Derive triple angle then solve equation

6 It is given that \(\tan 3 x = k \tan x\), where \(k\) is a constant and \(\tan x \neq 0\).

1. By first expanding \(\tan ( 2 x + x )\), show that $$( 3 k - 1 ) \tan ^ { 2 } x = k - 3$$
2. Hence solve the equation \(\tan 3 x = k \tan x\) when \(k = 4\), giving all solutions in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
3. Show that the equation \(\tan 3 x = k \tan x\) has no root in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\) when \(k = 2\).

8

1. By first expanding \(\sin ( 2 \theta + \theta )\), show that $$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$
2. Show that, after making the substitution \(x = \frac { 2 \sin \theta } { \sqrt { 3 } }\), the equation \(x ^ { 3 } - x + \frac { 1 } { 6 } \sqrt { } 3 = 0\) can be written in the form \(\sin 3 \theta = \frac { 3 } { 4 }\).
3. Hence solve the equation $$x ^ { 3 } - x + \frac { 1 } { 6 } \sqrt { } 3 = 0$$ giving your answers correct to 3 significant figures.

4

1. By first expanding \(\tan ( 2 x + x )\), show that the equation \(\tan 3 x = 3 \cot x\) can be written in the form \(\tan ^ { 4 } x - 12 \tan ^ { 2 } x + 3 = 0\).
2. Hence solve the equation \(\tan 3 x = 3 \cot x\) for \(0 ^ { \circ } < x < 90 ^ { \circ }\).

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 \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

8. (a) Using the trigonometric identity for \(\tan ( A + B )\), prove that $$\tan 3 x = \frac { 3 \tan x - \tan ^ { 3 } x } { 1 - 3 \tan ^ { 2 } x } , \quad x \neq ( 2 n + 1 ) 30 ^ { \circ } , \quad n \in \mathbb { Z }$$ (b) Hence solve, for \(- 30 ^ { \circ } < x < 30 ^ { \circ }\), $$\tan 3 x = 11 \tan x$$ (Solutions based entirely on graphical or numerical methods are not acceptable.)

6. (a) (i) By writing \(3 \theta = ( 2 \theta + \theta )\), show that $$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$ (ii) Hence, or otherwise, for \(0 < \theta < \frac { \pi } { 3 }\), solve $$8 \sin ^ { 3 } \theta - 6 \sin \theta + 1 = 0 .$$ Give your answers in terms of \(\pi\).
(b) Using \(\sin ( \theta - \alpha ) = \sin \theta \cos \alpha - \cos \theta \sin \alpha\), or otherwise, show that $$\sin 15 ^ { \circ } = \frac { 1 } { 4 } ( \sqrt { } 6 - \sqrt { } 2 )$$

7. (i) (a) Prove that $$\cos 3 \theta \equiv 4 \cos ^ { 3 } \theta - 3 \cos \theta$$ (You may use the double angle formulae and the identity $$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B )$$ (b) Hence solve the equation $$2 \cos 3 \theta + \cos 2 \theta + 1 = 0$$ giving answers in the interval \(0 \leqslant \theta \leqslant \pi\).
Solutions based entirely on graphical or numerical methods are not acceptable.
(ii) Given that \(\theta = \arcsin x\) and that \(0 < \theta < \frac { \pi } { 2 }\), show that $$\cot \theta = \frac { \sqrt { \left( 1 - x ^ { 2 } \right) } } { x } , \quad 0 < x < 1$$

12

1. By first writing \(\tan 3 \theta\) as \(\tan ( 2 \theta + \theta )\), show that \(\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }\).
2. Hence show that there are always exactly two different values of \(\theta\) between \(0 ^ { \circ }\) and \(180 ^ { \circ }\) which satisfy the equation \(3 \tan 3 \theta = \tan \theta + k\),
   where \(k\) is a non-zero constant. \section*{END OF QUESTION PAPER} \section*{OCR
   Oxford Cambridge and RSA}

12

1. Use the identity \(\tan 2 x \equiv \frac { 2 \tan x } { 1 - \tan ^ { 2 } x }\) to show that \(\tan 4 x \equiv \frac { 4 \left( 1 - \tan ^ { 2 } x \right) \tan x } { 1 - 6 \tan ^ { 2 } x + \tan ^ { 4 } x }\).
2. Hence, given that \(x = \frac { 1 } { 16 } \pi\) is a root of the equation \(\tan ^ { 4 } x + p \tan ^ { 3 } x - 6 \tan ^ { 2 } x - p \tan x + 1 = 0\) where \(p\) is a positive constant, find the value of \(p\).