1.05l Double angle formulae: and compound angle formulae

6

1. Use the Factor Theorem to show that \(4 x - 3\) is a factor of $$16 x ^ { 3 } + 11 x - 15$$
2. Given that \(x = \cos \theta\), show that the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ can be written in the form $$16 x ^ { 3 } + 11 x - 15 = 0$$
3. Hence show that the only solutions of the equation $$27 \cos \theta \cos 2 \theta + 19 \sin \theta \sin 2 \theta - 15 = 0$$ are given by \(\cos \theta = \frac { 3 } { 4 }\).

3

1. 1. Express \(3 \cos x + 2 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
      (3 marks)
   2. Hence find the minimum value of \(3 \cos x + 2 \sin x\) and the value of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) where the minimum occurs. Give your value of \(x\) to the nearest \(0.1 ^ { \circ }\).
2. 1. Show that \(\cot x - \sin 2 x = \cot x \cos 2 x\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).
   2. Hence, or otherwise, solve the equation $$\cot x - \sin 2 x = 0$$ in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).

2 The acute angles \(\alpha\) and \(\beta\) are given by \(\tan \alpha = \frac { 2 } { \sqrt { 5 } }\) and \(\tan \beta = \frac { 1 } { 2 }\).

1. 1. Show that \(\sin \alpha = \frac { 2 } { 3 }\), and find the exact value of \(\cos \alpha\).
   2. Hence find the exact value of \(\sin 2 \alpha\).
2. Show that the exact value of \(\cos ( \alpha - \beta )\) can be expressed as \(\frac { 2 } { 15 } ( k + \sqrt { 5 } )\), where \(k\) is an integer.

5 The polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } - 11 x - 3\).

1. Use the Factor Theorem to show that ( \(2 x + 3\) ) is a factor of \(\mathrm { f } ( x )\).
2. Write \(\mathrm { f } ( x )\) in the form \(( 2 x + 3 ) \left( a x ^ { 2 } + b x + c \right)\), where \(a , b\) and \(c\) are integers.
3. 1. Show that the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) can be written as \(4 x ^ { 3 } - 11 x - 3 = 0\), where \(x = \sin \theta\).
   2. Hence find all solutions of the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest degree.

5

1. 1. Express \(3 \sin x + 4 \cos x\) in the form \(R \sin ( x + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
   2. Hence solve the equation \(3 \sin 2 \theta + 4 \cos 2 \theta = 5\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
2. 1. Show that the equation \(\tan 2 \theta \tan \theta = 2\) can be written as \(2 \tan ^ { 2 } \theta = 1\).
   2. Hence solve the equation \(\tan 2 \theta \tan \theta = 2\) in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
3. 1. Use the Factor Theorem to show that \(2 x - 1\) is a factor of \(8 x ^ { 3 } - 4 x + 1\).
   2. Show that \(4 \cos 2 \theta \cos \theta + 1\) can be written as \(8 x ^ { 3 } - 4 x + 1\) where \(x = \cos \theta\).
   3. Given that \(\theta = 72 ^ { \circ }\) is a solution of \(4 \cos 2 \theta \cos \theta + 1 = 0\), use the results from parts (c)(i) and (c)(ii) to show that the exact value of \(\cos 72 ^ { \circ }\) is \(\frac { ( \sqrt { 5 } - 1 ) } { p }\) where \(p\) is an integer.
      [0pt] [3 marks]

8. (i) Given that \(\cos ( x + 30 ) ^ { \circ } = 3 \cos ( x - 30 ) ^ { \circ }\), prove that tan \(x ^ { \circ } = - \frac { \sqrt { 3 } } { 2 }\).
(ii) (a) Prove that \(\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta\).
(b) Verify that \(\theta = 180 ^ { \circ }\) is a solution of the equation \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).
(c) Using the result in part (a), or otherwise, find the other two solutions, \(0 < \theta < 360 ^ { \circ }\), of the equation using \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).

2. (a) Use integration by parts to find $$\int x \cos 2 x d x$$ (b) Prove that the answer to part (a) may be expressed as $$\frac { 1 } { 2 } \sin x ( 2 x \cos x - \sin x ) + C ,$$ where \(C\) is an arbitrary constant.

7 A particle is projected from a point on a plane which is inclined at an angle \(\alpha\) to the horizontal. The particle is projected up the plane with velocity \(u\) at an angle \(\theta\) above the plane. The motion of the particle is in a vertical plane containing a line of greatest slope of the inclined plane. \includegraphics[max width=\textwidth, alt={}, center]{daea0765-041a-4569-a535-f90fe4708313-5_401_748_516_644}

1. Using the identity \(\cos ( A + B ) = \cos A \cos B - \sin A \sin B\), show that the range up the plane is $$\frac { 2 u ^ { 2 } \sin \theta \cos ( \theta + \alpha ) } { g \cos ^ { 2 } \alpha }$$
2. Hence, using the identity \(2 \sin A \cos B = \sin ( A + B ) + \sin ( A - B )\), show that, as \(\theta\) varies, the range up the plane is a maximum when \(\theta = \frac { \pi } { 4 } - \frac { \alpha } { 2 }\).
3. Given that the particle strikes the plane at right angles, show that $$2 \tan \theta = \cot \alpha$$

7 A projectile is fired with speed \(u\) from a point \(O\) on a plane which is inclined at an angle \(\alpha\) to the horizontal. The projectile is fired at an angle \(\theta\) to the inclined plane and moves in a vertical plane through a line of greatest slope of the inclined plane. The projectile lands at a point \(P\), lower down the inclined plane, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{eed9842d-cd89-4eb7-b5ba-9380971be196-5_415_1098_495_463}

1. Find, in terms of \(u , g , \theta\) and \(\alpha\), the greatest perpendicular distance of the projectile from the plane.
2. 1. Find, in terms of \(u , g , \theta\) and \(\alpha\), the time of flight from \(O\) to \(P\).
   2. By using the identity \(\cos A \cos B + \sin A \sin B = \cos ( A - B )\), show that the distance \(O P\) is given by \(\frac { 2 u ^ { 2 } \sin \theta \cos ( \theta - \alpha ) } { g \cos ^ { 2 } \alpha }\).
   3. Hence, by using the identity \(2 \sin A \cos B = \sin ( A + B ) + \sin ( A - B )\) or otherwise, show that, as \(\theta\) varies, the maximum possible distance \(O P\) is \(\frac { u ^ { 2 } } { g ( 1 - \sin \alpha ) }\).
      (5 marks)

5 A particle is projected from a point \(O\) on a plane which is inclined at an angle \(\theta\) to the horizontal. The particle is projected down the plane with velocity \(u\) at an angle \(\alpha\) above the plane. The particle first strikes the plane at a point \(P\), as shown in the diagram. The motion of the particle is in a vertical plane containing a line of greatest slope of the inclined plane. \includegraphics[max width=\textwidth, alt={}, center]{3a1726d9-1b0c-41de-8b43-56019e18aac1-12_389_789_557_639}

1. Given that the time of flight from \(O\) to \(P\) is \(T\), find an expression for \(u\) in terms of \(\theta , \alpha , T\) and \(g\).
2. Using the identity \(\cos ( X - Y ) = \cos X \cos Y + \sin X \sin Y\), show that the distance \(O P\) is given by \(\frac { 2 u ^ { 2 } \sin \alpha \cos ( \alpha - \theta ) } { g \cos ^ { 2 } \theta }\).
   (6 marks)

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 \(t\).

3

1. Find the general solution, in degrees, of the equation $$\cos \left( 5 x + 40 ^ { \circ } \right) = \cos 65 ^ { \circ }$$
2. Given that $$\sin \frac { \pi } { 12 } = \frac { \sqrt { 3 } - 1 } { 2 \sqrt { 2 } }$$ express \(\sin \frac { \pi } { 12 }\) in the form \(\left( \cos \frac { \pi } { 4 } \right) ( \cos ( a \pi ) + \cos ( b \pi ) )\), where \(a\) and \(b\) are rational.
   (3 marks)

8

1. 1. Use de Moivre's theorem to show that $$\cos 4 \theta = \cos ^ { 4 } \theta - 6 \cos ^ { 2 } \theta \sin ^ { 2 } \theta + \sin ^ { 4 } \theta$$ and find a similar expression for \(\sin 4 \theta\).
   2. Deduce that $$\tan 4 \theta = \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta }$$
2. Explain why \(t = \tan \frac { \pi } { 16 }\) is a root of the equation $$t ^ { 4 } + 4 t ^ { 3 } - 6 t ^ { 2 } - 4 t + 1 = 0$$ and write down the three other roots in trigonometric form.
3. Hence show that $$\tan ^ { 2 } \frac { \pi } { 16 } + \tan ^ { 2 } \frac { 3 \pi } { 16 } + \tan ^ { 2 } \frac { 5 \pi } { 16 } + \tan ^ { 2 } \frac { 7 \pi } { 16 } = 28$$

9 A small ball P is projected with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation of \(( \alpha + \theta )\) from a point O at the bottom of a plane inclined at \(\alpha\) to the horizontal. P subsequently hits the plane at a point R , where OR is a line of greatest slope, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-07_456_862_406_242}

1. By deriving an expression, in terms of \(\theta\), \(\alpha\) and \(g\), for the time of flight of P , show that the distance OR, in metres, is $$\frac { 50 \sin \theta \cos ( \theta + \alpha ) } { g \cos ^ { 2 } \alpha }$$
2. By using the identity \(2 \sin \mathrm {~A} \cos \mathrm {~B} \equiv \sin ( \mathrm {~A} + \mathrm { B } ) - \sin ( \mathrm { B } - \mathrm { A } )\), determine, in terms of \(g\) and \(\sin \alpha\), an expression for the maximum range of P up the plane, as \(\theta\) varies.
3. Given that OR is the maximum range of P up the plane and is equal to 1.8 m , determine the value of \(\theta\). \includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-08_625_1180_255_239} A rigid wire ABC is fixed in a vertical plane. The section AB of the wire, of length \(b\), is straight and horizontal. The section BC of the wire is smooth and in the form of a circular arc of radius \(a\) and length \(\frac { 1 } { 2 } a \pi\). The centre of the arc is O , which is vertically above B . A bead P of mass \(m\) is threaded on the wire and projected from B with speed \(u\) towards C . The angle BOP when P is between B and C is denoted by \(\theta\), as shown in the diagram.

1. (a) Use the substitution \(t = \tan \left( \frac { x } { 2 } \right)\) to show that the equation

$$5 \sin x + 12 \cos x = 2$$ can be written in the form $$7 t ^ { 2 } - 5 t - 5 = 0$$ (b) Hence solve, for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\), the equation $$5 \sin x + 12 \cos x = 2$$ giving your answers to one decimal place.

1. 1. Use the substitution \(t = \tan \left( \frac { x } { 2 } \right)\) to prove that

$$\cot x + \tan \left( \frac { x } { 2 } \right) = \operatorname { cosec } x \quad x \neq n \pi , n \in \mathbb { Z }$$ (ii) \begin{figure}[h] \end{figure} An engineer models the vertical height above the ground of the tip of one blade of a wind turbine, shown in Figure 1. The ground is assumed to be horizontal. The vertical height of the tip of the blade above the ground, \(H\) metres, at time \(x\) seconds after the wind turbine has reached its constant operating speed, is modelled by the equation $$H = 90 - 30 \cos ( 120 x ) ^ { \circ } - 40 \sin ( 120 x ) ^ { \circ }$$

1. Show that \(H = 60\) when \(x = 0\) Using the substitution \(t = \tan ( 60 x ) ^ { \circ }\)
2. show that equation (I) can be rewritten as $$H = \frac { 120 t ^ { 2 } - 80 t + 60 } { 1 + t ^ { 2 } }$$
3. Hence find, according to the model, the value of \(x\) when the tip of the blade is 100 m above the ground for the first time after the wind turbine has reached its constant operating speed.

1. On a particular day, the depth of water in a river estuary at a specific location is modelled by the equation

$$D = 2 \sin \left( \frac { x } { 3 } \right) + 3 \cos \left( \frac { x } { 3 } \right) + 6 \quad 0 \leqslant x \leqslant 7 \pi$$ where the depth of water is \(D\) metres at time \(x\) hours after midnight on that day.

1. Write down the depth of water at midnight, according to the model. Using the substitution \(t = \tan \left( \frac { x } { 6 } \right)\)
2. show that equation (I) can be re-written as $$D = \frac { 3 t ^ { 2 } + 4 t + 9 } { 1 + t ^ { 2 } }$$
3. Hence determine, according to the model, the time after midnight when the depth of water is 5 metres for the first time. Give your answer to the nearest minute.

1. (a) Use \(t = \tan \frac { \theta } { 2 }\) to show that, where both sides are defined

$$\frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta } \equiv \frac { 5 t + 2 } { 2 t + 5 }$$ (b) Hence, again using \(t = \tan \frac { \theta } { 2 }\), prove that, where both sides are defined $$\frac { 20 + 21 \tan \theta } { 29 + 21 \sec \theta } \equiv \frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta }$$

1. (a) Use the substitution \(t = \tan \left( \frac { x } { 2 } \right)\) to show that the equation

$$3 \cos x - 2 \sin x = 1$$ can be written in the form $$2 t ^ { 2 } + 2 t - 1 = 0$$ (b) Hence solve, for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\), the equation $$3 \cos x - 2 \sin x = 1$$ giving your answers to one decimal place.

1. (a) Given that \(t = \tan \frac { X } { 2 }\) prove that

$$\cos x \equiv \frac { 1 - t ^ { 2 } } { 1 + t ^ { 2 } }$$ (b) Show that the equation $$3 \tan x - 10 \cos x = 10$$ can be written in the form $$( t + 2 ) \left( a t ^ { 2 } + b t + c \right) = 0$$ where \(t = \tan \frac { X } { 2 }\) and \(a , b\) and \(c\) are integers to be determined.
(c) Hence solve, for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\), the equation $$3 \tan x - 10 \cos x = 10$$

1. (a) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) to show that

$$\sec x - \tan x \equiv \frac { 1 - t } { 1 + t } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ (b) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) and the answer to part (a) to prove that $$\frac { 1 - \sin x } { 1 + \sin x } \equiv ( \sec x - \tan x ) ^ { 2 } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ \section*{Q uestion 1 continued}

1. (a) Given that \(| z | < 1\), write down the sum of the infinite series

$$1 + z + z ^ { 2 } + z ^ { 3 } + \ldots$$ (b) Given that \(z = \frac { 1 } { 2 } ( \cos \theta + \mathrm { i } \sin \theta )\),

1. use the answer to part (a), and de Moivre's theorem or otherwise, to prove that $$\frac { 1 } { 2 } \sin \theta + \frac { 1 } { 4 } \sin 2 \theta + \frac { 1 } { 8 } \sin 3 \theta + \ldots = \frac { 2 \sin \theta } { 5 - 4 \cos \theta }$$
2. show that the sum of the infinite series \(1 + z + z ^ { 2 } + z ^ { 3 } + \ldots\) cannot be purely imaginary, giving a reason for your answer.

5. $$y = \mathrm { e } ^ { 3 x } \sin x$$

1. Use Leibnitz's theorem to show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = 28 \mathrm { e } ^ { 3 x } \sin x + 96 \mathrm { e } ^ { 3 x } \cos x$$
2. Hence express \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\) in the form $$\operatorname { Re } ^ { 3 \mathrm { x } } \sin ( \mathrm { x } + \alpha )$$ where \(R\) and \(\alpha\) are constants to be determined, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)