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

9

1. Use de Moivre's theorem to show that $$\sec 6 \theta = \frac { \sec ^ { 6 } \theta } { 32 - 48 \sec ^ { 2 } \theta + 18 \sec ^ { 4 } \theta - \sec ^ { 6 } \theta }$$
2. Hence obtain the roots of the equation $$3 x ^ { 6 } - 36 x ^ { 4 } + 96 x ^ { 2 } - 64 = 0$$ in the form sec \(q \pi\), where \(q\) is rational.

1 \includegraphics[max width=\textwidth, alt={}, center]{a8e37fb1-14c7-4004-b186-d607878e200d-2_744_504_255_824} A uniform ladder \(A B\), of length \(3 a\) and weight \(W\), rests with the end \(A\) in contact with smooth horizontal ground and the end \(B\) against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point \(C\), where \(A C = a\). The other end of the rope is fixed to the point \(D\) at the base of the wall and the rope \(D C\) is in the same vertical plane as the ladder \(A B\). The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle \(\theta\) with the horizontal and the rope making an angle \(\alpha\) with the horizontal (see diagram). It is given that \(\tan \theta = 2 \tan \alpha\). Find, in terms of \(W\) and \(\alpha\), the tension in the rope and the magnitudes of the forces acting on the ladder at \(A\) and at \(B\).

8 A particle is projected with speed \(49 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\theta\) from a point \(O\) on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from \(O\) at time \(t\) seconds after projection are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(\theta\) and \(t\), and hence show that $$y = x \tan \theta - \frac { x ^ { 2 } \left( 1 + \tan ^ { 2 } \theta \right) } { 490 } .$$
   \includegraphics[max width=\textwidth, alt={}]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_627_1249_1699_447}
   The particle passes through the point where \(x = 70\) and \(y = 30\). The two possible values of \(\theta\) are \(\theta _ { 1 }\) and \(\theta _ { 2 }\), and the corresponding points where the particle returns to the plane are \(A _ { 1 }\) and \(A _ { 2 }\) respectively (see diagram).
2. Find \(\theta _ { 1 }\) and \(\theta _ { 2 }\).
3. Calculate the distance between \(A _ { 1 }\) and \(A _ { 2 }\).

4

1. Write down the expansion of \(\sin 3 x\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\).
2. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { 3 x \cos 2 x - \sin 3 x } { 5 x ^ { 3 } } \right]$$

6

1. It is given that \(y = \ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\).
   1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \tan x\).
   2. Find \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\).
2. Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\) are \(3 x - \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 12 } x ^ { 4 }\).
   (3 marks)
3. Write down the expansion of \(\ln ( 1 + p x )\), where \(p\) is a constant, in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\).
4. 1. Find the value of \(p\) for which \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) exists.
   2. Hence find the value of \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) when \(p\) takes the value found in part (d)(i).

7

1. Use the result \(\cos ( A + B ) = \cos A \cos B - \sin A \sin B\) to show that \(\cos ( A - B ) = \cos A \cos B + \sin A \sin B\). The function \(\mathrm { f } ( \theta )\) is defined as \(\cos \left( \theta + 30 ^ { \circ } \right) \cos \left( \theta - 30 ^ { \circ } \right)\), where \(\theta\) is in degrees.
2. Show that \(f ( \theta ) = \cos ^ { 2 } \theta - \frac { 1 } { 4 }\).
3. (i) Determine the following.

4 Prove that \(\sin ^ { 2 } ( \theta + 45 ) ^ { \circ } - \cos ^ { 2 } ( \theta + 45 ) ^ { \circ } \equiv \sin 2 \theta ^ { \circ }\).

6 It is given that the angle \(\theta\) satisfies the equation \(\sin \left( 2 \theta + \frac { 1 } { 4 } \pi \right) = 3 \cos \left( 2 \theta + \frac { 1 } { 4 } \pi \right)\).

1. Show that \(\tan 2 \theta = \frac { 1 } { 2 }\).
2. Hence find, in surd form, the exact value of \(\tan \theta\), given that \(\theta\) is an obtuse angle.

1. (a) Show that the equation

$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$ can be written in the form $$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$ (b) Hence solve, for \(0 \leqslant x < 90 ^ { \circ }\) $$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$ giving your answers, where appropriate, to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)

1. (a) Show that

$$\frac { 10 \sin ^ { 2 } \theta - 7 \cos \theta + 2 } { 3 + 2 \cos \theta } \equiv 4 - 5 \cos \theta$$ (b) Hence, or otherwise, solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$

9. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve with equation \(y = 3 \cos x ^ { \circ }\).
The point \(P ( c , d )\) is a minimum point on the curve with \(c\) being the smallest negative value of \(x\) at which a minimum occurs.

1. State the value of \(c\) and the value of \(d\).
2. State the coordinates of the point to which \(P\) is mapped by the transformation which transforms the curve with equation \(y = 3 \cos x ^ { \circ }\) to the curve with equation
   1. \(y = 3 \cos \left( \frac { x ^ { \circ } } { 4 } \right)\)
   2. \(y = 3 \cos ( x - 36 ) ^ { \circ }\)
3. Solve, for \(450 ^ { \circ } \leqslant \theta < 720 ^ { \circ }\), $$3 \cos \theta = 8 \tan \theta$$ giving your solution to one decimal place.
   In part (c) you must show all stages of your working.
   Solutions relying entirely on calculator technology are not acceptable.

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

Solutions relying entirely on calculator technology are not acceptable.

1. Show that $$\frac { 1 } { \cos \theta } + \tan \theta \equiv \frac { \cos \theta } { 1 - \sin \theta } \quad \theta \neq ( 2 n + 1 ) 90 ^ { \circ } \quad n \in \mathbb { Z }$$ Given that \(\cos 2 x \neq 0\)
2. solve for \(0 < x < 90 ^ { \circ }\) $$\frac { 1 } { \cos 2 x } + \tan 2 x = 3 \cos 2 x$$ giving your answers to one decimal place.

1. In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

1. Show that the equation $$4 \tan x = 5 \cos x$$ can be written as $$5 \sin ^ { 2 } x + 4 \sin x - 5 = 0$$
2. Hence solve, for \(0 < x \leqslant 360 ^ { \circ }\) $$4 \tan x = 5 \cos x$$ giving your answers to one decimal place.
3. Hence find the number of solutions of the equation $$4 \tan 3 x = 5 \cos 3 x$$ in the interval \(0 < x \leqslant 1800 ^ { \circ }\), explaining briefly the reason for your answer.

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

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

1. Solve, for \(0 < \theta \leqslant 450 ^ { \circ }\), the equation $$5 \cos ^ { 2 } \theta = 6 \sin \theta$$ giving your answers to one decimal place.
2. (a) A student's attempt to solve the question
   "Solve, for \(- 90 ^ { \circ } < x < 90 ^ { \circ }\), the equation \(3 \tan x - 5 \sin x = 0\) " is set out below. $$\begin{gathered} 3 \tan x - 5 \sin x = 0 \\ 3 \frac { \sin x } { \cos x } - 5 \sin x = 0 \\ 3 \sin x - 5 \sin x \cos x = 0 \\ 3 - 5 \cos x = 0 \\ \cos x = \frac { 3 } { 5 } \\ x = 53.1 ^ { \circ } \end{gathered}$$ Identify two errors or omissions made by this student, giving a brief explanation of each. The first four positive solutions, in order of size, of the equation $$\cos \left( 5 \alpha + 40 ^ { \circ } \right) = \frac { 3 } { 5 }$$ are \(\alpha _ { 1 } , \alpha _ { 2 } , \alpha _ { 3 }\) and \(\alpha _ { 4 }\) (b) Find, to the nearest degree, the value of \(\alpha _ { 4 }\)

1. (i) Solve, for \(- 90 ^ { \circ } \leqslant \theta < 270 ^ { \circ }\), the equation,

$$\sin \left( 2 \theta + 10 ^ { \circ } \right) = - 0.6$$ giving your answers to one decimal place.
(ii) (a) A student's attempt at the question
"Solve, for \(- 90 ^ { \circ } < x < 90 ^ { \circ }\), the equation \(7 \tan x = 8 \sin x\) " is set out below. $$\begin{gathered} 7 \tan x = 8 \sin x \\ 7 \times \frac { \sin x } { \cos x } = 8 \sin x \\ 7 \sin x = 8 \sin x \cos x \\ 7 = 8 \cos x \\ \cos x = \frac { 7 } { 8 } \\ x = 29.0 ^ { \circ } \text { (to } 3 \text { sf) } \end{gathered}$$ Identify two mistakes made by this student, giving a brief explanation of each mistake.
(b) Find the smallest positive solution to the equation $$7 \tan \left( 4 \alpha + 199 ^ { \circ } \right) = 8 \sin \left( 4 \alpha + 199 ^ { \circ } \right)$$

12. a. Show that $$\sec \theta - \cos \theta = \sin \theta \tan \theta \quad \theta \neq ( \pi n ) ^ { 0 } \quad n \in Z$$ b. Hence, or otherwise, solve for \(0 < x \leq \pi\) $$\sec x - \cos x = \sin x \tan \left( 3 x - \frac { \pi } { 9 } \right)$$

12. a. Prove that $$\frac { \sec ^ { 2 } x - 1 } { \sec ^ { 2 } x } \equiv \sin ^ { 2 } x$$ b. Hence solve, for \(- 360 ^ { \circ } < x < 360 ^ { \circ }\), the equation $$\frac { \sec ^ { 2 } x - 1 } { \sec ^ { 2 } x } = \frac { \cos 2 x } { 2 }$$

9. $$\mathrm { f } ( \theta ) = 4 \cos \theta + 5 \sin \theta \quad \theta \in R$$ a. Express \(\mathrm { f } ( \theta )\) in the form \(R \cos ( \theta - \alpha )\) where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\). Give the exact value of \(R\) and give the value of \(\alpha\), in radians, to 3 decimal places. Given that $$\mathrm { g } ( \theta ) = \frac { 135 } { 4 + \mathrm { f } ( \theta ) ^ { 2 } } \quad \theta \in R$$ b.find the range of \(g\).

15. The first three terms of a geometric series where \(\theta\) is a constant are $$- 8 \sin \theta , \quad 3 - 2 \cos \theta \quad \text { and } \quad 4 \cot \theta$$ a. Show that \(4 \cos ^ { 2 } \theta + 20 \cos \theta + 9 = 0\) Given that \(\theta\) lies in the interval \(90 ^ { \circ } < \theta < 180 ^ { \circ }\),
b. Find the value of \(\theta\).
c. Hence prove that this series is convergent.
d. Find \(S _ { \infty }\), in the form \(a ( 1 - \sqrt { 3 } )\)

1. Given that

$$y = \frac { 3 \sin \theta } { 2 \sin \theta + 2 \cos \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$ show that $$\frac { d y } { d \theta } = \frac { A } { 1 + \sin 2 \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$ where \(A\) is a rational constant to be found.

1. (a) Solve, for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\), the equation

$$5 \sin 2 \theta = 9 \tan \theta$$ giving your answers, where necessary, to one decimal place.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]
(b) Deduce the smallest positive solution to the equation $$5 \sin \left( 2 x - 50 ^ { \circ } \right) = 9 \tan \left( x - 25 ^ { \circ } \right)$$

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

Solutions relying entirely on calculator technology are not acceptable.

1. Show that $$\operatorname { cosec } \theta - \sin \theta \equiv \cos \theta \cot \theta \quad \theta \neq ( 180 n ) ^ { \circ } \quad n \in \mathbb { Z }$$
2. Hence, or otherwise, solve for \(0 < x < 180 ^ { \circ }\) $$\operatorname { cosec } x - \sin x = \cos x \cot \left( 3 x - 50 ^ { \circ } \right)$$

8. $$f ( x ) = \ln ( 2 x - 5 ) + 2 x ^ { 2 } - 30 , \quad x > 2.5$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval [3.5,4] A student takes 4 as the first approximation to \(\alpha\).
   Given \(\mathrm { f } ( 4 ) = 3.099\) and \(\mathrm { f } ^ { \prime } ( 4 ) = 16.67\) to 4 significant figures,
2. apply the Newton-Raphson procedure once to obtain a second approximation for \(\alpha\), giving your answer to 3 significant figures.
3. Show that \(\alpha\) is the only root of \(\mathrm { f } ( x ) = 0\)

1. Show that

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin 2 \theta } { 1 + \cos \theta } d \theta = 2 - 2 \ln 2$$