1.05l Double angle formulae: and compound angle formulae

4 It is given that \(A\) and \(B\) are angles such that $$\sec ^ { 2 } A - \tan A = 13 \quad \text { and } \quad \sin B \sec ^ { 2 } B = 27 \cos B \operatorname { cosec } ^ { 2 } B$$ Find the possible exact values of \(\tan ( A - B )\).

8 The functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = | 2 x + a | + 3 a \quad \text { and } \quad \mathrm { g } ( x ) = 5 x - 4 a$$ where \(a\) is a positive constant.

1. State the range of f and the range of g .
2. State why f has no inverse, and find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).
3. Solve for \(x\) the equation \(\operatorname { gf } ( x ) = 31 a\).
4. Show that \(\sin 2 \theta ( \tan \theta + \cot \theta ) \equiv 2\).
5. Hence
   1. find the exact value of \(\tan \frac { 1 } { 12 } \pi + \tan \frac { 1 } { 8 } \pi + \cot \frac { 1 } { 12 } \pi + \cot \frac { 1 } { 8 } \pi\),
   2. solve the equation \(\sin 4 \theta ( \tan \theta + \cot \theta ) = 1\) for \(0 < \theta < \frac { 1 } { 2 } \pi\),
   3. express \(( 1 - \cos 2 \theta ) ^ { 2 } \left( \tan \frac { 1 } { 2 } \theta + \cot \frac { 1 } { 2 } \theta \right) ^ { 3 }\) in terms of \(\sin \theta\).

6 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

3 By expressing \(\cos 2 x\) in terms of \(\cos x\), find the exact value of \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } \frac { \cos 2 x } { \cos ^ { 2 } x } \mathrm {~d} x\).

4 Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 - 2 \sin ^ { 2 } x } { 1 + 2 \sin x \cos x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

6

1. Use the quotient rule to show that the derivative of \(\frac { \cos x } { \sin x }\) is \(\frac { - 1 } { \sin ^ { 2 } x }\).
2. Show that \(\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 4 } \pi } \frac { \sqrt { 1 + \cos 2 x } } { \sin x \sin 2 x } \mathrm {~d} x = \frac { 1 } { 2 } ( \sqrt { 6 } - \sqrt { 2 } )\).

2 Use integration to find the exact value of \(\int _ { \frac { 1 } { 16 } \pi } ^ { \frac { 1 } { 8 } \pi } \left( 9 - 6 \cos ^ { 2 } 4 x \right) \mathrm { d } x\).

9 A curve has parametric equations \(x = 1 - \cos t , y = \sin t \sin 2 t\), for \(0 \leqslant t \leqslant \pi\).

1. Find the coordinates of the points where the curve meets the \(x\)-axis.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos 2 t + 2 \cos ^ { 2 } t\). Hence find, in an exact form, the coordinates of the stationary points.
3. Find the cartesian equation of the curve. Give your answer in the form \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a polynomial.
4. Sketch the curve.

4 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).

2 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ }$$

6 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h] \end{figure}

1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).

5 Show that \(\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta\).

8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

2 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

4 You are given that \(\mathrm { f } ( x ) = \cos x + \lambda \sin x\) where \(\lambda\) is a positive constant.

1. Express \(\mathrm { f } ( x )\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving \(R\) and \(\alpha\) in terms of \(\lambda\).
2. Given that the maximum value (as \(x\) varies) of \(\mathrm { f } ( x )\) is 2 , find \(R , \lambda\) and \(\alpha\), giving your answers in exact form.

9 The equation of a curve in polar coordinates is \(r = 2 \sin 3 \theta\) for \(0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi\).

1. Sketch the curve.
2. Find the area of the region enclosed by this curve.
3. By expressing \(\sin 3 \theta\) in terms of \(\sin \theta\), show that a cartesian equation for the curve is $$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } y - 2 y ^ { 3 } .$$ \section*{END OF QUESTION PAPER}

8 Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 10 \sin 3 x - 20 \cos 3 x$$ Show that, for large positive \(x\) and independently of the initial conditions, $$y \approx R \sin ( 3 x + \phi )$$ where the constants \(R\) and \(\phi\), such that \(R > 0\) and \(0 < \phi < 2 \pi\), are to be determined correct to 2 decimal places.

Use de Moivre's theorem to prove that $$\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }$$ State the exact values of \(\theta\), between 0 and \(\pi\), that satisfy \(\tan 3 \theta = 1\). Express each root of the equation \(t ^ { 3 } - 3 t ^ { 2 } - 3 t + 1 = 0\) in the form \(\tan ( k \pi )\), where \(k\) is a positive rational number. For each of these values of \(k\), find the exact value of \(\tan ( k \pi )\).

7 Expand \(\left( z + \frac { 1 } { z } \right) ^ { 4 } \left( z - \frac { 1 } { z } \right) ^ { 2 }\) and, by substituting \(z = \cos \theta + \mathrm { i } \sin \theta\), find integers \(p , q , r , s\) such that $$64 \sin ^ { 2 } \theta \cos ^ { 4 } \theta = p + q \cos 2 \theta + r \cos 4 \theta + s \cos 6 \theta$$ Using the substitution \(x = 2 \cos \theta\), show that $$\int _ { 1 } ^ { 2 } x ^ { 4 } \sqrt { } \left( 4 - x ^ { 2 } \right) \mathrm { d } x = \frac { 4 } { 3 } \pi + \sqrt { } 3$$

7 By considering the binomial expansion of \(\left( z - \frac { 1 } { z } \right) ^ { 6 }\), where \(z = \cos \theta + \mathrm { i } \sin \theta\), express \(\sin ^ { 6 } \theta\) in the form $$\frac { 1 } { 32 } ( p + q \cos 2 \theta + r \cos 4 \theta + s \cos 6 \theta ) ,$$ where \(p , q , r\) and \(s\) are integers to be determined. Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 6 } \theta \mathrm {~d} \theta\).

10 Use the identity \(2 \sin P \cos Q \equiv \sin ( P + Q ) + \sin ( P - Q )\) to show that $$2 \sin \theta \cos \left( \theta - \frac { 1 } { 4 } \pi \right) \equiv \cos \left( 2 \theta - \frac { 3 } { 4 } \pi \right) + \frac { 1 } { \sqrt { } 2 }$$ A curve has polar equation \(r = 2 \sin \theta \cos \left( \theta - \frac { 1 } { 4 } \pi \right)\), for \(0 \leqslant \theta \leqslant \frac { 3 } { 4 } \pi\). Sketch the curve and state the polar equation of its line of symmetry, justifying your answer. Show that the area of the region enclosed by the curve is \(\frac { 3 } { 8 } ( \pi + 1 )\).

5 State the sum of the series \(z + z ^ { 2 } + z ^ { 3 } + \ldots + z ^ { n }\), for \(z \neq 1\). By letting \(z = \cos \theta + \mathrm { i } \sin \theta\), show that $$\cos \theta + \cos 2 \theta + \cos 3 \theta + \ldots + \cos n \theta = \frac { \sin \frac { 1 } { 2 } n \theta } { \sin \frac { 1 } { 2 } \theta } \cos \frac { 1 } { 2 } ( n + 1 ) \theta$$ where \(\sin \frac { 1 } { 2 } \theta \neq 0\).

6 Use de Moivre's theorem to express \(\cot 7 \theta\) in terms of \(\cot \theta\). Use the equation \(\cot 7 \theta = 0\) to show that the roots of the equation $$x ^ { 6 } - 21 x ^ { 4 } + 35 x ^ { 2 } - 7 = 0$$ are \(\cot \left( \frac { 1 } { 14 } k \pi \right)\) for \(k = 1,3,5,9,11,13\), and deduce that $$\cot ^ { 2 } \left( \frac { 1 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 3 } { 14 } \pi \right) \cot ^ { 2 } \left( \frac { 5 } { 14 } \pi \right) = 7$$

7

1. Use de Moivre's theorem to prove that $$\tan 4 \theta = \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta } .$$
2. Hence find the solutions of the equation $$t ^ { 4 } - 4 t ^ { 3 } - 6 t ^ { 2 } + 4 t + 1 = 0$$ giving your answers in the form \(\tan k \pi\), where \(k\) is a rational number.

3

1. Use de Moivre's theorem to show that $$\cos 4 \theta = \cos ^ { 4 } \theta - 6 \cos ^ { 2 } \theta \sin ^ { 2 } \theta + \sin ^ { 4 } \theta$$
2. Hence find all the roots of the equation $$x ^ { 4 } - 6 x ^ { 2 } + 1 = 0$$ in the form \(\tan q \pi\), where \(q\) is a positive rational number.