Edexcel AEA 2009 June — Question 3 12 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2009
SessionJune
Marks12
PaperDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeExpand compound angle then solve
DifficultyChallenging +1.2 Part (a) requires expanding sin(π/3 - θ) using compound angle formula, then solving a linear trigonometric equation - straightforward for AEA level. Part (b) involves inverse trig manipulation and substitution, requiring more insight but still follows standard techniques. The multi-step nature and inverse function work elevate it above average, but it's accessible compared to typical AEA proof or extended reasoning questions.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05o Trigonometric equations: solve in given intervals
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 $$\begin{aligned} & \arcsin ( 1 - 2 x ) = \frac { \pi } { 3 } - \arcsin x , \quad 0 < x < 0.5 \\ & { \left[ \arcsin x \text { is an alternative notation for } \sin ^ { - 1 } x \right] } \end{aligned}$$ 
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

$$\begin{aligned}
& \arcsin ( 1 - 2 x ) = \frac { \pi } { 3 } - \arcsin x , \quad 0 < x < 0.5 \\
& { \left[ \arcsin x \text { is an alternative notation for } \sin ^ { - 1 } x \right] }
\end{aligned}$$

\hfill \mbox{\textit{Edexcel AEA 2009 Q3 [12]}}
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