Prove trigonometric identity

1. (i) Showing each step in your reasoning, prove that

$$( \sin x + \cos x ) ( 1 - \sin x \cos x ) \equiv \sin ^ { 3 } x + \cos ^ { 3 } x$$ (ii) Solve, for \(0 \leqslant \theta < 360 ^ { \circ }\), $$3 \sin \theta = \tan \theta$$ giving your answers in degrees to 1 decimal place, as appropriate.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

6 Given that the angle \(\theta\) is acute and \(\cos \theta = \frac { 3 } { 4 }\) find, without using a calculator, the exact value of \(\sin 2 \theta\) and of \(\cot \theta\).

Prove that $$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta.$$ [4]

Prove that $$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta$$ [6]

Show that \(\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta\). [3]

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_1}
   1. Show that \(\text{AB} = 2 \sin 15°\). [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 \(\sin 15° = \frac{1}{2}\sqrt{2 - \sqrt{3}}\). [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_2}
   1. Show that \(\text{DE} = 2 \tan 15°\). [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]

Prove that \(\sqrt{2} \cos(2\theta + 45°) = \cos^2 \theta - 2\sin \theta \cos \theta - \sin^2 \theta\), where \(\theta\) is measured in degrees. [3]

In this question you must show detailed reasoning.

1. Show that the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\) can be expressed in the form $$m \tan^2 \theta - 4 \tan \theta + (m + 3) = 0.$$ [3]
2. It is given that there is only one value of \(\theta\), for \(0 < \theta < \pi\), satisfying the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\). Given also that \(m\) is a negative integer, find this value of \(\theta\), correct to 3 significant figures. [5]

It is given that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\) and \(\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}\) Use these two expressions to show that \(\tan 15° = 2 - \sqrt{3}\) Fully justify your answer. [4 marks]

Prove the identity \(\cot^2 \theta - \cos^2 \theta = \cot^2 \theta \cos^2 \theta\) [3 marks]

Prove that \(\sqrt{2}\cos(2\theta + 45°) \equiv \cos^2\theta - 2\sin\theta\cos\theta - \sin^2\theta\), where \(\theta\) is measured in degrees. [3]

The function f is defined by \(f : t \mapsto 2 \sin t + \cos 2t\) for \(0 \leqslant t < 2\pi\).

1. Show that \(\frac{df}{dt} = 2 \cos t(1 - 2 \sin t)\). [2]
2. Determine the range of f. [5]

A curve \(C\) is given parametrically by \(x = 2 \cos t + \sin 2t\), \(y = f(t)\) for \(0 \leqslant t < 2\pi\).

1. Show that \(x^2 + y^2 = 5 + 4 \sin 3t\). [3]
2. Deduce that \(C\) lies between two circles centred at the origin, and touches both. [2]
3. Find the gradient of the tangent to \(C\) at the point at which \(t = 0\). [3]