Find exact trigonometric values

1. (i) Show that

$$\sin ( x + 30 ) ^ { \circ } + \sin ( x - 30 ) ^ { \circ } \equiv a \sin x ^ { \circ }$$ where \(a\) is a constant to be found.
(ii) Hence find the exact value of \(\sin 75 ^ { \circ } + \sin 15 ^ { \circ }\), giving your answer in the form \(b \sqrt { 6 }\).

2

1. Prove the identity $$\sin \left( x + 30 ^ { \circ } \right) + ( \sqrt { } 3 ) \cos \left( x + 30 ^ { \circ } \right) \equiv 2 \cos x$$ where \(x\) is measured in degrees.
2. Hence express \(\cos 15 ^ { \circ }\) in surd form.

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

1.(a)Write down the exact value of \(\cos 405 ^ { \circ }\) (b)Hence,using a double angle identity for cosine,or otherwise,determine the exact value of \(\cos 101.25 ^ { \circ }\) ,giving your answer in the form $$a \sqrt { b + c \sqrt { 2 + \sqrt { 2 } } }$$ where \(a\) ,\(b\) and \(c\) are rational numbers.

2.Find the value of $$\arccos \left( \frac { 1 } { \sqrt { 2 } } \right) + \arcsin \left( \frac { 1 } { 3 } \right) + 2 \arctan \left( \frac { 1 } { \sqrt { 2 } } \right)$$ Give your answer as a multiple of \(\pi\) . $$\text { (arccos } x \text { is an alternative notion for } \cos ^ { - 1 } x \text { etc.) }$$

14

1. Show that $$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) \Rightarrow \arctan \left( \frac { 1 } { 2 } \right) + \arctan \left( \frac { 1 } { 3 } \right) = \arctan 1 .$$
2. Use the arctan addition formula in line 23 to show that $$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) , \text { as given in line } 39 .$$

15 Prove that \(\arctan 1 + \arctan 2 + \arctan 3 = \pi\), as given in line 41 . \section*{END OF QUESTION PAPER} \section*{OCR
Oxford Cambridge and RSA}

15 Fig. 15 shows a unit circle and the escribed regular polygon with 12 edges. \begin{figure}[h] \end{figure}

1. Show that the perimeter of the polygon is \(24 \tan 15 ^ { \circ }\).
2. Using the formula for \(\tan ( \theta - \phi )\) show that the perimeter of the polygon is \(48 - 24 \sqrt { 3 }\).

6 In this question you must show detailed reasoning.

1. Use the formula for \(\tan ( A - B )\) to show that \(\tan \frac { \pi } { 12 } = 2 - \sqrt { 3 }\).
2. Solve the equation \(2 \sqrt { 3 } \sin 3 A - 2 \cos 3 A = 1\) for \(0 ^ { \circ } \leqslant A < 180 ^ { \circ }\).

3 It is given that \(3 \cos \theta - 2 \sin \theta = R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).

1. Find the value of \(R\).
2. Show that \(\alpha \approx 33.7 ^ { \circ }\).
3. Hence write down the maximum value of \(3 \cos \theta - 2 \sin \theta\) and find a positive value of \(\theta\) at which this maximum value occurs.

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]

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]