Edexcel AEA 2019 June — Question 4 17 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2019
SessionJune
Marks17
PaperDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeProve identity then solve equation only (no integral)
DifficultyChallenging +1.8 This AEA question requires proving a non-standard trigonometric identity through expansion of compound angles, then applying it cleverly to simplify a complex expression, and finally analyzing the uniqueness of solutions using calculus or monotonicity arguments. The multi-step nature, need for strategic insight in part (b), and rigorous proof in part (c) place it well above average difficulty but not at the extreme end for AEA material.
Spec1.01a Proof: structure of mathematical proof and logical steps1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae

4.(a)Prove the identity $$( \sin x + \cos y ) \cos ( x - y ) \equiv ( 1 + \sin ( x - y ) ) ( \cos x + \sin y )$$ (b)Hence,or otherwise,show that $$\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } \equiv \frac { 1 + \tan \theta } { 1 - \tan \theta }$$ (c)Given that \(k > 1\) ,show that the equation \(\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } = k\) has a unique solution in the interval \(0 < \theta < \frac { \pi } { 4 }\)

4.(a)Prove the identity

$$( \sin x + \cos y ) \cos ( x - y ) \equiv ( 1 + \sin ( x - y ) ) ( \cos x + \sin y )$$

(b)Hence,or otherwise,show that

$$\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } \equiv \frac { 1 + \tan \theta } { 1 - \tan \theta }$$

(c)Given that $k > 1$ ,show that the equation $\frac { \sin 5 \theta + \cos 3 \theta } { \cos 5 \theta + \sin 3 \theta } = k$ has a unique solution in the interval $0 < \theta < \frac { \pi } { 4 }$

\hfill \mbox{\textit{Edexcel AEA 2019 Q4 [17]}}