 10
2. The region enclosed by the curve \(y = \tan ^ { 2 } x\) and the horizontal line, which intersects the curve at \(x = - \frac { \pi } { 4 }\) and \(x = \frac { \pi } { 4 }\), is shaded in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-17_1059_967_461_539} Show that the area of the shaded region is $$\pi - 2$$ Fully justify your answer.
   Do not write outside the box \includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-19_2488_1716_219_153}

12 One of the rides at a theme park is a room where the floor and ceiling both move up and down for \(10 \pi\) seconds. At time \(t\) seconds after the ride begins, the distance \(f\) metres of the floor above the ground is $$f = 1 - \cos t$$ At time \(t\) seconds after the ride begins, the distance \(c\) metres of the ceiling above the ground is $$c = 8 - 4 \sin t$$ The ride is shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-16_448_766_932_635} 12

1. Show that the initial distance between the floor and ceiling is 8 metres.
   [0pt] [1 mark]
   \includegraphics[max width=\textwidth, alt={}]{6a03a035-ff32-4734-864b-a076aa9cbec0-17_2500_1721_214_148}

8

1. Given that \(\cos \theta \neq \pm 1\), prove the identity $$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } \equiv 2 \operatorname { cosec } ^ { 2 } \theta$$ 8
2. Hence, find the set of values of \(A\) for which the equation $$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } = A$$ has real solutions.
   Fully justify your answer.
   8
3. Given that \(\theta\) is obtuse and $$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } = 16$$ find the exact value of \(\cot \theta\)

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
Solve, for \(0 < \theta \leqslant 360 ^ { \circ }\), the equation $$3 \tan ^ { 2 } \theta + 7 \sec \theta - 3 = 0$$ giving your answers to one decimal place.