Trigonometric Functions

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Question 46

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

H(\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(\theta) = \sin \theta + \csc \theta

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

[4]

bi.

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

\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.

[1]

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]

Trigonometric Functions

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Question 46

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

H(\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(\theta) = \sin \theta + \csc \theta

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

[4]

bi.

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

\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.

[1]

bii.

[3]

Trigonometric Functions

Question 46

Markscheme

Trigonometric Functions

Question 46

Markscheme

Trigonometric Functions

Question 46

Trigonometric Functions Questions

Trigonometric Functions

Question 46

Trigonometric Functions Questions