3.4 Trigonometry (A-level only)

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Question 55

An acoustic engineer is modelling the resonant response, $R(θ)R(\theta)$ , of a signal filter where $θ\theta$ represents the phase angle in degrees.

Show that the response function

R(\theta) = \frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta

can be simplified to the form

R(\theta) = 2\csc \theta - 3\sin \theta

where $sin⁡θ≠0\sin \theta \neq 0$ and $cos⁡θ≠0\cos \theta \neq 0$ .

[3]

bi.

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

\frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta = 5

for $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ . They use the result from part (a) to produce the following steps:

Step 1: $2csc⁡θ−3sin⁡θ=52\csc \theta - 3\sin \theta = 5$

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

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

Step 4: $sin⁡θ=13\sin \theta = \frac{1}{3}$ or $sin⁡θ=−2\sin \theta = -2$

Step 5: $θ=19.5∘,160.5∘\theta = 19.5^{\circ}, 160.5^{\circ}$ (to 1 decimal place)

Explain why the value $sin⁡θ=−2\sin \theta = -2$ 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 $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ .

[3]

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

\frac{1}{2}\sin 2\theta \sec \theta + 2\cos 2\theta \csc \theta = 5

for $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ . They use the result from part (a) to produce the following steps:

Step 1: $2csc⁡θ−3sin⁡θ=52\csc \theta - 3\sin \theta = 5$

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

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

Step 4: $sin⁡θ=13\sin \theta = \frac{1}{3}$ or $sin⁡θ=−2\sin \theta = -2$

Step 5: $θ=19.5∘,160.5∘\theta = 19.5^{\circ}, 160.5^{\circ}$ (to 1 decimal place)

Explain why the value $sin⁡θ=−2\sin \theta = -2$ must be rejected in Step 4.

3.4 Trigonometry (A-level only)

Question 55

Markscheme

3.4 Trigonometry (A-level only)

Question 55

Markscheme

3.4 Trigonometry (A-level only)

Question 55

3.4 Trigonometry (A-level only) Questions

3.4 Trigonometry (A-level only)

Question 55

3.4 Trigonometry (A-level only) Questions