1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

2 A projectile is launched from a point \(O\) on top of a cliff with initial velocity \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\alpha\) and moves in a vertical plane. During the motion, the position vector of the projectile relative to the point \(O\) is \(( x \mathbf { i } + y \mathbf { j } )\) metres where \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal and vertical unit vectors respectively.

1. Show that, during the motion, the equation of the trajectory of the projectile is given by $$y = x \tan \alpha - \frac { 4.9 x ^ { 2 } } { u ^ { 2 } \cos ^ { 2 } \alpha }$$
2. When \(u = 21\) and \(\alpha = 55 ^ { \circ }\), the projectile hits a small buoy \(B\). The buoy is at a distance \(s\) metres vertically below \(O\) and at a distance \(s\) metres horizontally from \(O\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-04_601_935_964_548}
   1. Find the value of \(s\).
   2. Find the acute angle between the velocity of the projectile and the horizontal just before the projectile hits \(B\), giving your answer to the nearest degree.
      [0pt] [5 marks]

6

1. Given that $$x = \ln ( \sec t + \tan t ) - \sin t$$ show that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \sin t \tan t$$
2. A curve is given parametrically by the equations $$x = \ln ( \sec t + \tan t ) - \sin t , \quad y = \cos t$$ The length of the arc of the curve between the points where \(t = 0\) and \(t = \frac { \pi } { 3 }\) is denoted by \(s\). Show that \(s = \ln p\), where \(p\) is an integer.

1. (a) Use the standard summation formulae to show that, for \(n \in \mathbb { N }\),

$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = n \left( n ^ { 2 } - A n - B \right)$$ where \(A\) and \(B\) are integers to be determined.
(b) Explain why, for \(k \in \mathbb { N }\), $$\sum _ { r = 1 } ^ { 3 k } r \tan ( 60 r ) ^ { \circ } = - k \sqrt { 3 }$$ Using the results from part (a) and part (b) and showing all your working,
(c) determine any value of \(n\) that satisfies $$\sum _ { r = 5 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = 15 \left[ \sum _ { r = 6 } ^ { 3 n } r \tan ( 60 r ) ^ { \circ } \right] ^ { 2 }$$

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. (i) Use the substitution \(t = \tan \frac { X } { 2 }\) to prove the identity

$$\frac { \sin x - \cos x + 1 } { \sin x + \cos x - 1 } \equiv \sec x + \tan x \quad x \neq \frac { n \pi } { 2 } \quad n \in \mathbb { Z }$$ (ii) Use the substitution \(t = \tan \frac { \theta } { 2 }\) to determine the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { 5 } { 4 + 2 \cos \theta } d \theta$$ giving your answer in simplest form.

1. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(3 m g\).

The other end of the string is attached to a fixed point \(O\) on a ceiling.
The particle hangs freely in equilibrium at a distance \(d\) vertically below \(O\).

1. Show that \(d = \frac { 4 } { 3 } a\). The point \(A\) is vertically below \(O\) such that \(O A = 2 a\).
   The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).
2. Find, in terms of \(g\), the acceleration of \(P\) immediately after it is released from rest.
3. Find, in terms of \(g\) and \(a\), the maximum speed attained by \(P\) as it moves from \(A\) to \(B\).
4. Find, in terms of \(a\), the distance \(O B\).

17 In this question you must show detailed reasoning. Solve the equation \(2 \sin x + \sec x = 4 \cos x\), where \(- \pi < x < \pi\).

6 In this question you must show detailed reasoning. Solve the equation \(\tan x - 3 \cot x = 2\) for values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

3

1. Given that \(\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta\), show that \(6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0\).
2. In this question you must show detailed reasoning. Solve the equation $$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$ giving all values of \(\theta\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\) correct to 1 decimal place.
3. Explain why not all the solutions from part (ii) are solutions of the equation $$\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta$$

10 A small ball \(P\) is projected with speed \(5 \mathrm {~ms} ^ { - 1 }\) at an angle \(\theta\) above the horizontal from a point \(O\) and moves freely under gravity. The horizontal and vertically upwards displacements of the ball from \(O\) at any subsequent time \(t\) seconds are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively. The ball is modelled as a particle and the acceleration due to gravity is taken to be \(10 \mathrm {~ms} ^ { - 2 }\).

1. Show that the equation of the trajectory of \(P\) is $$y = x \tan \theta - \frac { x ^ { 2 } } { 5 } \left( 1 + \tan ^ { 2 } \theta \right)$$ It is given that \(\tan \theta = 3\).
2. Using part (i), find the maximum height above the level of \(O\) of \(P\) in the subsequent motion.
3. Find the values of \(t\) when \(P\) is moving at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 1 } { 3 }\).
4. Give two possible reasons why the values of \(t\) found in part (iii) may not be accurate. \includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-09_435_714_267_678} Two particles \(P\) and \(Q\), with masses 2 kg and 8 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at a point on the intersection of two fixed inclined planes. The string lies in a vertical plane that contains a line of greatest slope of each of the two inclined planes. Plane \(\Pi _ { 1 }\) is inclined at an angle of \(30 ^ { \circ }\) to the horizontal and plane \(\Pi _ { 2 }\) is inclined at an angle of \(\theta\) to the horizontal. Particle \(P\) is on \(\Pi _ { 1 }\) and \(Q\) is on \(\Pi _ { 2 }\) with the string taut (see diagram). \(\Pi _ { 1 }\) is rough and the coefficient of friction between \(P\) and \(\Pi _ { 1 }\) is \(\frac { \sqrt { 3 } } { 3 }\). \(\Pi _ { 2 }\) is smooth.
   The particles are released from rest and \(P\) begins to move towards the pulley with an acceleration of \(g \cos \theta \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
5. Show that \(\theta\) satisfies the equation $$4 \sin \theta - 5 \cos \theta = 1 .$$
6. By expressing \(4 \sin \theta - 5 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\), find, correct to 3 significant figures, the tension in the string.

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 The triangle \(A B C\), shown in the diagram, is such that \(B C = 6 \mathrm {~cm} , A C = 5 \mathrm {~cm}\) and \(A B = 4 \mathrm {~cm}\). The angle \(B A C\) is \(\theta\). \includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-3_442_652_452_678}

1. Use the cosine rule to show that \(\cos \theta = \frac { 1 } { 8 }\).
2. Hence use a trigonometrical identity to show that \(\sin \theta = \frac { 3 \sqrt { 7 } } { 8 }\).
3. Hence find the area of the triangle \(A B C\).

7

1. Sketch the graph of \(y = \tan x\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
2. Write down the two solutions of the equation \(\tan x = \tan 61 ^ { \circ }\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
3. 1. Given that \(\sin \theta + \cos \theta = 0\), show that \(\tan \theta = - 1\).
   2. Hence solve the equation \(\sin \left( x - 20 ^ { \circ } \right) + \cos \left( x - 20 ^ { \circ } \right) = 0\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
4. Describe the single geometrical transformation that maps the graph of \(y = \tan x\) onto the graph of \(y = \tan \left( x - 20 ^ { \circ } \right)\).
5. The curve \(y = \tan x\) is stretched in the \(x\)-direction with scale factor \(\frac { 1 } { 4 }\) to give the curve with equation \(y = \mathrm { f } ( x )\). Write down an expression for \(\mathrm { f } ( x )\).

8

1. Write down \(\int \sec ^ { 2 } x \mathrm {~d} x\).
2. Given that \(y = \frac { \cos x } { \sin x }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
3. Prove the identity \(( \tan x + \cot x ) ^ { 2 } = \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x\).
4. Hence find \(\int _ { 0.5 } ^ { 1 } ( \tan x + \cot x ) ^ { 2 } \mathrm {~d} x\), giving your answer to two significant figures.