In the analysis of a resonant electronic circuit, the phase shift α \alpha\,α across a specific component is related to the impedance through a series of trigonometric relationships.
Prove that
cotα−tanα≡2cot2α \cot \alpha - \tan \alpha \equiv 2 \cot 2\alpha cotα−tanα≡2cot2αfor α≠nπ2,n∈Z\displaystyle \alpha \neq \frac{n\pi}{2}, n \in \mathbb{Z}α=2nπ,n∈Z.
Using the identity in part (a), or otherwise, establish that
cot2α−tan2α≡4cot2αcsc2α \cot^2 \alpha - \tan^2 \alpha \equiv 4 \cot 2\alpha \csc 2\alpha cot2α−tan2α≡4cot2αcsc2αA particular resonance condition occurs when the operating phase ϕ \phi\,ϕ satisfies the equation
4cot2ϕcsc2ϕ=15tan2ϕ 4 \cot 2\phi \csc 2\phi = 15 \tan^2 \phi 4cot2ϕcsc2ϕ=15tan2ϕSolve this equation for −π2<ϕ<π2\displaystyle -\frac{\pi}{2} < \phi < \frac{\pi}{2}−2π<ϕ<2π, giving your answers to 2 decimal places.
Practise AQA A Level Maths 1.8 E: Trigonometry with exam-style questions for A Level Maths. 239 questions covering 1.8.1 Trigonometric definitions, rules and radians, 1.8.2 Small angle approximations (A-level only), 1.8.3 Trigonometric functions and exact values, 1.8.4 Reciprocal and inverse trigonometric functions (A-level only), 1.8.5 Trigonometric identities, 1.8.6 Compound and double angle formulae (A-level only), 1.8.7 Solving trigonometric equations, 1.8.8 Proofs with trigonometric identities, and 1.8.9 Trigonometry in context, matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
