1.05g Exact trigonometric values: for standard angles

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]

Using appropriate right-angled triangles, show that \(\tan 45° = 1\) and \(\tan 30° = \frac{1}{\sqrt{3}}\). Hence show that \(\tan 75° = 2 + \sqrt{3}\). [7]

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]

Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is \(40\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal. \includegraphics{figure_1}

1. Show that, according to the standard projectile model in which air resistance is neglected, the flight time, \(T\) s, and the range, \(R\) m, are given by $$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6] A company is designing a new type of ball and wants to model its flight.
2. Initially the company uses the standard projectile model. Use this model to show that when \(\alpha = 30°\) and the initial speed is \(40\text{ms}^{-1}\), \(T\) is approximately \(4.08\) and \(R\) is approximately \(141.4\). Find the values of \(T\) and \(R\) when \(\alpha = 45°\). [3] The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at \(40\text{ms}^{-1}\). When the angle of projection is set at \(30°\), the range is found to be \(125\) m.
3. Comment briefly on the accuracy of the standard projectile model in this situation. [1] The company refines the model by assuming that the ball has a constant deceleration of \(2\text{ms}^{-2}\) in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still \(4.08\) s when the angle of projection is \(30°\).
4. Using the new model, with \(\alpha = 30°\), show that the horizontal displacement from the point of projection, \(x\) m at time \(t\) s, is given by $$x = 40t\cos 30° - t^2.$$ Find the range and hence show that this new model is reasonably accurate in this case. [4] The company then sets the angle of projection to \(45°\) while retaining a projection speed of \(40\text{ms}^{-1}\). With this setting the range of the ball is found to be \(135\) m.
5. Investigate whether the new model is also accurate for this angle of projection. [3]
6. Make one suggestion as to how the model could be further refined. [1]

\includegraphics{figure_2} Fig. 7 shows a platform \(10\) m long and \(2\) m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O, just misses the other end, P. The ball is projected at \(68.5°\) to the horizontal with a speed of \(U\text{ms}^{-1}\). Air resistance may be neglected. At time \(t\) seconds after projection, the horizontal and vertical displacements of the ball from O are \(x\) m and \(y\) m.

1. Obtain expressions, in terms of \(U\) and \(t\), for
   1. \(x\),
   2. \(y\). [3]
2. The ball takes \(T\) s to travel from O to P. Show that \(T = \frac{U\sin 68.5°}{4.9}\) and write down a second equation connecting \(U\) and \(T\). [4]
3. Hence show that \(U = 12.0\) (correct to three significant figures). [3]
4. Calculate the horizontal distance of the ball from the platform when the ball lands on the ground. [5]
5. Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is $$y = x\tan 68.5° - \frac{4.9x^2}{U^2(\cos 68.5°)^2}.$$ Use this equation to show again that \(U = 12.0\) (correct to three significant figures). [4]

The matrix \(\mathbf{C} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}\), where \(a\) and \(b\) are positive real numbers, and \(\mathbf{C}^2 = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\) Use \(\mathbf{C}\) to show that \(\cos \frac{\pi}{12}\) can be written in the form \(\frac{\sqrt{m + n}}{2}\), where \(m\) and \(n\) are integers. [7 marks]

The complex number \(z\) is such that $$z = \frac{1 + \text{i}}{1 - k\text{i}}$$ where \(k\) is a real number.

1. Find the real part of \(z\) and the imaginary part of \(z\), giving your answers in terms of \(k\) [2 marks]
2. In the case where \(k = \sqrt{3}\), use part (a) to show that $$\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$$ [5 marks]

In this question you must show detailed reasoning. The diagram shows triangle \(ABC\). \includegraphics{figure_8} The angles \(CAB\) and \(ABC\) are each \(45°\), and angle \(ACB = 90°\). The points \(D\) and \(E\) lie on \(AC\) and \(AB\) respectively. \(AE = DE = 1\), \(DB = 2\). Angle \(BED = 90°\), angle \(EBD = 30°\) and angle \(DBC = 15°\).

1. Show that \(BC = \frac{\sqrt{2} + \sqrt{6}}{2}\). [3]
2. By considering triangle \(BCD\), show that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\). [3]