6.4 Trigonometric Identities

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A research team is modeling the vertical displacement, $H$ , of a specialized underwater sensor. The displacement is given by the function

$H(θ)=sin⁡2θsec⁡θ+cos⁡2θcsc⁡θ+sin⁡θH(\theta) = \sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta$

where $θ\theta$ is the tilt angle of the sensor.

Show that the expression for $H(θ)H(\theta)$ can be written as

$H(θ)=sin⁡θ+csc⁡θH(\theta) = \sin \theta + \csc \theta$

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

[4]

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A technician attempts to find the tilt angles where the displacement is exactly $4.25$ units by solving the equation

$sin⁡2θsec⁡θ+cos⁡2θcsc⁡θ+sin⁡θ=4.25\sin 2\theta \sec \theta + \cos 2\theta \csc \theta + \sin \theta = 4.25$

for $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ . They produce the following solution:

Step 1: $sin⁡θ+csc⁡θ=4.25\sin \theta + \csc \theta = 4.25$

Step 2: $sin⁡θ+1sin⁡θ=174\sin \theta + \frac{1}{\sin \theta} = \frac{17}{4}$

Step 3: $4sin⁡2θ−17sin⁡θ+4=04\sin^{2} \theta - 17\sin \theta + 4 = 0$

Step 4: $sin⁡θ=4\sin \theta = 4$ or $sin⁡θ=0.25\sin \theta = 0.25$

Step 5: $θ=14.5∘,165.5∘\theta = 14.5^{\circ}, 165.5^{\circ}$

Explain why the technician should reject the value $sin⁡θ=4\sin \theta = 4$ in Step 4.

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bii.

Determine if there are any other reasons, based on the original expression's domain, why solutions might need to be rejected, and state the final correct solutions for the technician's equation in the range $0∘≤θ≤360∘0^{\circ} \leq \theta \leq 360^{\circ}$ .

[3]

6.4 Trigonometric Identities Questions

Practise Edexcel A Level Maths 6.4 Trigonometric Identities with exam-style questions for A Level Maths. 12 questions, matched to the Edexcel A Level Maths (9MA0) 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.