Expand compound angle then solve

7

1. Find the possible values of \(x\) for which \(\sin ^ { - 1 } \left( x ^ { 2 } - 1 \right) = \frac { 1 } { 3 } \pi\), giving your answers correct to 3 decimal places.
2. Solve the equation \(\sin \left( 2 \theta + \frac { 1 } { 3 } \pi \right) = \frac { 1 } { 2 }\) for \(0 \leqslant \theta \leqslant \pi\), giving \(\theta\) in terms of \(\pi\) in your answers.

2 By first expanding \(\sin \left( \theta + 30 ^ { \circ } \right)\), solve the equation \(\sin \left( \theta + 30 ^ { \circ } \right) \operatorname { cosec } \theta = 2\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

2 Solve the equation \(\sec \theta \cos \left( \theta - 60 ^ { \circ } \right) = 4\) for \(- 180 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Solve the equation $$\cos \left( x + 30 ^ { \circ } \right) = 2 \cos x$$ giving all solutions in the interval \(- 180 ^ { \circ } < x < 180 ^ { \circ }\).

3 Solve the equation $$\cos \left( \theta + 60 ^ { \circ } \right) = 2 \sin \theta$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

2 Showing all necessary working, solve the equation \(\sin \left( \theta - 30 ^ { \circ } \right) + \cos \theta = 2 \sin \theta\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\). [4]

2 Solve the equation \(\cos \left( \theta - 60 ^ { \circ } \right) = 3 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$

3. (a) Solve, for \(0 \leqslant \theta < 2 \pi\), $$\sin \left( \frac { \pi } { 3 } - \theta \right) = \frac { 1 } { \sqrt { } 3 } \cos \theta$$ (b) Find the value of \(x\) for which $$\begin{aligned} & \arcsin ( 1 - 2 x ) = \frac { \pi } { 3 } - \arcsin x , \quad 0 < x < 0.5 \\ & { \left[ \arcsin x \text { is an alternative notation for } \sin ^ { - 1 } x \right] } \end{aligned}$$

5 Given the equation \(\sin \left( x + 45 ^ { \circ } \right) = 2 \cos x\), show that \(\sin x + \cos x = 2 \sqrt { 2 } \cos x\).
Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

1. Given that $$2 \sin \left( x - 60 ^ { \circ } \right) = \cos \left( x - 30 ^ { \circ } \right)$$ show that $$\tan x = 3 \sqrt { 3 }$$
2. Hence or otherwise solve, for \(0 \leqslant \theta < 180 ^ { \circ }\) $$2 \sin 2 \theta = \cos \left( 2 \theta + 30 ^ { \circ } \right)$$ giving your answers to one decimal place.

6

1. Show that the equation \(\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)\) can be written in the form \(\tan x = \frac { \sqrt { 2 } - 1 } { \sqrt { 3 } - \sqrt { 2 } }\).
2. Hence solve the equation \(\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)\) for \(0 \leqslant x \leqslant 2 \pi\).

7 The function f is defined for all real numbers by $$f ( x ) = \sin \left( x + \frac { \pi } { 6 } \right)$$

1. Find the general solution of the equation \(\mathrm { f } ( x ) = 0\).
2. The quadratic function g is defined for all real numbers by $$\mathrm { g } ( x ) = \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 4 } x ^ { 2 }$$ It can be shown that \(\mathrm { g } ( x )\) gives a good approximation to \(\mathrm { f } ( x )\) for small values of \(x\).
   1. Show that \(\mathrm { g } ( 0.05 )\) and \(\mathrm { f } ( 0.05 )\) are identical when rounded to four decimal places.
   2. A chord joins the points on the curve \(y = \mathrm { g } ( x )\) for which \(x = 0\) and \(x = h\). Find an expression in terms of \(h\) for the gradient of this chord.
   3. Using your answer to part (b)(ii), find the value of \(\mathrm { g } ^ { \prime } ( 0 )\).

3 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\). Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).