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Trigonometric Functions

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Question 25

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.

a.

Show that the response function

R(θ)=12sin⁡2θsec⁡θ+2cos⁡2θcsc⁡θ R(\theta) = \frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta R(θ)=21​sin2θsecθ+2cos2θcscθ

can be simplified to the form

R(θ)=2csc⁡θ−3sin⁡θ R(\theta) = 2\csc \theta - 3\sin \theta R(θ)=2cscθ−3sinθ

where sin⁡θ≠0\sin \theta \neq 0sinθ=0 and cos⁡θ≠0\cos \theta \neq 0cosθ=0.

[3]
bi.

The engineer needs to find the phase angles where the response is exactly 5. A student attempts to solve the equation

12sin⁡2θsec⁡θ+2cos⁡2θcsc⁡θ=5 \frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta = 5 21​sin2θsecθ+2cos2θcscθ=5

for 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: 2csc⁡θ−3sin⁡θ=52\csc \theta - 3\sin \theta = 52cscθ−3sinθ=5

Step 2: 2sin⁡θ−3sin⁡θ=5\frac{2}{\sin \theta} - 3\sin \theta = 5sinθ2​−3sinθ=5

Step 3: 3sin⁡2θ+5sin⁡θ−2=03\sin^{2} \theta + 5\sin \theta - 2 = 03sin2θ+5sinθ−2=0

Step 4: sin⁡θ=13\sin \theta = \frac{1}{3}sinθ=31​ or sin⁡θ=−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 sin⁡θ=−2\sin \theta = -2sinθ=−2 must be rejected in Step 4.

[1]
bii.

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∘.

[3]

Trigonometric Functions Questions

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