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

4 It is given that \(\tan ^ { 2 } x = \sec x + 11\).

1. Show that the equation \(\tan ^ { 2 } x = \sec x + 11\) can be written in the form $$\sec ^ { 2 } x - \sec x - 12 = 0$$
2. Hence show that \(\cos x = \frac { 1 } { 4 }\) or \(\cos x = - \frac { 1 } { 3 }\).
3. Hence, or otherwise, solve the equation \(\tan ^ { 2 } x = \sec x + 11\), giving all values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).

3

1. Solve the equation \(\sec x = 5\), giving all the values of \(x\) in the interval \(0 \leqslant x \leqslant 2 \pi\) in radians to two decimal places.
2. Show that the equation \(\tan ^ { 2 } x = 3 \sec x + 9\) can be written as $$\sec ^ { 2 } x - 3 \sec x - 10 = 0$$
3. Solve the equation \(\tan ^ { 2 } x = 3 \sec x + 9\), giving all the values of \(x\) in the interval \(0 \leqslant x \leqslant 2 \pi\) in radians to two decimal places.

2

1. Solve the equation \(\sec x = 3\), giving the values of \(x\) in radians to two decimal places in the interval \(0 \leqslant x < 2 \pi\).
2. Show that the equation \(\tan ^ { 2 } x = 2 \sec x + 2\) can be written as \(\sec ^ { 2 } x - 2 \sec x - 3 = 0\).
3. Solve the equation \(\tan ^ { 2 } x = 2 \sec x + 2\), giving the values of \(x\) in radians to two decimal places in the interval \(0 \leqslant x < 2 \pi\).

7

1. Given that \(y = \frac { \sin \theta } { \cos \theta }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \sec ^ { 2 } \theta\).
2. Given that \(x = \sin \theta\), show that \(\frac { x } { \sqrt { 1 - x ^ { 2 } } } = \tan \theta\).
3. Use the substitution \(x = \sin \theta\) to find \(\int \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x\), giving your answer in terms of \(x\).

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}

1. (a) Use the derivatives of \(\sin x\) and \(\cos x\) to prove that

$$\frac { \mathrm { d } } { \mathrm {~d} x } ( \tan x ) = \sec ^ { 2 } x$$ The tangent to the curve \(y = 2 x \tan x\) at the point where \(x = \frac { \pi } { 4 }\) meets the \(y\)-axis at the point \(P\).
(b) Find the \(y\)-coordinate of \(P\) in the form \(k \pi ^ { 2 }\) where \(k\) is a rational constant.

4. (a) Use the identities for ( \(\sin A + \sin B\) ) and ( \(\cos A + \cos B\) ) to prove that $$\frac { \sin 2 x + \sin 2 y } { \cos 2 x + \cos 2 y } \equiv \tan ( x + y ) .$$ (b) Hence, show that $$\tan 52.5 ^ { \circ } = \sqrt { 6 } - \sqrt { 3 } - \sqrt { 2 } + 2 .$$

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 \(0 \leq x \leq 180 ^ { \circ }\).

2. Giving your answers to 1 decimal place, solve the equation $$5 \tan ^ { 2 } 2 \theta - 13 \sec 2 \theta = 1 ,$$ for \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\).

1. (a) Given that \(\cos x = \sqrt { 3 } - 1\), find the value of \(\cos 2 x\) in the form \(a + b \sqrt { 3 }\), where \(a\) and \(b\) are integers.
   (b) Given that

$$2 \cos ( y + 30 ) ^ { \circ } = \sqrt { 3 } \sin ( y - 30 ) ^ { \circ }$$ find the value of \(\tan y\) in the form \(k \sqrt { 3 }\) where \(k\) is a rational constant.

6. (a) Prove the identity $$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z } .$$ (b) Solve, for \(0 \leq x < \pi\), the equation $$2 \cot 2 x + \tan x = \operatorname { cosec } ^ { 2 } x - 7 ,$$ giving your answers to 2 decimal places.

6

1. 1. Given that \(\tan 2 x + \tan x = 0\), show that \(\tan x = 0\) or \(\tan ^ { 2 } x = 3\).
   2. Hence find all solutions of \(\tan 2 x + \tan x = 0\) in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
      (l mark)
2. 1. Given that \(\cos x \neq 0\), show that the equation $$\sin 2 x = \cos x \cos 2 x$$ can be written in the form $$2 \sin ^ { 2 } x + 2 \sin x - 1 = 0$$
   2. Show that all solutions of the equation \(2 \sin ^ { 2 } x + 2 \sin x - 1 = 0\) are given by \(\sin x = \frac { \sqrt { 3 } - 1 } { p }\), where \(p\) is an integer.

2 Angle \(\alpha\) is acute and \(\cos \alpha = \frac { 3 } { 5 }\). Angle \(\beta\) is obtuse and \(\sin \beta = \frac { 1 } { 2 }\).

1. 1. Find the value of \(\tan \alpha\) as a fraction.
      (1 mark)
   2. Find the value of \(\tan \beta\) in surd form.
2. Hence show that \(\tan ( \alpha + \beta ) = \frac { m \sqrt { 3 } - n } { n \sqrt { 3 } + m }\), where \(m\) and \(n\) are integers.
   (3 marks)

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.

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 \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).

5 A ball is projected with speed \(u \mathrm {~ms} ^ { - 1 }\) at an angle of elevation \(\alpha\) above the horizontal so as to hit a point \(P\) on a wall. The ball travels in a vertical plane through the point of projection. During the motion, the horizontal and upward vertical displacements of the ball from the point of projection are \(x\) metres and \(y\) metres respectively.

1. Show that, during the flight, the equation of the trajectory of the ball is given by $$y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)$$
2. The ball is projected from a point 1 metre vertically below and \(R\) metres horizontally from the point \(P\).
   1. By taking \(g = 10 \mathrm {~ms} ^ { - 2 }\), show that \(R\) satisfies the equation $$5 R ^ { 2 } \tan ^ { 2 } \alpha - u ^ { 2 } R \tan \alpha + 5 R ^ { 2 } + u ^ { 2 } = 0$$
   2. Hence, given that \(u\) and \(R\) are constants, show that, for \(\tan \alpha\) to have real values, \(R\) must satisfy the inequality $$R ^ { 2 } \leqslant \frac { u ^ { 2 } \left( u ^ { 2 } - 20 \right) } { 100 }$$
   3. Given that \(R = 5\), determine the minimum possible speed of projection.

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)

3 A player projects a basketball with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal. The basketball travels in a vertical plane through the point of projection and goes into the basket. During the motion, the horizontal and upward vertical displacements of the basketball from the point of projection are \(x\) metres and \(y\) metres respectively. \includegraphics[max width=\textwidth, alt={}, center]{3a1726d9-1b0c-41de-8b43-56019e18aac1-06_737_937_513_550}

1. Find an expression for \(y\) in terms of \(x , u , g\) and \(\tan \theta\).
2. The player projects the basketball with speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point 0.5 metres vertically below and 5 metres horizontally from the basket.
   1. Show that the two possible values of \(\theta\) are approximately \(63.1 ^ { \circ }\) and \(32.6 ^ { \circ }\), correct to three significant figures.
   2. Given that the player projects the basketball at \(63.1 ^ { \circ }\) to the horizontal, find the direction of the motion of the basketball as it enters the basket. Give your answer to the nearest degree.
3. State a modelling assumption needed for answering parts (a) and (b) of this question.
   (1 mark)