1.05o Trigonometric equations: solve in given intervals

The angle \(\theta\) is such that \(0° < \theta < 90°\).

1. Given that \(\theta\) satisfies the equation \(6 \sin 2\theta = 5 \cos \theta\), find the exact value of \(\sin \theta\). [3]
2. Given instead that \(\theta\) satisfies the equation \(8 \cos \theta \cosec^2 \theta = 3\), find the exact value of \(\cos \theta\). [5]

The value of \(\tan 10°\) is denoted by \(p\). Find, in terms of \(p\), the value of

1. \(\tan 55°\), [3]
2. \(\tan 5°\), [4]
3. \(\tan \theta\), where \(\theta\) satisfies the equation \(3 \sin(\theta + 10°) = 7 \cos(\theta - 10°)\). [5]

\includegraphics{figure_1} Figure 1 shows the curve \(y = f(x)\) which has a maximum point at \((-45, 7)\) and a minimum point at \((135, -1)\).

1. Showing the coordinates of any stationary points, sketch on separate diagrams the graphs of
   1. \(y = f(|x|)\),
   2. \(y = 1 + 2f(x)\). [6]

Given that $$f(x) = A + 2\sqrt{2} \cos x^{\circ} - 2\sqrt{2} \sin x^{\circ}, \quad x \in \mathbb{R}, \quad -180 \leq x \leq 180,$$ where \(A\) is a constant,

1. show that f(x) can be expressed in the form $$f(x) = A + R \cos (x + \alpha)^{\circ},$$ where \(R > 0\) and \(0 < \alpha < 90\), [3]
2. state the value of \(A\), [1]
3. find, to \(1\) decimal place, the \(x\)-coordinates of the points where the curve \(y = f(x)\) crosses the \(x\)-axis. [4]

Find the values of \(x\) in the interval \(-180 < x < 180\) for which $$\tan (x + 45)^{\circ} - \tan x^{\circ} = 4,$$ giving your answers to 1 decimal place. [9]

Find all values of \(\theta\) in the interval \(-180 < \theta < 180\) for which $$\tan^2 \theta^\circ + \sec \theta^\circ = 1.$$ [6]

\includegraphics{figure_7} The diagram shows the curve \(y = \text{f}(x)\) which has a maximum point at \((-45, 7)\) and a minimum point at \((135, -1)\).

1. Showing the coordinates of any stationary points, sketch the curve with equation \(y = 1 + 2\text{f}(x)\). [3]

Given that $$\text{f}(x) = A + 2\sqrt{2} \cos x° - 2\sqrt{2} \sin x°, \quad x \in \mathbb{R}, \quad -180 \leq x \leq 180,$$ where \(A\) is a constant,

1. show that f\((x)\) can be expressed in the form $$\text{f}(x) = A + R \cos (x + \alpha)°,$$ where \(R > 0\) and \(0 < \alpha < 90\), [3]
2. state the value of \(A\), [1]
3. find, to 1 decimal place, the \(x\)-coordinates of the points where the curve \(y = \text{f}(x)\) crosses the \(x\)-axis. [4]

By forming and solving a suitable quadratic equation, find the solutions of the equation $$3 \cos 2\theta - 5 \cos \theta + 2 = 0$$ in the interval \(0° < \theta < 360°\), giving your answers to the nearest \(0.1°\). [5 marks]

Solve, correct to 2 decimal places, the equation \(\cot 2\theta = 3\) for \(0° < \theta < 180°\). [4]

Express \(3\sin x + 2\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\). Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 < x < \pi.$$ [7]

Express \(4\cos\theta - \sin\theta\) in the form \(R\cos(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{1}{2}\pi\). Hence solve the equation \(4\cos\theta - \sin\theta = 3\), for \(0 \leq \theta \leq 2\pi\). [7]

Given that \(\cos\text{ec}^2\theta - \cot\theta = 3\), show that \(\cot^2\theta - \cot\theta - 2 = 0\). Hence solve the equation \(\cos\text{ec}^2\theta - \cot\theta = 3\) for \(0° \leq \theta \leq 180°\). [6]

Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).

1. Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre O. AB is one of the sides of the polygon. C is the midpoint of AB. Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \includegraphics{figure_8.1}
   1. Show that AB = \(2\sin 15°\). [2]
   2. Use a double angle formula to express \(\cos 30°\) in terms of \(\sin 15°\). Using the exact value of \(\cos 30°\), show that \(\sin 15° = \frac{1}{4}\sqrt{2 - \sqrt{3}}\). [4]
   3. Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6\sqrt{2 - \sqrt{3}}\). [2]
2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \includegraphics{figure_8.2}
   1. Show that DE = \(2\tan 15°\). [2]
   2. Let \(t = \tan 15°\). Use a double angle formula to express \(\tan 30°\) in terms of \(t\). Hence show that \(t^2 + 2\sqrt{3}t - 1 = 0\). [3]
   3. Solve this equation, and hence show that \(\pi < 12(2 - \sqrt{3})\). [4]
3. Use the results in parts (i)(C) and (ii)(C) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form. [2]

Solve the equation \(\cosec^2 \theta = 1 + 2 \cot \theta\), for \(-180° \leqslant \theta \leqslant 180°\). [6]

Given the equation \(\sin(x + 45°) = 2\cos x\), show that \(\sin x + \cos x = 2\sqrt{2}\cos x\). Hence solve, correct to 2 decimal places, the equation for \(0° < x < 360°\). [6]

Show that the equation \(\cos ec x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos^2 x + 5 \cos x - 2 = 0.$$ Hence solve the equation for \(0° \leq x \leq 360°\), giving your answers to 1 decimal place. [7]