A high-precision sensor measures the output voltage V V\,V of a microscopic resonator as a function of its deflection angle θ\thetaθ (in radians). The sensor output is modeled by the equation:
V(θ)=3cos(2θ)−3cos(2θ)cos(6θ) V(\theta) = 3\cos(2\theta) - 3\cos(2\theta)\cos(6\theta) V(θ)=3cos(2θ)−3cos(2θ)cos(6θ)Show that for small values of θ\thetaθ, V(θ)≈54θ2V(\theta) \approx 54\theta^2V(θ)≈54θ2.
The energy E E\,E dissipated during an oscillation cycle is given by E=∫00.1154V(θ) dθ\displaystyle E = \int_{0}^{0.1} \sqrt{\frac{1}{54}V(\theta)} \, d\thetaE=∫00.1541V(θ)dθ. Show that the energy E E\,E can be approximated by E≈2m×5nE \approx 2^m \times 5^nE≈2m×5n, where m m\,m and n n\,n are integers to be determined.
Explain why ∫12.612.7θ dθ\int_{12.6}^{12.7} \theta \, d\theta∫12.612.7θdθ is not a suitable approximation for ∫12.612.7154V(θ) dθ\displaystyle \int_{12.6}^{12.7} \sqrt{\frac{1}{54}V(\theta)} \, d\theta∫12.612.7541V(θ)dθ.
Explain how ∫12.612.7154V(θ) dθ\displaystyle \int_{12.6}^{12.7} \sqrt{\frac{1}{54}V(\theta)} \, d\theta∫12.612.7541V(θ)dθ may be approximated by ∫abθ dθ\int_{a}^{b} \theta \, d\theta∫abθdθ for suitable values of a a\,a and bbb.
317 exam-style questions on AQA A Level Maths 1.8 E: Trigonometry, 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. Each one has a worked solution and a mark scheme showing where the marks go.