1.05l Double angle formulae: and compound angle formulae

4

1. Show that the equation $$\sin \left( x + 30 ^ { \circ } \right) = 2 \cos \left( x + 60 ^ { \circ } \right)$$ can be written in the form $$( 3 \sqrt { } 3 ) \sin x = \cos x$$
2. Hence solve the equation $$\sin \left( x + 30 ^ { \circ } \right) = 2 \cos \left( x + 60 ^ { \circ } \right)$$ for \(- 180 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

4

1. Show that the equation \(\sin \left( 60 ^ { \circ } - x \right) = 2 \sin x\) can be written in the form \(\tan x = k\), where \(k\) is a constant.
2. Hence solve the equation \(\sin \left( 60 ^ { \circ } - x \right) = 2 \sin x\), for \(0 ^ { \circ } < x < 360 ^ { \circ }\).

5

1. Express \(\cos ^ { 2 } 2 x\) in terms of \(\cos 4 x\).
2. Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 8 } \pi } \cos ^ { 2 } 2 x \mathrm {~d} x\).

4

1. Find \(\int \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x\).
2. Express \(\sin ^ { 2 } 3 x\) in terms of \(\cos 6 x\) and hence find \(\int \sin ^ { 2 } 3 x \mathrm {~d} x\).

8

1. By first expanding \(\cos ( 2 x + x )\), show that $$\cos 3 x \equiv 4 \cos ^ { 3 } x - 3 \cos x$$
2. Hence show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \left( 2 \cos ^ { 3 } x - \cos x \right) d x = \frac { 5 } { 12 }$$

4

1. Express \(\cos ^ { 2 } x\) in terms of \(\cos 2 x\).
2. Hence show that $$\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \left( \cos ^ { 2 } x + \sin 2 x \right) \mathrm { d } x = \frac { 1 } { 8 } \sqrt { } 3 + \frac { 1 } { 12 } \pi + \frac { 1 } { 4 }$$

3 Solve the equation $$2 \cos 2 \theta = 4 \cos \theta - 3$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

8

1. Given that \(\tan A = t\) and \(\tan ( A + B ) = 4\), find \(\tan B\) in terms of \(t\).
2. Solve the equation $$2 \tan \left( 45 ^ { \circ } - x \right) = 3 \tan x$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

3

1. Find \(\int 4 \cos ^ { 2 } \left( \frac { 1 } { 2 } \theta \right) \mathrm { d } \theta\).
2. Find the exact value of \(\int _ { - 1 } ^ { 6 } \frac { 1 } { 2 x + 3 } \mathrm {~d} x\).

7 The angle \(\alpha\) lies between \(0 ^ { \circ }\) and \(90 ^ { \circ }\) and is such that $$2 \tan ^ { 2 } \alpha + \sec ^ { 2 } \alpha = 5 - 4 \tan \alpha$$

1. Show that $$3 \tan ^ { 2 } \alpha + 4 \tan \alpha - 4 = 0$$ and hence find the exact value of \(\tan \alpha\).
2. It is given that the angle \(\beta\) is such that \(\cot ( \alpha + \beta ) = 6\). Without using a calculator, find the exact value of \(\cot \beta\).

5 \includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-2_554_689_1354_726} The diagram shows part of the curve $$y = 2 \cos x - \cos 2 x$$ and its maximum point \(M\). The shaded region is bounded by the curve, the axes and the line through \(M\) parallel to the \(y\)-axis.

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. Find the exact value of the area of the shaded region.

7 \includegraphics[max width=\textwidth, alt={}, center]{250b4df9-2646-4246-bb6d-2be92bf29598-3_553_689_258_726} The parametric equations of a curve are $$x = 6 \sin ^ { 2 } t , \quad y = 2 \sin 2 t + 3 \cos 2 t$$ for \(0 \leqslant t < \pi\). The curve crosses the \(x\)-axis at points \(B\) and \(D\) and the stationary points are \(A\) and \(C\), as shown in the diagram.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 3 } \cot 2 t - 1\).
2. Find the values of \(t\) at \(A\) and \(C\), giving each answer correct to 3 decimal places.
3. Find the value of the gradient of the curve at \(B\).

6 \includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-3_453_650_258_744} The diagram shows the curve with parametric equations $$x = 3 \cos t , \quad y = 2 \cos \left( t - \frac { 1 } { 6 } \pi \right)$$ for \(0 \leqslant t < 2 \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 } ( \sqrt { } 3 - \cot t )\).
2. Find the equation of the tangent to the curve at the point where the curve crosses the positive \(y\)-axis. Give the answer in the form \(y = m x + c\).

7 The equation of a curve is \(y = \frac { \sin 2 x } { \cos x + 1 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \left( \cos ^ { 2 } x + \cos x - 1 \right) } { \cos x + 1 }\).
2. Find the \(x\)-coordinate of each stationary point of the curve in the interval \(- \pi < x < \pi\). Give each answer correct to 3 significant figures.

6

1. Show that \(\frac { \cos 2 \theta } { 1 + \cos 2 \theta } \equiv 1 - \frac { 1 } { 2 } \sec ^ { 2 } \theta\).
2. Solve the equation \(\frac { \cos 2 \alpha } { 1 + \cos 2 \alpha } = 13 + 5 \tan \alpha\) for \(0 < \alpha < \pi\).
3. Find the exact value of \(\int _ { - \frac { 1 } { 2 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \cos x } { 1 + \cos x } \mathrm {~d} x\).

2 Solve the equation \(5 \cos \theta ( 1 + \cos 2 \theta ) = 4\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7 \includegraphics[max width=\textwidth, alt={}, center]{6bf7ba66-8362-4ac0-8e5c-3f88a3ccdf86-12_424_488_260_826} The diagram shows the curve with equation \(y = \sin 2 x + 3 \cos 2 x\) for \(0 \leqslant x \leqslant \pi\). At the points \(P\) and \(Q\) on the curve, the gradient of the curve is 3 .

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. By first expressing \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in the form \(R \cos ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the \(x\)-coordinates of \(P\) and \(Q\), giving your answers correct to 4 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7

1. Use the factor theorem to show that ( \(2 x + 3\) ) is a factor of $$8 x ^ { 3 } + 4 x ^ { 2 } - 10 x + 3$$
2. Show that the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) can be expressed as $$8 \cos ^ { 3 } \theta + 4 \cos ^ { 2 } \theta - 10 \cos \theta + 3 = 0 .$$
3. Solve the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

6

1. Showing all necessary working, solve the equation $$\sec \alpha \operatorname { cosec } \alpha = 7$$ for \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Showing all necessary working, solve the equation $$\sin \left( \beta + 20 ^ { \circ } \right) + \sin \left( \beta - 20 ^ { \circ } \right) = 6 \cos \beta$$ for \(0 ^ { \circ } < \beta < 90 ^ { \circ }\).

3 Express the equation \(\tan \left( \theta + 60 ^ { \circ } \right) = 2 + \tan \left( 60 ^ { \circ } - \theta \right)\) as a quadratic equation in \(\tan \theta\), and hence solve the equation for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4 A curve has equation \(y = \cos x \sin 2 x\).
Find the \(x\)-coordinate of the stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\), giving your answer correct to 3 significant figures.

5 By first expressing the equation $$\tan \theta \tan \left( \theta + 45 ^ { \circ } \right) = 2 \cot 2 \theta$$ as a quadratic equation in \(\tan \theta\), solve the equation for \(0 ^ { \circ } < \theta < 90 ^ { \circ }\).

3

1. Given that \(\cos \left( x - 30 ^ { \circ } \right) = 2 \sin \left( x + 30 ^ { \circ } \right)\), show that \(\tan x = \frac { 2 - \sqrt { 3 } } { 1 - 2 \sqrt { 3 } }\).
2. Hence solve the equation $$\cos \left( x - 30 ^ { \circ } \right) = 2 \sin \left( x + 30 ^ { \circ } \right)$$ for \(0 ^ { \circ } < x < 360 ^ { \circ }\).

6

1. Prove that \(\operatorname { cosec } 2 \theta - \cot 2 \theta \equiv \tan \theta\).
2. Hence show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } ( \operatorname { cosec } 2 \theta - \cot 2 \theta ) \mathrm { d } \theta = \frac { 1 } { 2 } \ln 2\).

5

1. By first expanding \(\tan ( 2 \theta + 2 \theta )\), show that the equation \(\tan 4 \theta = \frac { 1 } { 2 } \tan \theta\) may be expressed as \(\tan ^ { 4 } \theta + 2 \tan ^ { 2 } \theta - 7 = 0\).
2. Hence solve the equation \(\tan 4 \theta = \frac { 1 } { 2 } \tan \theta\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).