An acoustic engineer is modelling the resonant response, R(θ)R(\theta)R(θ), of a signal filter where θ\thetaθ represents the phase angle in degrees.
Show that the response function
R(θ)=12sin2θsecθ+2cos2θcscθ R(\theta) = \frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta R(θ)=21sin2θsecθ+2cos2θcscθcan be simplified to the form
R(θ)=2cscθ−3sinθ R(\theta) = 2\csc \theta - 3\sin \theta R(θ)=2cscθ−3sinθwhere sinθ≠0\sin \theta \neq 0sinθ=0 and cosθ≠0\cos \theta \neq 0cosθ=0.
The engineer needs to find the phase angles where the response is exactly 5. A student attempts to solve the equation
12sin2θsecθ+2cos2θcscθ=5 \frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta = 5 21sin2θsecθ+2cos2θcscθ=5for 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘. They use the result from part (a) to produce the following steps:
Step 1: 2cscθ−3sinθ=52\csc \theta - 3\sin \theta = 52cscθ−3sinθ=5
Step 2: 2sinθ−3sinθ=5\frac{2}{\sin \theta} - 3\sin \theta = 5sinθ2−3sinθ=5
Step 3: 3sin2θ+5sinθ−2=03\sin^{2} \theta + 5\sin \theta - 2 = 03sin2θ+5sinθ−2=0
Step 4: sinθ=13\sin \theta = \frac{1}{3}sinθ=31 or sinθ=−2\sin \theta = -2sinθ=−2
Step 5: θ=19.5∘,160.5∘\theta = 19.5^{\circ}, 160.5^{\circ}θ=19.5∘,160.5∘ (to 1 decimal place)
Explain why the value sinθ=−2\sin \theta = -2sinθ=−2 must be rejected in Step 4.
Determine if there are any further reasons, based on the domain constraints of the original filter model, why specific solutions might need to be rejected, and state the final correct solutions for 0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}0∘≤θ≤360∘.
38 exam-style questions on AQA A Level Maths 1.8.8 Proofs with trigonometric identities. Each one has a worked solution and a mark scheme showing where the marks go.